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• CHAPTER 09 .:

Ray Optics andOptical Instruments

Chapter Analysis w.r.t: Lost 3 Yeor's Boord ExomsThe analysis given here gives you an analytical picture of this chapter and will helpyou to identify theconcepts of the chapter that are to befocussed more from exam point of view.

Number of Questions asked in last 3 years2015 2016 2017

Delhi All India Delhi All India Delhi All IndiaVery Short Answer (l mark) 1 Q lQ 1 Q

Short Type I Answer (2 marks) lQ 1 QShort Type II Answer 13 marks) lQ lQ lQ lQ

Long Answer 15 marks) lQ lQValue Based Questions (4 marks)

• In 2015, in Delhi set, one question of 1marks based on Scattering of Light , one numericalquestion of 2 marks based on Magnifying Power of Telescope and one question of 5 marksbased on Angle of Deviation, Dispersion of Prism and Reflecting Index of Glass were asked.In All India set, one question of 1 mark based on Refractive Index, one question of 2 marks I

based on Mirror Formula were asked.• In 2016, in Delhi set, only one numerical question of 3 marks based on Prism Formula was

asked. In All India set, one question of 3 marks based on Prism Formula and one question of 5marks based on Spherical Refracting Surface and Its Assumptions, Lens Maker's Formulaand numerical based on it were asked.

• In 2017, in Delhi set only one numerical question of 3 marks based on Prism Formula wasasked. In All India set, one question of 1mark based on Angular Deviation, one numericalquestion of 3 marks based on Critical Angle were asked.

On the basis of above analysis, it can be said that from exam point of view Angle of Deviation,Dispersion of Prism, Reflecting Index of Glass, Prism Formula and Lens Maker's Formula are mostimportant concepts of the chapter.

[TOPIC 1] Reflection, Refroctionand Dispersion of Light

Ray Optics or GeometricalOpticsIn this branch of optics, the light isconsidered as a ray which travels in astraight line.It states that for each and every object, thereis an image. It deals with the phenomena ofreflection and refraction of light by ordinarygeometrical methods.

1.1ReflectionReflection is the phenomenon of changingthe path of light after incidenting on aboundary separating two media without anychange in the medium.

Reflection of LightThe returning back of light in the samemedium from which it has come afterstriking a surface is called reflection of light.

Laws of ReflectionTwo laws of reflection are given as below:

;N(Normal)IIIIIII

i Ir

rrm7T1Tl7T1Tlmn:rmr.~7T1TI7T1TImn:rmr.rmn Mirroro

(i) The angle of incidence i is equal to theangle of reflection r.i.e. Li = Zr.

(ii) The incident ray, reflected ray andnormal to the reflecting surface at thepoint of incidence all lie in the sameplane.

Types of MirrorPlane MirrorIn a plane mirror, image formed is always virtual, erectequal in size as that of the object and at the same distancebehind the mirror as the object is in front of the mirror.Image in a plane mirror is always laterally inverted.

Spherical MirrorA type of mirror whose reflecting surfaces is part of ahollow sphere. Spherical mirrors are of two types.

(i) Concave spherical mirror(ii) Convex spherical mirror

Polished

convex(~p: Surfacemirror Concave

mirror

, E /, /

<,----Retlectinq surface

1.2 Sign ConventionAll measurements should be taken from pole of mirror.• All measurements along the direction of incident ray will

be positive and opposite to incident ray are negative.• All the measurements for the distances above the

principal axis are taken as positive and below theprincipal axis are taken as negative.

• For a real object, u is negative whereas v is negative forreal image and positive for virtual image.

Mirror

Heightsupwardpositive

Distances alongincident light

positive

P(Pole)

X-axisHeightsdownwards

negative

CHAPTER 9 : Roy Optics And Optical Instruments 253

1.3 Mirror FormulaMirror formula is a relation between focal lengthof the mirror, distances of objects and image fromthe mirror.

1 1 1-+-=-v u f

where, f = focal length, u = distance of the objectfrom mirror, v = distance of the image from mirror.

Focal length of mirror(f) = Radius of curvature (R) => f = ~

2 2

1.4 Linear MagnificationThe ratio of the size of the image formed by aspherical mirror I to the size of the object 0 iscalled the linear magnification produced by thespherical mirror.

I v f f-vm=-=--=--=--o u f-u f

where, I = height of image and 0 = height ofobject

Magnification (m)It is negative corresponding to real image andpositive for virtual image.

1.5 RefractionThe phenomenon of changing in the path of lightas it goes from one transparent medium toanother is called refraction.

Laws of RefractionIncident ray Normal Incident ray

(Rarer)Air

Air(Rarer)

Refracted ray(b)

Normal

Glass(Denser)

(Rarer)Air

(c)Two laws of refraction are given as below:

(i) The incident ray, refracted ray and thenormal to the refracting surface at the pointof incidence lie in the same plane.

(ii) The ratio of the sine of the angle ofincidence to the sine of the angle ofrefraction is constant for the two givenmedia. This constant is denoted by ~ and iscalled the relative refractive index.

1~2 = s~ i (Snell's law)smr

where, ~ is refractive index of the secondmedium with respect to first medium.

Refractive Index in Terms ofWavelengthThe refractive index (u) of a material is the ratio ofthe speed of light (c) in vacuum to the speed oflight in the medium (v).

Mathematically, if refractive index is given by the

I . Speed of light in the vacuum cre ation ~ = --0.. __ ---'''-- _

Speed of light in the medium vIt is also referred to as absolute refractive indexof the substance.

Principle of Reversibility of LightNIIIIIIII

• IZ I

Principle of reversibilityof light states that whenfinal path of a ray oflight after suffering anynumber of reflectionsand refractions isreversed, the ray retracesits path, exactly.

=> 1~2 X 2~1 = 11

2~1=-1~2

A

L..---N~,;---"-",,aRemirror

Principle of reversibility of light

254

Critical Angle (ic)It is the angle of incidence in denser medium forwhich angle of refraction in rarer medium is 90° iscalled the critical angle of the denser medium.

111Critical ic Densermediumangle

RarermediumIl2

[

w~e~ i = ic,r = 90° 1sinr, 1

-'--0 =21J.1=-sm90 lIJ.2

where, IJ.= refractive index of denser mediumw.r.t. rarer medium.

i.e. 1IJ.=-sin t,

1.5 Total InternalReflection (TIR)

When a ray of light travelling from densermedium to rarer medium is incident at theinterface of two medium at an angle greater thanthe critical angle for the two media, the ray istotally reflected back to denser medium. Thisphenomenon is called Total Internal Reflection. Itoccurs only when angle of incidence in densermedium is greater (not equal) than critical angle,i.e. i» i..

Rarermedium

(Air)

Densermedium(Water)

APartially

C reflected rays

Total Internal Reflection

1.6 Refraction at aSpherical Surface

(i) Refraction formula for refraction by convex orconcave spherical refracting surface is givenby

[21 Chapterwise CSSE Solved Papers PHYSICS

1J.2_h=1J.2-1J.1V u R

where IJ.I' IJ.2 are refractive index of rarerand denser media and u, v and R are to betaken with their proper signs.

(ii) When refraction takes place from denser torarer medium, then

IJ.I _1J.2 = IJ.I -1J.2v Ru

1.7 LensLens is a transparent medium bounded by twosurfaces of which one or both surfaces arespherical.

Convex or Converging LensA lens which is thicker at the centre and thinnerat its end is called convex lens. Convex lenses areof three types which are given as below:

~ ~ ))Double convex Plano-convex Concavo-convex

lens lens lens

Concave or Diverging LensA lens which is thinner at the centre and thickerat its ends is called a concave lens. Concave lensesare of three types which are given as below:

Double Plano-concave Convexo-concaveconcave lens lens lens


Lens Maker's Formula for aConvex LensThis formula relates the focal length of a lens tothe refractive index oflens and radii of curvatureof two surfaces.

.!. = (11 -1) (~ -~)f RJ R2

11= refractive index of material of lens w.r.t.surrounding media and RJ, R2 = radii of curvaturesof two surfaces.

A

o

. (c) c

Some Important PointsWhen lens of refractive index 11 is immersed in amedium of refractive index 11', then

(i) When lens is taken in another medium,then focal length changes to fm which isgiven by

(ii) If 11' =1, l.e. medium is air, the focal lengthof lens (i.e. fa) is given by

~ = (11-1) (~-~) ... (ii)fa RJ R2

(iii) fm = ((11 -ll) [dividing Eq. (ii) by Eq. (i)]fa ~-1

11'

(iv) If 11 = 11' => fm = 00

=> lens behaves like a glass slab.(v) If 11 < 11' => fm> fa and nature reversed.

(vi) If 11 > 11' => t;> fa and nature remainssame.

1.7 Linear MagnificationProduced by a Lens

Linear magnification of a lens is defined as, theratio of the height of the image formed by the lensand height of the object.

Lin nificati Height of image (I)ear mag icanon, m = ---=-----=-"'-'-Height of object (0)

For Convex LensCASE I When image is real, m = -I = ~

o -uWhen image is real, it is inverted andforms on the other side of object.

CASE II When image is virtual, m = !... = ~o u

When image is virtual, it is erect andforms on the same side of object .Thus, it can be said convex lens givespositive linear magnification for virtualimage and negative linear magnificationfor real image.

For Concave LensConcave lens always forms virtual image, linearmagnification of concave lens,

I vm=-=-.o u

256

Concave lens always gives positive linearmagnification. Other formula for linearmagnification are

v I -v Im=-=--=--u I I+u

loB Thin Lens FormulaIt is a relation between focal length of a lens anddistances of objects and image from optical centreof the lens.

1• Lens formula = - - -I v u

B'

A'I+---V I

• Lateral or transverse magnification,I v I I-vm=-=-=--=--,o u I+u I

where, angular magnification = ~8

where, 80 = angle made by image and 8= anglemade by object.

• Hlml> 1 ::::}image is magnified.

• Iflml < 1 ::::}image is diminished.

• If Im I = 1 ::::}image is of same size as the object.

1.9 Power of LensThe ability of a lens to converge or diverge the raysof light incident on it is called the power of thelens.

Thus, P = 1I (in m)

S1 unit of power oflens = dioptre (D) = m-I

Power of combination lenses in contact is given byP =PI +P2 +...+Pn

o ehapterwise eBSE Solved Papers PHYSICS

1 1 1 1-=-+-+ ...+-111/2 In

NOTE Magnification by combination of lensesm = mi X m2 X m3 .

1.10 PrismPrism have got the property of bending theincident light towards its base.A prism is a wedged shaped portion of atransparent medium bounded by two plane facesinclined to each other at a suitable angle.We get, 1j +'2 = A and 0 = il + i2 - A

A Refracting edge

When the prism is adjusted at angle of minimumdeviation, then

(a) angle of incidence is equal to the angle ofemergence

(b) il = i2, 1j = '2' 0 = Om' A + Om = 2i and 2, = A.

(c) ~ =Sin( A:Om )/Sin~

::::}Om'" (u. -l)A

(for small angle of prism)

Dispersion by a PrismThe phenomenon of splitting of white light intoits component colours on passing through arefracting medium is known as dispersion.

Whitelight beam White light

spectrumV

Glass prism

CHAPTER 9 : Roy Optics And Opticol Instruments 257

Terms Related to Dispersion ofLight by PrismAngular DispersionAngular dispersion produced by a prism forwhite light is the difference in the angles ofdeviation for two extreme colours i.e. violet andred.

Angular DeviationsFor violet Ov = QJ.v-1)AFor red, OR= (Il R - I)Aand for yellow Oy = QJ.y -1)A

Dispersive PowerDispersive power of a prism is defined as theratio of angular dispersion to the meandeviation produced by the prism.

II -II 0-0Dispersive power, co = /""'"v /""'"R = V R

Ily -1 0Angular dispersion

Mean deviationAs A. red > A. violet

. . Il red < Il violet

Hence, Ored < Oviolet

where, 0 = minimum deviation.Angular dispersion

= Ov -OR = (Ilv -1)A- (IlR -1)A

= (Ilv -IlR)A

= Ov - OR= (Ilv -IlR)A

Combining Two Thin PrismsWe can study two conditions HA

(i) Dispersion without average 0:/deviation . (0

Ov - OR= QJ.y -1) A(co - co') A

(ii) Average deviation without dispersion

0= (Ily -l)A[I- :,]

Scattering of LightIts a phenomena in which light is deflected fromits path due to its interaction with the particlesof the medium through which it passes.

Rayleigh Law of ScatteringIt states that the intensity of light of wavelength A.present in the scattered light is inverselyproportional to the fourth power of A., provided thesize of scattering particles are much smaller than A.,Mathematically scattering oc -;. [for a < < A.]. The

A.bluishness of sky and reddishness of sunrise andsunset could be explained by this law.

PREVIOUS YEARS'!EXAMINATION QUESTIONSTOPIC 1o 1 Mark Questions

1. How does the angle of minimum deviation ofa glass prism vary, if the incident violet lightis replaced by red light? Give reason.All India 2017

2. Why does sun appear red at sunrise andsunset? All India 2016

3. Why does bluish colour predominate in aclear sky? Delhi 2015

4. A concave lens of refractive index 1.5 isimmersed in a medium of refractive index1.65. What is the nature of the lens?All India 2015

5. When an object is placed between f and 2fof a concave mirror, would the imageformed be (i) real or virtual and (ii)diminished or magnified? Delhi 2015C

6. A biconvex lens made of a transparentmaterial of refractive index 1.25 isimmersed in a water of refractive index1.33. Will the lens behave as a convergingor a diverging lens? Give reason.All India 2014

7. A biconvex lens made of a transparentmaterial of refractive index 1.5 is immersed,in a water of refractive index 1.33. Will thelens behave as a converging or a diverginglens? Give reason. All India 2014

258

8. A convex lens is placed in contact with aplane mirror. A point object at a distanceof 20 em on the axis of this combinationhas its image coinciding with itself.What is the focal length of the lens?AllIndia2014

9. Write the relationship between angle ofincidence i, angle of prism A and angle ofminimum deviations from a triangularprism. Deihl 2013

10. When red light passing through a convexlens is replaced by light of blue colour,40w will the focal length of the lenschange? AllIndia2013C

11. How does focal length of a lens changewhen red light incident on it is replaced byviolet light? Give reason for your answer.Foreign 2012

12. Name the physical quantity whichremains same for microwaves ofwavelength 1 mm and UV-radiation of1600 A in vacuum. Deihl2012

13. Under what condition, does a biconvexlens of glass having a certain refractiveindex act as a plane glass sheet whenimmersed in a liquid? Deihl 2012

14. For the same value of angle of incidence,the angles of refraction in three media A,Band Care 15°, 25° and 35° respectively.In which medium, would the velocity oflight be minimum? AllIndia2012

15. When monochromatic light travels fromone medium to another, its wavelengthchanges but frequency remains the same.Explain.Delhi 2011

16. The refractive index of diamond is muchgreater than that of glass. How does adiamond cutter make use of this fact?HOTS;AllIndia2011C

17. If a ray of light propagates from a rarerto a denser medium, how does itsfrequency change? AllIndia2011C

18. State the criteria for the phenomenon oftotal internal reflection of light to takeplace. Delhi 2011. 2010. 200BC

o Chopterwise C8SE Solved Papers PHYSICS

19. A lens behaves as a converging lens in airand a diverging lens in water (u = 4/ 3).What will be the condition on the value ofrefractive index <Il) of the material of thelens? Delhi 2011C

20. A converging lens axially in contact with adiverging lens; both the lenses being ofequal focal lengths. What is the focallength of the combination? AllIndia2010

21. A glass lens of refractive index 1.45disappears when immersed in a liquid.What is the value of refractive index of theliquid? Deihl 2010

22. Calculate the speed of light in a mediumwhose critical angle is 30°. Deihl2010

23. Why does the sky appear blue? Foreign 2010

24. Under what condition does the formation ofrainbow occur? AllIndia2010C

25. Two thin lenses of power + 6D and - 2Dare in contact. What is the focal length ofthe combination? AllIndia2010

26. Two thin lenses of power + 4 D and - 2 Dare in contact. What is the focal length ofthe combination? AllIndia2010

27. Two thin lenses of power + 5D and - 2.5 Dare in contact. What is the focal length ofthe combination? AllIndia2010

28. Why are convex mirrors used as side viewmirrors in cars? Delhi 2009C

o 2 Marks Questions29. A ray PQ incident normally

on the refracting face BA isrefracted in the prism BACmade of material ofrefractive index 1.5.Complete the path of raythrough the prism. Fromwhich face will the ray Bemerge? Justify your answer. AllIndia2016

30. Use the mirror equation to show that onobject placed between F and 2F of aconcave mirror produces a real imagebeyond 2F. AllIndia2015

Ap

Q

CHAPTER 9 : Roy Optics And Optical Instruments

31. How does the refractive index of atransparent medium depend on thewavelength of incident light used? Velocityof light in glass is 2 x 108ml s and in air is3 x 108ml s. If the ray of light passes fromglass to air, calculate the value of criticalangle. Foreign 2015

32. An equiconvex lens of focal length f is cutinto two identical plane convex lenses. Howwill the power of each part be related to thefocal length of the original lens?A double convex lens of + 5D is made ofglass of refractive index 1.55 with bothfaces of equal radii of curvature. Find thevalue of its radius of curvature. Foreign 2015

33. Two monochromatic rays of light areincident normally on the face AB of anisosceles right-angled prism ABC. Therefractive indices of the glass prism for thetwo rays 1 and 2 are respectively 1.35 and1.45. Trace the path of these rays afterentering through the prism.

A 450

B C All Indio 2014

34. A ray of light falls on a transparent spherewith centre C as shown in the figure. Theray emerges from the sphere parallel to theline AB. Find the angle of refraction of A ifthe refractive index of material of sphere isJi Foreign 2014

35. Figure shows a ray of light passing througha prism. If the refracted ray QR is parallelto the base BC, show that

(i) rl = r2 = A and. 2

259

(ii) Angle of minimum deviation,Dm = 2i - A Foreign 2014

ps

B C36. Write the conditions for observing a

rainbow. Show by drawing suitablediagram how one understands theformation of a rainbow. All India 2014C

37. A ray PQ is incident A I

normally on the face PAB of a triangularprism of refractingangle of 60° made ofa transparent B 2 Cmaterial of refractive index ./.i as shownin the figure. Trace the path"tJIthe ray asit passes through the prism. Also,calculate the angle of emergence andangle of deviation. Delhl2014C

38. A convex lens of focal length it is kept incontact with a concave lens of focallength /2. Find the focal length of thecombination. All India 2013

39. When monochromatic light travels froma rarer to a denser medium, explain thefollowing giving reasons.

(i) Is the frequency of reflected andrefracted light same as thefrequency of incident light?

(ii) Does the decrease in speed imply areduction in the energy carried bylight wave? Deihl 2013

40. (i) Write the necessary conditions forthe phenomenon of total internalreflection to occur.

(ii) Write the relation between therefractive index and critical anglefor a given pair of optical media.

Deihl 2013

260

41. A convex lens of focal length 25 em isplaced coaxially in contact with a concavelens of focal length 20 cm. Determine thepower of the combination. Will the systembe converging or diverging in nature?Delhi 2D13

42. Trace the path of a ray of Alight passing through aglass prism ABC asshown in the figure. If therefractive index of glass is

r{ ./3, then find out thevalue of the angle of

fIR I mergence from the B Cprism. HDTS; Foreign 2D12

4~. A ray of light incident on an equilateralglass prism (/J. g = ./3) moves parallel to

the base line of the prism inside it. Findthe angle of incidence for this ray. Deihl 2012

44. An object AB is kept in front of a concavemirror as shown in the figure.

A

B C F

(i) Complete the ray diagram showingthe image formation of the object.

(ii) How will the position and intensity ofthe image be affected if the lower _half of the mirror's reflecting surfaceis painted black? All Indio 2012

45. (i) Plane and convex mirrors are knownto produce virtual images of theobjects. Draw a ray diagram to showhow, in the case of convex mirrors,virtual objects can produce realimages.

(ii) Why are convex mirrors used as side- view mirrors in vehicles?

Delhi 2012C

o Chapterwise CBSE Solved Papers PHYSICS

46. (i) Draw a ray diagram for a convexmirror showing the image formationof an object placed anywhere in frontof the mirror.

(ii) Use this ray diagram to obtain theexpression for its linearmagnification. All Indio 2012C

47. How does focal length of a lens changewhen red light incident on it is replacedby violet light? Give, reason for youranswer. Foreign 2D12

48. Two thin lenses of power - 4 D and 2 Dare placed in contact coaxially. Find thefocal length of the combination. All India 2012C

49. Draw a ray diagram to show the imageformation by a concave mirror.When the object is kept between its focusand the pole.Using this diagram, derive themagnification formula for the imageformed. Delhi 2011

50. A beam of light converges at a point P. Aconcave lens of focal length 16 em isplaced in the path of this beam 12em fromP. Draw a ray diagram and find thelocation of the point at which the beamwould now converge. All India 2011C

51. Draw a diagram showing the formation ofprimary rainbow and explain at whatangles the primary rainbow is visible.Delhi 2010C

52. The radii of curvature of the faces of adouble convex lens are lOcm and 15 ern. Iffocal length of the lens is 12 em, find therefractive index of the material of thelens. Delhi 2010

53. (i) The bluish colour predominates inclear sky.

(ii) Violet colour is seen at the bottom ofthe spectrum when white light isdispersed by a prism. State reasonsto explain these observations.

Delhi 2010


54. A biconvex lens has a focal length 2/3times the radius of curvature of eithersurface. Calculate the refractive index oflens material. Delhi 2D10

55. (i) Why does the sun appear reddish atsunset or sunrise?

(ii) For which colour, the refractiveindex of prism material is maximumand minimum? Delhi 2010

56. Find the radius of curvature of the convexsurface of a plano-convex lens, whose focallength is 0.3 m and the refractive index ofthe material of the lens is 1.5. Deihl 2010

57. (i) Out of blue and red light, which isdeviated more by a prism? Givereason.

(ii) Give the formula that can be used todetermine refractive index ofmaterial of a prism in minimumdeviation condition. Deihl 2010

58. The following table gives the values of theangle of deviation for different values ofthe angle of incidence for a triangularprism.

Angle of 33° 38° 4r 52° 60° 71°incidenceAngle of 60° 50° 46° 40° 43° 50°deviation

(i) For what value of the angle ofincidence, is the angle of emergencelikely to be equal to the angle ofincidence itself?

(ii) Draw a ray diagram showing thepassage of a ray of light through thisprism when the angle of incidencehas the above value. Delhi 2010C

59. (i) State the principle on which theworking of an optical fibre is based.

(ii) What are the necessary conditions forthis phenomenon to occur? All India 2009

60. (i) What is the relation between criticalangle and refractive index of amaterial?

(ii) Does critical angle depend on thecolour of light? Explain. All India 2009

o 3 Marks Questions61. (i) Monochromatic light of wavelength

589 nm is incident from air on awater surface. If u for water is 1.33,find the wavelength, frequency andspeed of the refracted light.

(ii) A double convex lens is made of aglass of refractive index 1.55 withboth faces of the same radius ofcurvature. Find the radius ofcurvature required, if the focal lengthis 20 cm. All India 2017

62. (i) A ray of light incident of face AB of anequilateral glass prism, showsminimum deviation of 30°. Calculatethe speed of light through the prism.

aLe(ii) Find the angle of incidence at face

AB, so that the emergent ray grazesalong the face AC. Delhi 2017

63. (i) Calculate the distance of an object ofheight h from a concave mirror ofradius of curvature 20cm, so as toobtain a real image of magnification2. Find the location of image also.

(ii) Using mirror formula, explain whydoes a convex mirror always producea virtual image? Delhi 2016

64. In the following diagram, an object '0' isplaced 15cm in front of a convex lens ~ offocal length 20 em and the final image isformed at I at a distance of 80 cm from thesecond lens L2. Find the focal length of thelens L2•

20 em so em

15em....-

262

65. (i) A mobile phone lies along theprincipal axis of a concave mirror.Show with the help of a suitablediagram, the formation of its image.Explain why magnification is not auniform.

(ii) Suppose the lower half of theconcave mirror's reflecting surfaceis covered with an opaque material.What effect this will have on theimage of the object? Explain.Delhl2D14

66. A convex lens of focal length 20 em isplaced coaxially with a convex mirror ofradius of curvature 20 cm. The two arekept at 15 em from each other. A pointobject lies 60 em in front of the convexlens. Draw a ray diagram to show theformation of the image by thecombination. Determine the nature andposition of the image formed. All India 2014

67. A convex lens of focal length 20 cm isplaced coaxially with a concave mirror offocal length 10 em at a distance of 50 cmapart from each other. A beam of lightcoming parallel to the principal axis isincident on the convex lens. Find theposition of the final image formed bythis combination. Draw the ray diagramshowing the formation of the image.All India 2014

68. Define the term 'critical angle' for a pairof media.A point source of monochromatic light '8'is kept at the centre of the bottom of acylinder of radius 15.0 cm. The cylindercontains water (refractive index 4/3) to aheight of 7.0 em. Draw the ray diagramand calculate the area of water surfacethrough which the light emerges in air.Delhl2013C

69. Define power of a lens. Write its units.Deduce the relation.!. = ~ + ~ for two

f it f2thin lenses kept in contact coaxially.Farelgn 2012

o Chapterwise CSSE Solved Papers PHYSICS

70. Youare giventhree lenses ~ , L2 and Laeach offocallength 10cm.An objectis kept at15em in front of~ as shown.The final realimage is formed at the focusI ofLa.Find the separation between ~ , L2 and La. AllIndia 2012

~----~+----4-4----~-+----I

71. A small bulb (assumed to be a point source)is placed at the bottom of a tank containingwater to a depth of 80 cm. Find out the areaof the surface of water through which lightfrom the bulb can emerge. Take the value ofthe refractive index of water to be 4/3.Delhl2013C

72. Draw a ray diagram to show the formationof the image of an object placed on the axisof a convex refracting surface of radius ofcurvature 'R', separating the two media ofrefractive indices 'ni' and 'n2' (n2> ni).Use this diagram to deduce the relationn2 _ ni = n2 - ni, where u and v representv u R

respectively the distance of the object andthe image formed. Deihl 2012 C

73. A convex lens made up of a glass ofrefractive index 1.5 is dipped in turn in

(i) a medium of refractive index 1.65(ii) a medium of refractive index 1.33

(a) Will it behave as a converging lens ora diverging lens in the two cases?

(b) Howwill its focal length change in thetwo media? All India 2011

74. Use the mirror equation to show that(i) An object placed between f and 2f of a

concave mirror produces a real imagebeyond 2f.

(ii) A convex mirror always produces avirtual image independent of thelocation of the object.

(iii) An object placed between the pole andfocus of a concave mirror produces avirtual and enlarged image. All India 2011


75. Find the position of the image formed ofthe object 0 by the lens combination givenin the figure. Foreign 2011

f= +10cm-10cm 30cm

o1-30cm

I 5cm I 10cm I

76. A converging lens has a focal length of20 em in air. It is made of a material ofrefractive index 1.6. It is immersed in aliquid of refractive index 1.3. Calculate itsnew focal length. All Indio 2011

77. State the necessary conditions forproducing total internal reflection of light.Draw ray diagrams to show how speciallydesigned prisms make use of totalinternal reflection to obtain invertedimage of the object by deviation of rays

(i) through 90° and(ii) through 180°. All Indio 2011

78. With the help of a suitable ray diagram,derive a relation between the object-distance (u), image distance (v) and radiusof curvature R for the convex sphericalsurface when a ray of light travels from ararer to denser medium. OelhI2011C, 2008 C

79. A ray of light is incident on one face of aglass prism and emerges out from theother face. Trace the path of the ray andderive an expression for refractive indexof the glass prism. Deihl 2011C

80. The image obtained with a convex lens iserect and its length is four times thelength of the object.If the focal length of the lens is 20 em,calculate the object and image distances.All India 2010

81. A convex lens is used to obtain amagnified image of an object on a screen10 em from the lens. If the magnificationis 19, find the focal length of the lens.All India 2010

82. An illuminated object and a screen areplaced 90 em apart. Determine the focallength and nature of the lens required toproduce a clear image on the screen, twicethe size of the object. All India 2010•

83. (i) How is the focal length of a sphericalmirror affected when the wavelengthof the light used is increased? '

(ii) A convex lens has 20 em focal lengthin air. What is its focal length inwater? (Refractive index of air-water= 1.33, refractive index of air-glass= 15). Foreign 2010

84. (i) How is the focal length of a sphericalmirror affected when it is immersedin water?

(ii) A convex lens has 10 em focal lengthin air. What is its focal length inwater? (Refractive index of air-water= 1. 33, refractive index of air-glass= 15). Foreign 2010

85. An object of 3 em height is placed at adistance of 60 cm from a convex mirror offocal length 30 cm. Find the nature,position and size of the image formed.All India 2010C

86. An object of 2 em height is placed at adistance of 30 em from a convex mirror offocal length 15 cm. Find the nature,position and size ofthe image formed.All India 2010C

A87. Three light rays,red (R), green (G)and blue (B) areincident on aright angledprism ABC atface AB. The B'----, ----'---"'Crefractive indices of the material of theprism for red, green and blue wavelengthsare 1.39, 1.44 and 1.47 respectively. Outof the three, which colour of ray willemerge out of face AC ? Justify youranswer. Trace the path of these rays afterpassing through face AB. HOTS, Dllhl 2009

BGR

88. State the conditions under which totalinternal reflection occurs. One face of aprism with a refracting angle of 30° iscoated with silver. A ray incident onanother face at an angle of 45° is refractedand reflected from the silver coated faceand retraces its path. Find the refractiveindex of the material of the prism.Foreign 2009

89. A tank is filled with water to a height of12.4 cm. The apparent depth of a needlelying at the bottom of the tank is measuredby a microscope to be 9.3 cm. What is therefractive index of water? If water isreplaced by a liquid of refractive index 1.6up to the same height, by what distancewould the microscope have to be moved tofocus on the needle again?Delhi 2009C

90. When a ray of light passes through atriangular glass prism, find out the relationfor the total deviation, 0 in terms of theangle of incidence, i and angle ofemergence, e. Plot a graph showing thevariation of angle of deviation with theangle of incidence and obtain the conditionfor the angle of minimum deviation.All Indio 2009C

91. Light incident at an angle of incidence of45° in a certain medium goes grazing alongits surface of separation from air afterrefraction.What is

(i) the velocity of light in this medium?(ii) the angle of incidence at which light

from air must be incident on thismedium so that the refracted ray isnormal to the reflected ray? Also,name the special property exhibited bylight for this angle of incidence.

All India 2009C


o 5 Marks Questions92. (i) Derive the Mathematical relation

between refractive indices n1 and n2

of two radii and radius of curvatureR for refraction at a convexspherical surface. Consider theobject to be a point since lying onthe principal axis in rarer mediumof refractive index nl and a realimage formed in the denser mediumof refractive index n2' Hence, derivelens maker's formula.

(ii) Light from a point source in air fallson a convex spherical glass surfaceof refractive index 1.5 and radius ofcurvature 20 cm. The distance oflight source from the glass surfaceis 100 cm. At what position is theimage formed. All India 2016

93. (i) Plot a graph to show variation ofthe angle of deviation as a functionof angle of incidence for lightpassing through a prism. Derive anexpression for refractive index ofthe prism in terms of angle ofminimum deviation and angle ofprism.

(ii) What is dispersion of light? What isits cause?

(iii) A ray of light incident normally onone face of a right isosceles prism istotally reflected as shown in figure.What must be the minimum valueof refractive index of glass? Giverelevant calculations.

All India 2015


94. (i) A point object 0 is n1 (iE2kept in a medium ----O_--u-_- ---R---~--of refractive indexn1 in front of aconvex spherical surface of radius ofcurvature R which separates thesecond medium of refractive index n2from the first one, as shown in thefigure.Draw the ray diagram showing theimage formation and deduce therelationship between the objectdistance and the image distance interms of n1, n2 and R.

(ii) When the image formed above acts as avirtual object for a concave sphericalsurface separating the medium n2 fromn1(n2> n1), draw this ray diagram andwrite the similar [similar to (i)]relation. Hence, obtain the expressionfor the lens maker's formula.All India 2015

(i) A ray PQ of light is incident on the faceABof a glassprism ABC (as Ashown in thefigure) andemerges out ofthe face AC.Trace the path ofthe ray. Show that Zi + Le = LA + Lowhere, 0 and e denote the angle ofdeviation and angle of emergencerespectively.Plot a graph showing the variation ofthe angle of deviation as a function ofangle of incidence. State the conditionunder which Lo is minimum.

(ii) Find out the relation between therefractive index (11) of the glass prismand LA for the case, when the angle ofprism (A) is equal to the-angle ofminimum deviation (om)' Hence, obtainthe value of the refractive index forangle of prism A = 60°. Delhi 2015

95.

26196. (i) A point object is placed in front of a

double convex lens (or refractiveindex 11 = n2 with respect to air)

n1

with its spherical faces of radii ofcurvature ~ and R2• Show the pathof rays due to refraction at first andsubsequently at the second surfaceto obtain the formation of the realimage of the object.Hence, obtain the lens maker'sformula for a thin lens.

(ii) A double convex lens having bothfaces of the same radius ofcurvature has refractive index 1.55.Find out the radius of curvature ofthe lens required to get the focallength of 20 cm.All India 2014 C

97. Draw a ray diagram showing theformation of the image by a point objecton the principal axis of a sphericalconvex surface separating two media ofrefractive indices n1 and n2' when apoint source is kept in rarer medium ofrefractive index n1. Derive the relationbetween object and image distance interms of refractive index of the mediumand radius of curvature of the surface.Hence, obtain the expression for lensmaker's formula in the case of thinconvex lens. Delhi 2014C

98. (i) Draw a ray diagram to showrefraction of a ray ofmonochromatic light passingthrough a glass prism. Deduce theexpression for the refractive indexof glass in terms of angle of prismand angle of minimum deviation.

(ii) Explain briefly how thephenomenon of total internalreflection is used in fibre optics.Oelhi 2011

266

99. (i) Obtain lens maker's formula using theexpression

~ _ n1= (~ - n1)

v u RHere, the ray of light propagatingfrom a rarer medium of refractiveindex (n1) to a denser medium ofrefractive index (n2) is incident on theconvex side of spherical refractingsurface of radius of curvature R.

(ii) Draw a ray diagram to show thatimage formation by a concave mirrorwhen the object is kept between itsfocus and the pole. Using thisdiagram, derive the magnificationformula for the image formed. Delhl2D11

100. Figure shows a convex spherical surfacewith centre of curvature C separating thetwo media of refractive indices nl and n2.Draw a ray diagram showing the formationof the image of a point object 0 lying on theprincipal axis. Derive the relationshipbetween the object and image distance interms of refractive indices of the media andthe radius of curvature R of the surface.Delhi 2011

on1

B

G

R

45°b L-----.l..-~c

101. Trace the rays of light showing theformation of an image due to a point objectplaced on the axis of a spherical surfaceseparating the two media of refractiveindices n1 and n2. Establish the relationbetween the distance of the object, thedistance of image and the radius of

o Chapterwise CSSE Solved Papers PHYSICS

curvature from the central point of thespherical surface. Hence, derive theexpression of the lens maker's formula.Delhi 2009

102. (i) Draw a ray diagram for formation ofimage of a point object by a thindouble convex lens having radii ofcurvatures Rt and R2 and hence,derive lens maker's formula.

(ii) Define power of a lens and give itsSI unit. If a convex lens of length50 em is placed in contact coaxiallywith a concave lens of focal length20 em, what is the power of thecombination? Foreign 2009

103. (i) Draw the ray diagram for theformation of image of an object by aconvex mirror and use it (along withthe sign convention) to derive themirror formula.

(ii) Use the mirror formula to show thatfor an object kept between the poleand focus of a concave mirror theimage appears to be formed behindthe mirror. All India 2009C

o Explanations1. Wavelength of violet light is smaller than that of

red light. Also, angle of minimum deviation,om =(J.J. -l)A =>0m oc Il

As, IlR < Ilv =>(Om)R < (Om)VAs, deviation is less for red light, hence, angle ofdeviation decreases. rn

2. During sunrise and sunset, the rays have totravel a larger part of the atmosphere becausethey are very close to the horizon. According toRayleigh's law of scattering, scattering oc ~,

Awavelength of red is large, hence it is leastscattered.Therefore, light rays other than red is mostlyscattered away. Most of the red light, which isthe least scattered, enters our eyes. Hence, thesun appears red at sunrise and sunset. (1l

CHAPTER 9 Roy Optics And Optical Instruments

3. The blue colour of the sky is due to the scatteringof sunlight by the molecules of the atmosphere.According to Rayleigh's law of scattering, theintensity of scattered light, I ec -;., blue colour

Ahaving the short wavelength in the visiblespectrum scatter the most. When we look at thesky, the scattered light enters our eyes and thislight contains blue colour in a large proportion, sothe sky appears blue. (1)

4. A concave lens is made up of certain materialbehaves as a diverging lens, when it is placed in amedium of refractive index less than therefractive index of the material of the lens andbehave .s a converging lens, when it is placed ina mcc.um of refractive index greater than ther, ,-.cnve index of the material of the lens. (1)

In the given case, concave lens is immersed inmedium having refractive index greater than therefractive index of the material of the lens(1· 65> 1· 5). Therefore, it will behave as aconverging lens.

5. When an object is placed between / and 2/ of aconcave mirror, the image formed is real. invertedand magnified.

2F ,6. When a lens is placed in a liquid, where rei ctive

index is more than that of the material of lens,then the nature of the lens changes. So, when abiconvex lens of refractive index 1.25 is immersedin water (refractive index 1.33), i.e. in the liquidof higher refractive index, its nature will change.So, biconvex lens will act as biconcave lens ordiverging lens. (1)

7. A biconvex lens acts as a converging lens in airbecause the refractive index of air is less thanthat of the material of the lens. The refractiveindex of water is less than the refractive index ofthe material of the lens (1.5). So, its nature willnot change, it behaves as a converging lens. (1)

8. The adjacent figure shows a convex lens L incontact with a plane mirror M. P is the pointobject kept in the front of this combination at adistance of 20 em from it. (112)

Since, the image is coinciding with the objectitself, the rays from the object after refractionfrom the lens fall normally on the mirror M md

267

form an image coinciding with the object itself.So, the image is formed at the focus of the lens.So, focal length of the lens is 20 cm.

L M

(1/2)

9. The relation between the angle of incidence i,angle of prism A and the angle of minimumdeviation, ~m for a triangular prism is given by

i= A + ~m.

2 m10. Focal length of the lens decrease when red light is

replaced by blue light. (1)

11. This question can be answered by considering thelens maker's formula.From the formula, we can identify which factor willchange on changing the wavelength.

The refractive index of the material of a lensincreases with the decrease in wavelength of theincident light. So, focal length will decrease withdecrease in wavelength according to the formula.

~ = (J.I. -1) (~ - ~)/ R\ R2

12.

Thus, when we replace red light with violet lightthen due to increase in wavelength the focallength of the lens will decrease.Both microwave and UV-rays are a part of theelectromagnetic spectrum. Thus, the physicalquantity that remains for both types of radiationwill be their speeds equal to c.c = 3xl08 m/s

(1)

(1)

13. When refractive index of lens is equal to therefractive index of liquid.

r sin i cFrom Snell slaw, J.I. = -- = -sin r v

~ v cc sin r for given value of i~ Smaller angle of refraction, smaller thevelocity of light in medium.Velocity of light is minimum in medium A as angleof refraction is minimum, i.e. 15°. (1)

(1)

14.

268 o Chopterwlse eBSE Solved Papers PHYSICS

15. Because refractive index for a given pair of mediadepends on the ratio of wavelengths and velocityof light in two medium and not on frequency. (1)

16. The refractive index of diamond is much higherthan that of glass. Due to high refractive index, thecritical angle for diamond air interface is low. Thediamond is cut suitably so that the light enteringthe diamond from any face suffers multiple totalinternal reflections at the various surfaces. (1)

17. Frequency remains unchanged when light travelsfrom one transparent medium to anothertransparent medium. (1)

18. Following are the criteria for total internalreflection

(i) Light must pass from a optically denser to aoptically rarer medium.

(ii) Angle of incidence in denser medium is must begreater than critical angle for two media. (1)

19. When a lens is immersed in a liquid whose refractiveindex is more than that of the material of lens, thennature of lens changes, i.e. converging lens behaveslike diverging lens and vice-versa.

Refractive index of the material of lens is less thanthe refractive index of water. (1)

20. Combined focal length of a lens combination.!.. = 2.. + 2.. (For two thin lenses in contact)1 It 12

As, /2 = - It(focal lengths are equal, one is convex and other isconcave)

1~ -= 0 ~ /=00.1 rn

21. When a lens immersed in a liquid disappears then,Illiquid = Il 9 = 1.45 (1)

22. Critical angle is the angle of incidence for whichangle of refraction becomes 90°. Here, in this case

f . . d 1re ractive In ex, 1.1= -,-,-sin 'c

Rf lve ind c 1.: e racnve m ex, Il = - = -.-.-v sm1c

~ v= c sin ic = 3xl08 x sin300

= 3X108 x ~ = 1.5 X 108 m/s2

23. Due to large scattering of visible light of smallerwavelength (blue colour) as intensity of scatteredlight oc-;.~ rn

(1)

24. Availability of rain drops causes refraction,dispersion and total internal reflection of sunlight results in the form of rainbow and the backof the observer should be towards the sun. (1)

25. Resultant power of the combination,p = ~ + P2 = 6 - 2 = 4D

1 1- = 4 ~ / = - m = 25 em/ 4 (1)

(1)

(1)

26. Refer to Ans. 25, (Ans. / = 50 em) .

27. Refer to Ans. 25, (Ans.f = 40 em) .

28.

29.

Because convex mirror forms virtual. erect andsmaller image of object irrespective of relativeposition of object from mirror and therefore, itsfield of view is very wide. '(1)

Given, refractive index of Athe material of the prism,Il = 1.5:. Critical angle for thematerialsin C = .!.. = _1_ = 2/3

Il 1.5

~ C = sin'? (~) =4~.

From the ray diagram, itis clear that angle of incidence i= 30°< C.Therefore the ray incident at the face AC will notsuffer total internal reflection and merges outthrough this face. (2)

According to the mirror equation, we have1 1 1-+-=-V u /

p

8L.....l.=----JC

30.

where, u = distance of the object from the mirrorv = distance of the image from the mirror

/ = focal length of the mirrorFrom the mirror equation, we have

v=~ ...(i)u- /

Applying new cartesian sign convention, we get/ = - ve and u = - ve

Given, / < u < 2/~ v=-ve [fromEq.(i))Magnification is given by

m=-(=:)=-veHence, the image formed is real. (1)

CHAPTER 9 : Ray Optics And Optical Instruments 269

31.

From the mirror formula, when u = - 2/.1 1 1

~ --+-=--2/ v -/

1 1 -1~ -=---=-v 2/ / 2/

Therefore, when the object is at [, then image isformed at infinity. Shows that when/ < u < 2/, 00< v < 2/ (1)

(i) The refractive index of a transparent mediumis inversely proportional to the wavelength ofincident light. The relationship between thetwo is given by

AIl=~

A (1)

where,Il = Refractive index of medium

AO = Wavelength of incident light in vacuumA = Wavelength of incident light in medium

(ii) Given,Velocity of light in air, va = 3xl08m/s

Velocity of light in glass, V9 = 2xl08 m/s

The refractive index of glass is given by,119 = .£.Where c is speed of light in vacuum.

v9

The refractive index of air is given by, Ila = !2va

:. The refractive index of glass W.r.t. air will bea _1l9 a _Va_3X108_15119 - ~ ~ 119 -;- - 2x108 - .

"'a 9

We know a 1119 = sine

where, c is the critical angle for the interface:. _1_ = 1.5~ sine = 2.-

sine 1.5~ e = sin " (0.66) ==) c = 41.3°

:. Critical angle c = 41.3° (1)

32. (i) The focal length of original equiconvex lens isf.Let the focal length of each part after cuttingbeF.

1 1 1 1 2 FHere, - = - + - ~ - = - ~ / = - ==) F = 2// F F / F 2

Power of each part will be given byP=.!..==)P=~

F ~ m

(ii) From lens maker formula, we have

p = (1l-1)(2.- -~)Rl R2

5= ~.55-1)H-CR)}5= 0.55X2

R

R = 0.55 x 2 = 0.22m = 22cm5

33. The paths are shown as below:

(1)

45°

2'

From the figure, it is clear that angle of incidencefor ray 1 is 4SO.For ray 1,

sin i = sin 450 = 1.J2 1.414For ray 1, the refractive index of theIl = 1.35

1 . 1 1Il = -- ~ sme = - = --

sine Il 1.35

Here, _1_ < 11.414 1.35

prism is

(1)

i.e. sin i < sin e or i < eSo, ray 1 will be refracted by the prism.For ray 2, angle of incidence, i = 45°

sin i = sin 4SO= ~ = _1_.J2 1.414

For ray 2, the refractive index, Il = 1.451

Il=-sine

. 1 1==) sme = - = --Il 1.45

Here, _1_ > _1_1.414 1.45

i.e. sini> sine or i » eSo, ray 2 will get total internally reflected. (1)

270

34. Given, i = 600, ~ = .f3From Snell's law, we have

sini-=~sinr

sin60° =.f3sinr

. sin 600 1 .f3smr=--=-x-.f3 .f3 2

sinr = 0.5r = sin'? (0.5)

r = 300

35. (i) When QR is parallel to the base BC, we havei = e (prism is in the position of minimumdeviation)

A

p

BL...L-----~,(1)

... (i)(let)

36.

~ r1=r2=rWe know that

r1+r2=AFrom Eq. (i), we get

2r = A. r = ~2

Arl=r2=-2

(ii) Also, we haveA+D=i+e

Substituting, D = Dm and e = iA+ Dm = i+ i

. . Dm = 2i - A (1)

The conditions for observing a rainbow are givenbelow:

(i) The sun comes out after a rainfall or shower.(ii) The observer stands with his/her back towards

the sun.

... (ii)

o ehapterwise eSSE Solved Papers PHYSICS

(bJ Secondary Rainbow

(1)

(1)

1:~-------+--...p.~:.-.....c::~ R'-_----+-+---+-+-+-~- Primary

V"

s

Formation of rainbowThe rays of light reach the observer through arefraction followed by a reflection, followed by arefraction.Figure shows red light from drop 1 and violetlight from drop 2, reaching the observer eye.

37. If the i, is the critical angle for the prism.1Now, 11 =--

sinie.. 1 1 .f3

.. sin Ie = - = - = -11 2 2

.f3

~ . i, = sin'" ( ~) ~ ie = 600

Angle of incidence at face AC of the prism is 600•

Hence, refracted ray grazes the surface AC. Angleof emergence = 900

Angle of deviation = 300 (2)

CHAPTER 9 : Roy Optics And Opticol Instruments

38. Focal length for convex lens = fiFocal length for concave lens = - 12The equivalent focal length of a combination ofconvex lens and concave lens is given by

~=~+_1_ ~ F= fil2F fi - 12 12 - fi (2)

39. (i) The frequency of reflected and refracted lightremains same as the frequency of incidentlight because frequency only depends on thesource of light. (1)

(ii) Since, the frequency remains same, hencethere is no reduction in energy. (1)

40. (i) Refer to ADs.18. (1)

(ii) al1b = _1_, where a and b are the rarer andsin C

denser media, respectively. C is the criticalangle for the given pair of optical media. (1)

41. Given, focal length of convex lens,fi = + 25cm = + 0.25 m and focal length ofconcave lens, 12 = - 20 em = - 0.20 mEquivalent focal length of convex and concavelens,

F=~+ ~=~+ _1_= __ 1_fi 12 25 - 20 100

.. F=-lOOcm=-lm

Now, the power of lens, P = ~I

1 1For convex lens, P, = - = -fi 0.25

1 1For concave lens, P2 = - = --12 -0.20

Hence, power of the combinationP = P, + P = _1_ + _1_ = 100 + 100

2 0.25 - 0.20 25 - 20= 400 - 500 = - 100 = _ I D

100 100Here, the focal length of the combination=100 cm= -I mSince, the focal length is in negative, so the systemwill be diverging in nature. (2)

42. While tracing the path of the ray, we shouldremember that prism bends the incident raystowards its base.

Refractive index of glass,119= ,J3Since, i = 0

(1)

271

At the interface AC, we have (according to Snell'slaw)

A

sini=119sin r l1a

But, sin i = sin 0° = 0Thus, sin r = l1asin i = 0

119Hence, r = 0This ray pass unrefracted at AC interface andreaches AB interface. Here, we can see angle ofincidence becomes 30°.Thus, applying Snell's law

sin 30° _ l1a _ 1sine - 119- ,J3

sin e = ,J3 x sin 300= ,J32

Thus, e = 60°Hence, angle of emergence is 60°. (1)

43. To draw the ray diagram for the refraction fromthe prism. FOllowing things should be kept in themind.

(i) Draw normal to the point of incidence.(ii) Consider each boundary of the prism as separate

interface and draw the ray diagram for therefraction taking place.

The reflection of light through prism is shown asbelow:

A

272

By geometry Angle of refraction, r = 30°

Given, refractive index, Ii = .J3Using Snell's law, Ii = sin j

sin r

(112)

(112)

=> sin j = Ii sin r

= (.J3) sin (300) = .J32 (1/2)

Angle of incidence, { = 60° = ~3

. 1tz=-

3(1/2)

44. To draw the ray diagram for image formation, thefollowing are the rules to form the image fromspherical mirror.

(i) The ray parallel to principal axis passes throughthe focus after reflection.

(ii) The ray passing through the focus becomesparallel to principal axis after reflection.

(iii) The ray passing through the centre of curvaturereturns on the same path after reflection.

(i) The ray diagram showing the imageformation of the object (1)

AB = ObjectA'B' = Image

45.

(ii) The position of the image remains samewhereas intensity of image reduces.

(112+1/2 = 1)

(i) If a plane or a convex mirror is placed in thepath of rays converging to a point the rays getreflected to a point in front of the mirror.Real image can be obtained on a screen.

Realimage Virtual

object .•0

...>:::.::>:"

(ii) The convex mirror is used as side viewmirrors in vehicles as it gives a wide field ofview of the traffic. (1)

IZI ehopterwise eBSE Solved Papers PHYSICS

46. (i)

c

vFigure shows the formation of image A' B' of afinite object AB by a convex mirror, virtual,erect and diminished. (1)

(ii) Now, MBP -M'B'PA'B' PB'

:.--=-AB PB

Applying the new cartesian sign convention,A'B'= h» AB = hI' PB'= v, PB =-u

.. h2 =.!.....hI -u

Linear magnification m = h2 = -~hI u (1)

47. We know.!.. = (IJ.-l)(~-~)f RI R2

1f oc -- and liv > IiR(IJ.-1)

The increase in refractive index would result indecrease of focal length of lens. Hence, we can saythat replacing red light with violet light, decreasesthe focal length of the lens used.

48. Net power P = 1l. + P2 = -4+ 2=-2D1 1Focal length f = - = -m = -O.5m = -SOcmP -2

49. Ray diagram of image formation by a concavemirror. M"

(1) c p N"

In t,M'N'P and M" N"PM"N" N" P------M'N' N'P

By sign convention, PN' = - u, PN" = + vM'N' = hI and M"N" = h2

h2=~hI - U

:.Linear magnification, m = h2hI

vm=--u (2)


50. Ray diagram

51.

Given, U =+ 12 cm , f =- 16 cm , v =?Using lens equation,

1 1 1 1 1 1-=--- ~ --=---f v U 16 v 12

~ 1 1 _=4-3=2..v 12 16 48 48v=+48cm

The image of virtual object at P forms at P' at adistance 48 cm from the lens. (1)

For ray diagram for the formation of rainbowRefer to Ans. 36. (1)

The primary rainbow is formed by those rayswhich suffer one total internal reflection and tworefraction and comes out of the rain drop at angleof minimum deviation.The violet and red light colours emerge as cone ofrays at 410 and 430 respectively and can be viewedby observer. (1)

Given, RJ =+ 10 ern R2=-15cm, f =+ 12cm, Il =?Lens maker's formula, (1)

2.. = (Il -1) (2.. - 2..)f RJ R2

~ = (Il -1) (~ + ~) = (Il -1) (2)12 10 15 30

1~ 1l-1=-

23

1l="2

52.

~

53.

(1)

(i) Refer to Ans. 3. (1)

(ii) Violet colour can be seen at bottom as itundergoes largest deviation in the visiblespectrum because of highest value ofrefractive index, Il of prism for this colourbeing of smaller wavelength. (1)

(1)

54. Given, f = 3 R, RJ = + R, R2 = - R3

Using Lens maker's formula,

2.. = (Il - 1) (2.. - 2..)f RJ R2

~ = (Il -1) (2.. + 2..)2R R R3 m~ = (1l-1) (i)

1l-1=~4

3 7Il = 1 + - ~ Il = - = 1.75 (1)4 4

55. (i) Reddishness at sunset and sunrise Whenthe sun is near the horizon at sunset orsunrise, the light rays have to traverse a largerthickness of the atmosphere than when thesun is overhead at noon. In accordance withRayleigh's scattering law, the lowerwavelengths in the blue region are almostcompletely scattered away by the airmolecules.The higher wavelengths in the redregion are least scattered and reach our eyes.Hence, the sun appears almost reddish atsunset and sunrise. (1)

(ii) Refractive index Il of prism is maximum forviolet and minimum for red colour. (1)

56. For a plano-convex lens, RJ = 00

R2 = - R; f = 0.3 m= 30 cmIl = 1.5R=?

Lens maker's formula,

2.. = (Il -1) (2.. - 2..)f RJ R2 (1)

~ 2.. = (Il -1) (~ __ 1_)30 00 - R2..=(1.5-1)30 R

~ R=15cm (1)

57. (i) Blue light suffers more deviation by a prismthan red light.This happens due to high value of refractiveindex of material of prism for blue light becauseof its smaller wavelength in visible spectrum(1)

(ii) om = (Il -1),1. where om = angle of minimumdeviationIl = refractive index, A = prism angle. (1)

274

58. (i) i = 52°. When prism is adjusted at an angle ofminimum deviation, then angle of inddence isequal to the angle of emergence becausei+ e = A + o. (1)

(ii) The ray diagram in the condition of minimumdeviation is shown as below:

A

sp

tAngleo!deviation

i= i' i'

59. (i) Optical fibre works on the principle of totalinternal reflection.When a light ray travelling from denser to ararer medium is incident at an angle greaterthan the critical angle, then it is reflected backinto the same medium. This phenomenon iscalled total internal reflection. (1)

(ii) Refer to Ans. 18. (1)

(i) Relation between critical angle and refractive. d f ial i 1ID ex 0 a maten IS Ii =--

sin i,

60.

61.

where, i, = critical angleIi = refractive index of denser medium w.r.t.rarer medium. (1)

(ii) Yes, critical angle depends on colour of lightbecause colour of light is associated withwavelength. Smaller the wavelength, higherthe refractive index and lower the critical angleand vice -W7!12. Like A,.ed> Aviolet' hencelired > Ii violet (1)

(i) In refraction, frequency remains same so,

frefracted beam = fincident beam

[": V = IA]

IZI ehapterwise eSSE Solved Papers PHYSICS

62.

VI 3 X 1088-1=> v2 = - = --- = 225 x 10 ms

1i21 1.33

.. A2 = ~ = 589 = 442.85 = 443nrn1i21 1.33

So, wavelength of reflected beam = 443 nrn andits speed = 2.25 x 108 ms'"

(ii) For a biconvex lens, using lens maker s formula,

.!. =/J! -1) (..!.. -..!..)I RI R2

Here, 1=20 em, Ii = 1.55 => RI = + R andR2 =-R

1 2:. We have, - =/J! -1)-I R

=> R = 2(1i -1)1 = 2 x (1.55-1) x 20= 22cm:. Radius of 22 em is required.

(i) Given, angle of minimum deviation, om = 30°:. Angle of prism, A = 60°By prism formula, reflected index

sin om + A sin 30° + 60°___ 2",--_ 2 __SID_·_4_SO_

sin A I 2 sin 30°

=...!..X2=1212

AI _ speed of light in vacuum (e)so, Ii - ....o..-_-:--~-,- -'c..:..speed of light in prism (v)

=> v = elli = (3XI08 112) mJs

Hence, speed of light through prism is(3 X 108 112) m/s

(1)

(ii) Critical angle i, is given as,.. 1

SID I =-c12=> ic = 45°

A = r + i, = 60°=> r = 60° - 45° = 15°Using Snell's law, sin i = 12

sin r=> sin i = 12 sin r = 12 x sin 15°

i = sin'" (12sin 15°)

63. (i) According to question,

+ht

_--u----c


Given, magnification (m) = - 2, R = - 20 cm

1= -10cmi.e. h2 = _ 2 = - v ~ U = ~ or v = 2u

hi U 2 (1)

Now, using mirror formula1 1 1 1 1 1-+-=- ~ -+-=-V U I zu u -10

1;u2= - 110 C I = %)

2=-~ ~ U= -10x3=_15cm2u 10 2

v=2xu=2x-15=-30cm (1)

Hence, the object distance and image distance are-15 ern and -30 ern respectively in front of themirror.

1 1 1(ii) According to mirror formula, i.e. - + - =-v u I

~=~-~ ~ v=-.!!Lv I u u-I

And we know, the value of u and I for a convexmirror are always negative and positiverespectively. So, the value of v will always bepositive it means convex mirror always forms avirtual image. (1)

64. As per the figure,The virtual image formed by lens LI is at P.

Therefore, using lens formula ~ = ~ - ~.I v u

As per the parameters given in the questionu = -15cm, ILl = 20 ernSo, the image distance will be

I 1 1- ----v (-15) 20

v = -60cmNow, this image is acting as an object for the lensL2. We can again use the lens formula and otherparameters given in the question and questionfigure to find the focal length of lens L2• (1)

1 1 1---=-VL> uL> IL>

Here, uL> = v+ (-20) = -60 - 20 = -80cm

VL> = 80cm1 1 1- ----80 (-80) IL>

IL> = 40cmSo, the focal length of the lens L2 = 40cm (1)

275

65. (i) The ray diagram for the formation of the imageof the phone is shown as below.The image of the part which is on the planeperpendicular to the principal axis will be onthe same plane. It will be of the same size,i.e. B'C = BC (112)

(1)

(ii) We may think that the image will now showonly half of the object, but considering thelaws of reflection to be true for all points of theremaining part of the mirror, the image will bethat of the whole object. However, as the areaof the reflecting surface has been reduced, theintensity of the image will be low, i.e. half. (1)

v

A ....,;::::-----------*.

,8,,,,

(112)

66. The ray diagram showing the image formation isshown as below:

(1)

(1)

o is at 21 of lens so it will form image at 2/, i.e.60 cm from lens so position of object for mirror isat (60 - 15) cm = 45 em behind the mirror. (1)

For mirror1=+10ernU= + 45cmv=?

1 1 1-+-=- ~V u I

90v=+-cm7

1 1 1-+-=-v 45 10

(behind the mirror)(1)

276n. Image formed by the lens will be 1at focus.

o

50 cm (1)

For mirror, U = - 30, 1= - 10According to lens formula,

111111111-=-+- ~ -=--- ~ -=---1 v U -10 v 30 v 30 10

~ ~=1-3 ~ v=-15cmv 30 (2)

A. The angle of incidence in denser medium forwhich the angle of refraction in rarer medium is90° is called the critical angle (icl for the pair ofmedia. The light rays emerge through a circle ofradius r.Because radius r= htani, = h- sini., = hn;1 III

COSI, 1-2..112

Hence, area of water surface

= 1th2 = 22 X (7)2 = 200.28cm2Il 2-1 7 (1.33)2-1

•••S (Source)

The power of a lens is equal to the reciprocal of itsfocal length when it is measured in metre. Power

of a lens, P = 1 Its SI unit is dioptre (D).I(metre)

o Chopterwise CBSE Solved Papers PHYSICS

Consider two lenses A and B of focal lengths, h and12 placed in contact with each other. An object isplaced at a point 0 beyond the focus of the firstlens A

The first lens produces an image (real image) at II'which serves as a virtual object for the second lensB producing the final image at I. (112)

Since, the lenses are thin, we assume the opticalcentres P of the lenses to be coincident. For theimage formed by the first lens A we obtain

1 1 1VI U h

For the image formed by the second lens B, weobtain

... (i)

1

V VI 12Adding Eqs. (i) and (ii). we obtain

1 1 1 1---=-+-V U h 12

... (ii)

... (ill)(1)

If the two lenses system is regarded as equivalentto a single lens of focal length I, we have

1 1 1v 1

... (iv)U

From Eqs. (iii) and (iv). we obtain1 I I-+-=-II 12 1

I 1(1)

70. For lens LI,

1 Uv (112)

Given, U = -15 em,v =7,1 =+ 10 em1 I I-=-+-10 v 15

Distance of image from lens LI,

~ V= 30 em1 I IFor lens L3, - = - - -t: v" u"

(112)

Distance of image from lens L3,

v"=10cm

~=~+~ ~ un=oo10 10 u" (1)

The refracted rays from lens LI become parallel toprincipal axis. It is possible only, when imageformed by LI lies at first focus of L2 i.e. at adistance of 10 cm from L2•

:. Separation between LI and L2 is= 30 + 10 = 40 em

The distance between L2 and L3 may take anyvalue. (1)


71. The light rays starting from bulb can pass throughthe surface if angle of incidence at surface is lessthan or equal to critical angle (C) for water airinterface. If h is the depth of bulb from thesurface, the light will emerge only through acircle of radius r (1J

As r = htani = h- sini, = h.~, cosi, ~1- 1

112 (1J

.................$ .. b hgiven yr = r2-:

v1l2-11th2

Area of water surface = -2-Il -1 (1J

=22x (0.80)2 =2.6m27 {1.33)2-1

72. Refraction at convex spherical surface. Whenobject is in rarer medium and image formed isreal. (1J

In l10AC, i = a + 'YandInl1AIC,'Y = r+~or r = 'Y-~

'1 1 sini i a + 'Y:. By Snell s aw n2 = -- = - = --sinr r 'Y-~

or n2 =a+'Y orn2'Y-n2~=l1:ta+nl'Yn1 'Y - ~

or (n2 - nd'Y = n1a + n2~ ... (i) (1J

As a, ~ and 'Yare small and P and N lie close toeach other.

So, a = tann = AN = ANNO PO

~ = tan~ = AN = ANNI PIAN AN'Y = tanv = - = -NC PC

271· .~•.•it

On using them in Eq. (i), we getAN AN AN(n2 - n1)- = n1 - + n2-PC PO PI

or n2 - n1 =.5... + n2PC PO PI

where, PC = +R radius of curvaturePO = -u, object distancePI = +V, image distanceSo, n2 - n1 =.5... + n2

R -u v

...(ii)

73.

n2-n1 =n2_~R v u

This gives formula for refraction at sphericalsurface, when object is in rarer medium. (1)

(i) From lens makers formula,

.!..=( Il -1)(~-~)1 m s Rl R2

or

As, ails < allm (1.65) for the first medium withrefractive index 1.65. (1J

(ii) And ails> allm{1.33) for the second mediumwith refractive index 1.33.Hence, the value of focal length 1 will benegative in the first medium.

(a) So, the convex lens will behave as thediverging lens for first medium and willbehave as the converging lens for the secondmedium as the sign of the focal length will notchange in second case: (1J

Given, ails = 1.5,allw = ~3

454

wlls = ails = 1. 5

allw 4/3

As, .!..=(1l-1)(~-~)1 Rl R2

(b):. 12 = ( ails - 1) (1.5 - 1 )ft wlls - 1 (:s - 1)

, .~

0.5 = 40.5

4

12 =4 ~ 12=4ftft

Change in focal length = 4ft - ft = 3ft (1)

Change in focal length is equal to thrice of itsoriginal focal length.

278 (ZJ Chapterwise eSSE Solved Papers PHYSICS

74 (.) F . 1 1 1• 1 or a muror, - = - + -I v u

For concave mirror, 1< 0

Also, u « 0 (by sign convention)

1 1 (take u and Iwith sign)-=----III v lul

1 1 1 ... (i)v lul III

where,12/1>lul> III ~~<O ~ v-c Ov

i.e. image forms on the same side as that ofobject, i.e. real image forms for extreme value.

Also, P/I> lul > III

1 1 1-<-<-1211 lul III

_l__ ~<-.!..._~<o12/1 III lul III

__ 1_<~<0 ~ v>-12/112/1 Ivl

Image is formed beyond 21 on the same side ofobject. (1)

(ii) For convex mirror, I> 0

Also, u « 0But, .!.. = ~ + !. = ~ - -.!...

I v u v lul(taking uwith sign)

1 1 1-=-+-V I lul

For I and lul to be positive,

~>o ~ v>Ov

~ Virtual image formed correspondingto the object.

(iii) For concave mirror,

1 1->-lul III

~>ov

v> 0Image is formed on RHS of mirror, i.e. virtualimage .

Also, 1 1 1-=---

I lvl lulFor concave mirror I is negative.

1 1-<-Ivl lul

M>1 ~ m>1lul

75.

Enlarged, virtual image formed on the other side ofmirror. (1)

According to the diagram,Given, for lens of focal length 10 cm.

I = + 10 em, u = - 30 emUsing lens formula,

1 1

I v u1 1~ -=----10 v (- 30)

~ v = 15 cm (1)

The image formed by rust lens acts as an virtualobject for plano-concave lens.For plano-concave lens,

u = + 10 em, I = - 10 em, v = ?Using lens formula,

1 1 1

I v u1 1~ --=---

10 v 10(1)

~=O ~v= 00.

v (1)

1< 0, u « 0

III> lul> 0!.=~+!.I v u1 1~ --=---

III v lul1 1 1-=--v lul III

lul< III

The refracted ray becomes parallel to prirlcipal axisfor convex lens of focal length 30 em.

u = - 00, v =?, 1= 30 em1 1 1-=---I v u

-.!... = _ - _1_ ~ v = 30 cm30 v (- 00)

So, final image is formed at a distance of 30 emfrom second convex lens on the other side of it. (1)


Using lens maker's formula (in water) forconverging lens,

~ = (wllg -1) (~+~)12 RI Rl

.l. = (.Ilg -1) (~ + ~)1; RI R2

76. Given, 1; = + 20 ern,

allg = 1.6,allw = 1.3

•Ilg 1.6~ wllg =-= -

allw 1.3

In air,

Dividing Eq. (ii) by Eq. (i), we get12_(.1l9-1) (1.6-1) 0.6x1.3

1; - (wllg -1) (~6_1) 031.3

12 = 2.61;

New focal length,12= 2.6 x 1; = 26 x 20

12= 52 ern

77. Refer to Am. 18.(i) Deviation of light rays through 900

R

L--_+--+-_~Q

(H) Deviation of light rays through 1800

R

... (i)

... (ii)(1)

p

(1x 2 = 2)

78. Let an object 0 is placed at a distance u fromconvex spherical refracting surface whose virtualimage formed at I at a distance v from surface. LetR is the radius of curvature of surface .

(1)

Rarer -111

1---- v -----IB

(1)

IndONC, i =a + yIn dINC, r = ~ + YAlso, for small angles o, ~ and y

... (i)... (H)

(1)

NM NM hn = tan n=- = - =-OM PO -u

[Minimum close to P]

where, h= NMNM NM h~ =tan~=-=-- =-1M PI - v

NM NM hAlso, y=tany=-=-=-MC PC + R

1

(1)

... (iii)

(1)

b S II' I sin i 112But Y ne saw, -- = -sin r III

where 112' III are the refractive indices of densermedium and rarer medium, respectively ..,' Angles iand r are small

(1)

sin i '" i, sin r '" ri = 112r III

III i =112 rIII (a + y) = 112 (l3 + y)

[From Eqs. (i) and (H)]~ Illn-1l2~=y(j..l2-1l1)

III (~)-1l2 (~)-u -v

= (+\}1l2 -IlIl [From Eq. (Hi)]

112 _~=1l2-1l1v u R (1)

This is the required expression known as Lensmakers formula.

280

79. When a ray after passing through a prism sufferst, minimum deviation, the ray will travel parallel to the

base of the prism inside the prism.)(1 : L...,,- -'- -'- .....I

Normal at Q -_.•.... -.•.•

Ns

8

Let PQ and RS are incident and emergent rays. Letincident ray get deviated by 0 by prism, i.e.

LTMS =0Suppose 01 and O2 are deviation produced atrefractors taking place at AB and AC, respectively... 0 = 01 + O2

o = (il - T1) + (i2 - T2)

0= (il + i2) - (T1 + T2) •.• (i) (1/2)

Also, in quadrilateral AQNR.

A + LQNR = 1800

[.; QN and RN are normal on two surfaces]Also, In 6.QNR.

LQNR + TI + T2 = 1800

~ A = TI + T2 ••• (ii)From Eqs. (i) and (ii), we get

0= (il + i2) - A ... (ill) (1/2)

Angle of deviation produced by prism varies withangle of incidence. When prism is adjusted atangle of minimum deviation, then

[say]il = i2 = iAt 0 = Om~ ~=~=T [g~From Eqs. (i) and (ii), we have

om = 2i - 2T and 2T = A. A+ Om~ 1=---

2AT=-2

. . Refractive index of material of prism is

sin (.4 + Om)Sill 1 2

~=SinT= . ASill-

2This is the required expression.

[21 Chapterwise CBSE Solved Papers PHYSICS

80. For convex lens: For erect image u = -ve, v = +ve

ifi I-vMagru icanon, m = - = -o u

where, 0 = length of objectI = length of image

Given, 1= + 20 ern I = 4 x length of object

~ i=4 ~ ~=4o -uv = - 4u (1)

Using lens formula,1 1 1

(1)1 v u (1)

1 1----1 (-4u) (- u)

1 1 1-=--+-1 4u u1 4 -1 3~ -=--=-

20 4u 4u

81.

u=20x3=15cm4

u = 15em, v = 4 u = 15 x 4 = 60 ern

Distance of the object, u = 15cm

Distance of the image, v = 60 em

The image is on the same side of the object.(1x 2 = 2)

Since, real and inverted image of an object can betaken on the screen.Given, v = + 10 emand magnification, U = -ve (for real image)

m = -19, 1=?

.. m=i=~ ~ -19=~ ~ v=-19uo u u

vU= --

19 (1)

(1)

Using lens formula,1 1 1-----

,I v u1 1

1 v -C~)1 1 19 20-=-+- ~1 v v 1 vv=lOcm

1.. I=-cm2

I=O.5crrr (2)


82. As the image of the object is formed by the lenson the screen, therefore the image is real.Let the object is placed at a distance x from thelens. As the distance between the object and thescreen is 90 cm. Therefore, the distance of theimage from the lens is (90 - x).

According to new cartesian sign conventions,U = -x, v = +(90 - x)

Magnification m = ~U

-2= (90-x) :=:)x=30cm-x

:. U = -30 cm , v = 60 ernLet f be focal length of the lens.According to thin lens formula

1 1 1 1 1 1~ - ;. = -f 60 - -(30) = f

1 1 1- + - = - :=:) f = + 20cm60 30 f

A convex lens of focal length 20 cm is required.(i) Focal length of spherical mirror does not get

affected with the increase of wavelength. (1)

(ii) Using lens maker's formula,

.!.. = (~-1)(~-~)f Rl R2

83.

fw = (.~9 - 1)

fa (w~9 - 1)

fw =20

1.5 -1

(~ -1)1.33

U5 x 1.33 x 200.17

fw = 78.2 cm

84. (i) No change as f of mirror depends only on itsradius of curvature. (1)

(ii) Refer to Ans. 86 (ii), fw = 39.11 em (2)

85. Given, length of object 0 = + 3cm

u=-60cm,

f= + 30cm1 1 1-=-+-f v U

[mirror formula 1(1)

or 1 1 1-=-+--30 v (- 60)1 1 1 2+1-=-+-=--v 30 60 60

:=:) V= 20cmor

(1)

(1)

(1)

281

~ = _ ~ :=:) _1_ = _ (+ 20) :=:) 1 = 1cmo u (+ 3) (- 60) (1)

86.

So, the virtual, erect and diminished image will beformed on the other side of the mirror.

Given ho (height of object = 2cm, u = -30 ern,f = 15cm)

U· . f 1111smg muror ormu a - = - + -f v u

1 1 1 1 1 1:=:) -=--- :=:) -=-+-

15 v 30 v 15 30

:=:) ~=2. :=:) v=lOcmv 30

L· ifi hj vmear magm icanon m = - =--ho U

or !!i =_~2 -30

2h, = - = 0.6cm3

87. To emerge from the prism, total internal reflectionwill take place on the surface hence incidence angleshould be greater than critical angle.

By geometry, angle of incidence (i) of all threerays is 45°. Light suffers total internal reflectionfor which this angle of incidence is greater thancritical angle.

i » ic :=:) sin i > sin ic1 1or sin 45° > sin ic or --- < --

sin 45° sin ic

(1)

Total internal reflection takes place on AC for rayswith ~ > .J2 = 1.414., i.e. green and blue coloursuffers. Total internal reflection whereas redundergoes refraction.

A

B--+-...,......,~

G --f-L--f--~

R--t--+--f-----». _

RB '--'-_+-_+- __ -'-..>0.

B G(2)

282 o ehopterwise eBSE Solved Papers PHYSICS

88. Refer to Ans. 18. (1)

The ray P retraces its path at silvered face when itfalls on silvered face normally.-n A

V'.1 Ill,!. r

p

(1)

For second face of the prism,

i2 ='2 = 0

But for prism,

"I + '2 = A = 30° ~ 1j = 30°For first face of the prism,

il = 45", 'I = 30°

11 = sin i = sin 45" = ~ x 2sin r sin 30° .J2

11 =.J2

11 = 1414 (1)

89 Real depth 12.4• As, 11 = --.-.:..:..:..:...--.:....::.!..-~Apparent depth 9.3·

4 " ~-1',. 'J":'11=-

3 rnRefractive index of water is ~. • ,

3. Again for another liquid, 11 = 1.6

(1lJ I .1.6= 12.4

Apparent depth

A d h 12.4~ pparent ept =-1.6= 7.75 cm (1)

Distance moved by microscope

= 9.3- 7.75= 1.55 em (1)

90. Let PQ and RS are incident and emergent rays. Letincident ray get deviated by 0 by the prism.i.e. LTMS =0

Let 01 and O2 are deviation produced at refractionstaking place at AB and AC, respectively.

A

p(1)

s

o = 01 + O2 = (i - 1j) + (e - '2)

= (i + e) - ('I + '2)But inMNR,

LFNR + LRFN + LFRN = 180°

.. ,(i)

or LFNR = 180° - ('I + '2) ... (ii)In DFARNF, LAFN and LARN are right angles.So, LFNR = 180° - A ... (ill)where, A is angle of prism.From Eqs. (ii) and (ill), we have

A ='1 + r2 •.• (iv)From Eqs. (i) and (iv), we have

0= (i + e) - A ... (v)i-0 graph is shown in the figure

r€Q)Ol

IV~ : om

Angle of incidence (i)---+ (1)

The conditions for the angle of minimum deviationare given us below:

(i) Angle of incidence (i) and angle of emergence(e) are equal.i.e. Li = Le

(ii) In equilateral prism, the refracted ray isparallel to base of prism.

(ill) The incident and emergent rays are bent onsame angle from refracting surfaces of theprism.i.e. L'I = L'2For minimum deviation position,

putting, = 'I = '2and i = e in Eq. (iv)

2r=A ~ ,=~ ...(vi)2


From Eq. (i) and l)m = 2i-A

i = A + l)m .•. (vii)2

:. Refractive index of material of prism issin i

11=-.-smr

From Eqs. (vi) and (vii), we get

_ sin (~_~/!R)11- sin Al2

91. Here, ic = 45"

(i) By definition of critical angle, i.e.~ 11=_1_=_1_=../2

sin ic sin 45" (1/2)

11=../2=:'v

c 3 X lOBv=-=---../2 1.414

v=212xlOB mls

92.

(ii) By Brewster's law,11= tan ip ~ t, = tan-Ill = tan " (.fi:)

ip = tan'" (../2) = 54.650'" 55 (1/2)

This particular angle of incidence is known aspolarising angle. (1)

(i) A refracting surface which forms a part of asphere of transparent refracting material iscalled a spherical refracting surface.

u v

n1

Io______I~~---------

U vThe above figure shows the geometry offormation of image I of an object 0 and theprincipal axis of a spherical surface with centreof curvature C and radius of curvature R .

283

(1)

Assumptions(a) The aperture of the surface is small as "

compared to other distance involved. "(b) NM will be taken to be nearly equal to the

length of the perpendicular from the point Non the principal axis.

tan LNOM = MNOM

~ tan LNCM = MNMC

tan LNIM = MNMI

For small angles, tanO '" sinO", 0

So, LNOM= MNOM

LNCM= MNMC

LNIM= MNMI

For ANOC, i is the exterior angle... i = L NOM + L NCM

= MN + MN ... (i)OM MC

For ANIC, LNCM is the exterior angle... LNCM = r+ LNIM

or r = L NCM - L NIMMN MNi.e. r = - - - ... (ii)MC Ml

By Snell's law, nl sin I = n2sin rFor small angles, 1\ i= n2r

Substituting the values of iand r from Eqs. (i) and(ii), we obtain

nl(MN + MN)=n2(MN _ MN)OM MC MC MI

(1)

or .!!L + .!!L = n2 - nlOM MI MC

j(;g-... (iii)

Applying new cartesian sign conventions,OM=-u

Ml=+ v

MC=+ R

Substituting these values in Eq. (iii), we obtainn2 _ nl = n2 - nl

v u RThis equation holds for any curved sphericalsurface.

''1:'' c.;

... (iv)(3)

284

Lens Maker's formulaIf a convex lens is made up of two convexspherical refracting surfaces. The final imagesformed after two refractions. Let n2 be therefractive index of the material of the lens and nl

be the refractive index of the rarer mediumaround the lens.

Let RI be the radius of curvature of second surfaceof the lens, II would have been a real image of 0formed after refraction, then from Eq. (iv),

... (v)

Let R2 be the radius of curvature of the secondsurface of the lens. Refraction is now taking placefrom denser to rarer medium

~ n2 = n2 - nl

v VI R2

Adding Eqs. (v) and (vi), we get

~ - ~ = (n2 - nd (~ - ~)v U RI R2

Put n2 = n = refractive index of material of thenl

lens with respect to surrounding medium

1 _ 1 _ (n2 -1) [1 _ 1] ... (viii)v u nl RI R2

... (vi)

... (vii)

When object on the left of lens is at infinity, thenimage is formed at the principal focus of the lens.

:. When u = 00, v = f = focal length of the lens.

:1' ~ = (n -1) [~ - ~]f RI R2

This is the lens maker's formula.(ii) According to question,

nl =1n2 = 1.5

R = 20cmu = -100cm

[Given]

IZl Chapterwise C8SE Solved Papers PHYSICS

93.

So, from surface formulan2 ~ = n2 - nl

v u R1.5 1 1.5-1-+-=--v 100 20

1.5 0.5 1-=---v 20 1001.5 5 1-=---v 200 1001.5 = 5-2 = _3_v 200 200

1.5 3---v 200

V= 200x1.5 = 300 =lOOcm3 3 (2)

(i) If the angle of incidence is increased gradually,then the angle of deviation first decreases,attains a minimum value (om)and then againstarts increasing.

co~.~ 0'0-o , ,~ om -----~c---- '~ ! I =e!, ,

, I

i1 i IzAngle of incidence

When angle of deviation is minimum, theprism is said to be placed in the minimumdeviation position.There is only one angle of incidence for whichthe angle of deviation is minimum.When o=om

[prism in minimum deviation position]e = i and r2 = 1j .•. (i)

rl+r2=A

From Eq. (i), r + r = A or Ar=-

2

Also, we haveA + 0 = i+ e ... (ii) (1)

Putting 0 = om and e = i in Eq. (ii), we get

A + om = i + i ~ i = ( A +2Om)

. (A + Om)sm ---.. Il= 2

. Asm-2

sin iIl=-

sin r(1)


(ii) The phenomenon of splitting of light into itscomponent colours is known as dispersion.The pattern of colour components of light iscalled the spectrum of the sunlight. (1)

R

White lightbeam

--I"'-~~~g~~~White light/ spectrum

v

Glass prismThe different colours of the white light havedifferent wavelengths. The wavelength ofviolet light is smaller than that of red light.The refractive index of a material in terms ofthe wavelength of the light is given byCauchy's expression

b c~==a+-+- ...(i)")..} A.4 (1/2)

where, a, b, c are constants for the material.Refractive index of material of prism ismaximum for violet colour (minimumwavelength) and minimum for red colour(maximum wavelength).i.e. u, >~,For a small angle prism, we have0==A (J.1 -1)

Since u, > Ilr' the violet light will suffergreater deviation than red light, i.e. Ov > 0,. (1)

(iii) Applying Snell's law at surface BC

Ar----r __ -+-__ ----jC

BIl xsin i == sin 900 x1. . 1 1~ sm 1 ==- ~ ~ ==---

Il sin 450

~ Il ==J2 (1/2)

So, J2 be the minimum refractive index ofglass so that the incident light undergoes totalinternal reflection.

94. Let a spherical surface separate a rarer medium ofrefractive index n1 from the second medium. ofrefractive index n2• Let Cbe the centre ofcurvature and R == MCbe the radius of the ' 1

surface. J

Consider a point object 0 lying on the principalaxis of the surface. Let a ray starting from oincident normally on the surface along OMandpass straight. Let another ray of light incident onNM along ON and refract along NI.From M , draw MN perpendicular to 01.The above figure shows the geometry of theformation of image I of an object 0 and theprincipal axis of a spherical surface with centre ofcurvature C and radius of curvature R . (2)

Here, we have to make following assumptions,(i) the aperture of the surface is small as

compared to the other distance involved.(ii) NM will be taken as nearly equal to the length

of the perpendicular from the point N on theprincipal axis. ('10 x 2)

tanLNOM == MNOM

tanLNCM == MNMC

tanLNIM == MNMI

For !:!NOC, is the exterior angle... Li == LNOM + LNCMFor small angles,

. MN MN1==--+-

OM NCSimilarly, r == LNCM - LNIM

MN MN~ r==----

NC NIBy Snell's law, we getn1sini ==n2sinrFor small angles, n1i == n2 rPut the values of i and r from Eqs. (i) and (ii},we get

nl(MN + MN) ==n2(MN _ MN)OM MC MC MI

... (i)

...(ii)

...(iii)

Applying new cartesian sign conventions, wegetOM ==- u, MI ==+ v and MC ==+ RSubstituting this in Eq. (ill), we get

286 o ehapterwise eBSE Solved Papers PHYSICS

n2 _ /12 = n2 - n, ...(iv)v u R

Now, the image I' acts as a virtual object for thesecond surface that will form a real at 1. As,refraction takes place from denser to rarermedium,

I

- n2 + !!L = n2 - nl

v y' -R'

On adding Eqs. (iv) and (v). we get

7 = (n21 - l{~-~,){.: n2, = :; , 7 = ~ - ;} (2)

...(v)

95. (i) A

The angles of incidence and refraction at first faceAB are il and Ii.

The angle of refraction at second face AC is r2 andthe angle of emergence e.The angles between the emergent ray RS andincident rays PQ is called angle of deviation (5)Here, LPQN = i, LSRK = e

LRQO = r" LQRO = r2LKIS = 1I, LTQO = iLTQR= i - r,

or LTRQ =e - r2In tJ.TQR, the side QT has been produced outwards.Therefore, the exterior angle 1I should be equal tothe sum of the interior opposite angles.i.e. 1I = LTQR + LTRQ

= (i - rl) + (e - r2)

1I = (i + e) - (rl + r2) ••• (i)

IntJ.QRO,rl + r2 + LROQ = 1800

••• (ii)from quadrilateral AROQ, we have sum of theangles

« AQO+ < ARO = 180°)This means that the sum of remaining two anglesshould be 180°.i.e. LA + LROQ = 1800 ••• (iii)

[LA is called the angle of prism]From Eqs. (ii) and (iii), we get

rl + r2 = A ... (iv)Substituting the value from Eq. (iv) to Eq. (I). weget

0= (i + e) - ALi + Le = Lo + LA (1%)

If the angle of incidence is increased gradually,then the angle of deviation first decreases, attains aminimum value (Om) and then again startsdecreasing.

;1 ; ;2Angle of incidence -

When angle of deviation is minimum the prism issaid to be placed in the minimum deviationposition.There is any only angle of incidence for whichangle of deviation is minimum.When °= om ~ e = i and r2 = rl(ii) Since, according to prism formula, we have

sin (A + 011I)

~ = 2

sin (1)As per the question,Angle of prism (A) = Angle of minimum deviation(lIm)

i.e. LA = LlIlII ... (Ii)Substituting the value of LOm from Eq. (ii) toEq. (i), we get

. (A + A)Sin --

~ ~= sin (4) ~

(1)

... (i)

sinA~=--sin (1)


~ Il = 2sin (4)- cos ( 4) ~ Il = 2COS(~). (A) 2SID -

2This is a required relation between refractive indexof the glass prism and angle of prism. (1112)

Since. LA = 6ri' (given)

(600

)Il = 2cos 2 ~Il= 2cos300

Il = 2x.J3 ~ Il =.J3 ~Il = 1.7322

96.(1)

(i) The incident rays coming from the object 0kept in the rarer medium of refractive indexnJ • incident on the refracting. Surface NMproduces the real image at I.

u c v

cFrom the diagram.

Li = LNOM + LNCM= NM + NMOM MC

Lr = LNCM _ LNIM= NM _ NMOM NI

From Snell's law.n2 = sini _~nJ sinr r

(for small angle. sin 9 '" 9)

(1)

n2 - nJ = n2 _ !iR v v (1J

The first refracting surface ABC forms the image IIof the object o. The image II acts as a virtual objectfor the second refracting surface ADC which formsthe real image I as shown in the diagram.For refraction at ABC.

n2 -!i = n2 - nl ••• (i)vJ U RI

For refraction of ADC.

n2 -!i = nJ - n2 ... (ii)v U RJ2

Adding Eq. (i) and Eq. (It). we get

n2 _ !i= (n2 - nJ) (~ - ~)v U RJ R2.~ ~ - ( :; 1)UI ~J

We know that. if LI = 00. V = f1 1 1~ ---=-v u f

~ = (n2 -I) (~ -~)f RI R2

(ii) Given. Il = 1.55f= 20cm

We know that

~ = (Il - I) (~ - ~)f RJ R2

.l.= (1.55-1) [~- (~)]20 R-R

~ = 0.55 x ~20 R

R = 0.55x z x 20R = 22cm

(2)

(1)

97.9B.

Refer to Ans. 96 (i). (5)

(i) Refer to Ans. 79. (2)

(ii) When light is incident on one end ofthe opticalfibre at an angle of incidence greater than thecritical angle for the glass cladding pair ofmedia.The light suffers repeated total internalreflection and light travels through the

2BB

optical fibre without any loss of energy fromone place to other inside the optical fibre. (2)

CladdingLow 11=1.48

Glass coreHigh 11=1.52

99. (i) Refer to Ans. 78.

(ii) Refer to Ans. 50.A'B' PB' - vAB PB - u

A'B' = - I

AB=+ 0I v

m=--=-o uI v

m=-=--o u

100. Let, NM = hThe convex spherical refracting surface formsthe image of object 0 at I. The radius of curvatureis R (1)

Rarer n1Denser n2

Br--u v ---+I

PC=+ RPI = + v

PO=-u

In mco, i = Y + aIn mCI, y = r + ~~ r=y-~

... (i)

... (ii)

o ehapterwise eBSE Solved Papers PHYSICS

For small angles a, ~ and y, we have

a=tana= MN = MN =~MO PO - u (1)

~=tan~=MN=MN=~ ... (ill)MI PI - v

MN MN hy=tany=-=-=-MC PC + R

(1)

(3)

Assuming M is very close to P.

By Snell's law,n2 sin i-=11=-nl sin r

For small i and r,n2 i .- =- or m2 = Inlnl r

(1)

(2)

n2 (y -~) = (a + y) nl [From Eqs. (i) and (ii)](n2 - nd y = nla + n2~(n2 - nd(i) =nl (_hu )+n2(~)

[From Eq. (iii)]n2 _ ~ = n2 - nl

v u R (1)

101. Refer to Ans. 100. (3)

Lens maker's formula refer to sol. 96. (2)

102. (i) Refer to Ans. 98. (2)(ii) The reciprocal of focal length of lens is known

as power of lens when focal length is taken inmetre.

(1)

P = 1f (in metre)

SI unit of power oflens is dioptre (D).1; = + 50 ern, f2 = - 20 cm

.: ~ = ~ + ~ = ~ _ ~ = 2 - 5 = ~f 1; f2 50 20 100 100

100 1f=-- cm ~ f=--m3 3

p= __I_=_I_=_3Df(in m) (-1/3) (1)

(1)

(1)


103. (i) Let the convex mirror form virtual, erectdiminished image A' B' on the other side ofmirror of an object AB as shown in the figure.Let PC = + 2f = R, PB' = + v, PB = - u (1)

A

B

'I' v 1I+-- f --+I

I- 2f ---+1'1

M'B'Cis similar to MECA' B' CB' PC - PB' R - v~--=-= =--AB CB PC + PB R - u

Also, M' B' P - MEP.A'B' AB' v~ ------AB PB - u

From Eqs. (i) and (ii). we getR-v v--=--R-u uuR - uv = - vR + uv

... (i)

... (ii)

289

2uv= uR + vROn dividing by uvk; we get

2 1 1-=-+-R v u

1 1 1R = 2f or - = - + -f v u

(1)

(1)

This is the required expression for mirrorformula.

(ii) For concave mirror, f < o.Object distance, u < 0 (1)

But, (f) > lul> 0By mirror formula, we have

1 1 1 1- = - + - or - = - - -f v u v f u

f=-Iflu=-Iul1 1 1 1 1-=--+- ~v -If I lul v lul If I

~ (v> 0)But, lul< (f)

~ ~<~ ~ ~-~>Olul If I lul If I

~ ~>o.v (1)

The image form on the other side of mirror as vis positive.

[TOPIC 2] Optirol tnstrumentsNear Point and Far Point of Human Eye. Near Point (0) The minimum distance from the eye at which an object can be seen most distinctly

without any strain. For a healthy normal eye, it is 25 em, Its also known as least distance of distinctvision.

. Far Point The farthest point from the eye, at which an object can be seen clearly by the eye is calledthe far point of the eye. For a normal eye, the far point is at infinity.

Simple Microscope (Magnifying Glass)It is a converging lens of short focal length, held close to the eye.

Case I When image is formed at the near point. Then,M=l+~

fIn the case of when eye is placed behind the lens at adistance a, then

M=l+(D-a)f

Case II When the image is formed at, the infinity.D

M=-f

Compound Microscope

A'"~..•.",'...,.............•.", L....•.." .•..•...•.....

A" ---------

8' F-o----\+<

Its an optical device which consists of two convex lenses, one objective of very small focal length withshort aperture and one eyepiece, E of moderate focal length and large aperture.

Eye lens,E,,,,,,,,,, ,

A

8 Fo

j+-uo----+f.r++-,,

" A',,," ,~ ..,,, ,", ,.,.-:::",

,~:::-:-:.•. ""

A" wI~""1----0 I

Compound microscope, finol image at D.

---u---oj

Magnifying power, M = me x m,where, me and mo are the individual magnifying powers of objective and eye lens.


M = Vo (1 + E.), when final image is at near point M = Vo x D, when final image is at the infinity», t, ». Ie t

= -~ x E., where L = Distance between the two lenses = tube lengthh h .Eyepiece

~qO)201JIM9JQmi2

Compound microscope, final image at infinity,

i.e. when the final image is formed at infinity, the length of the compound microscope, L = Vo + Ie

Reflecting Astronomical TelescopeThese telescopes form image free from chromatic aberration and spherical aberration.

Cossegrainion reflecting telescopeThe two types of reflecting telescopes are

(i) Newtonian reflecting telescope.

Parallel raysfrom distant

object

(ii) Cassegrainian reflecting telescope.

Magnification, m = 10 (1 + Ie)Ie D

292 o Chopterwise CBSE Solved Papers PHYSICS

For final image formed at infinity,

where, I. = ~ and m = fa = R/2o 2 I, t,

and R is the radius of curvature of objective mirror.

Refracting Astronomical TelescopeIt consists of an objective lens of a large focal length (fo) and large aperture, also an eye lens of smallaperture and focal length.

(i) Magnification when final image is formed at D,

=> m = - fo (1 + fe) and length of telescope, L = II,I + fe DI, D fe+D

Objective lens---fa

Eye

---- L------

(ii) Magnification when final image is formed at infinity (normal adjustment)

m=- fofe

L = Ifol + Ifel

Resolving PowerI . f . 1 2Jlsin e R I· 1• Reso vmg power 0 rrucroscope = d = A ' eso vmg power oc i

where, Jl sin e = numerical aperture, e is the semi-vertical angle of the cone formed by object at objective,Jl = R1 of the medium between object and lens and d = minimum distance between two objects whoseimage can be seen through microscope vivid and clear. A is a wavelength of light used to illuminate theobject.

= Resolving power of a telescope = _1_ = _D_ => Resolving power oc ~~e 1.221. Awhere, D is the diameter or aperture of objective lens and A. is the wavelength of light used.

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 2o 2 Marks Questions

1. Define the magnifying power of acompound microscope when the finalimage is formed at infinity. Why mustboth the objective and the eyepiece of acompound microscope has short focallengths? Explain. Delhi 2017

2. A small telescope has an objective lens offocal length 150 em and eyepiece of focallength 5 cm. What is the magnifyingpower of the telescope for viewingdistance objects in normal adjustments.If this telescope is used to view a 100 mtall tower 3 km away, then what is theheight of the tower formed by theobjective lens. Delhi 2015

3. You are given two converging lenses offocal lengths 1.25 em and 5 ern to designto compound microscope. If it is desired tohave a magnification of 30, find out theseparation between the objective and theeyepiece. All India 2015

4. Draw a ray diagram of a reflecting typetelescope. State two advantages of thistelescope over a refracting telescope.Delhi 2014 C; All India 200BC; Delhi 200B

5. Draw a schematic diagram of refractingtelescope. Write its two importantlimitations. Delhi 2014C

6. Draw a ray diagram for the formation ofimage by a compound microscope. Write theexpression for total magnification when theimage is formed at infinity. Delhi 2014C

7. Draw a schematic arrangement of areflecting telescope (Cassegrain) showinghow rays coming from a distant object arereceived at the eyepiece. Write its twoimportant advantages over a refractingtelescope. Delhi 2013C

8. Two convex lenses of same focal lengthbut of apertures Ar and ~ (~ < A 1)' areused as the objective lenses in twoastronomical telescopes having identicaleyepieces. What is the ratio of theirresolving power? Which telescope will youprefer and why? Give reason. Delhi 2011

9. Define the resolving power of a telescope.Write any two advantages of a reflectingtelescope over a refracting telescope.Delhi 2D10C J t,

10. Define the magnifying power.of.a =compound microscope. Why should boththe objective and the eyepiece have smallfocal lengths in a microscope? Delhi 2010C

11. The near point of a hypermetropic personis 50 em from the eye. What is the powerof the lens required to enable the personto read clearly a book held at 25 cm fromthe eye? All India 2009C

o 3 Marks Questions12. (i) Draw a ray diagram depicting the

formation of the image by anastronomical telescope in normaladjustment.

(ii) You are given the following threelenses. Which two lenses will you useas an eyepiece and as an objective toconstruct an astronomical telescope?Give reason.

Lenses811

Power (D) Aperture (em)3610 .

All India 2017

13. Draw a schematic ray diagram of reflectingtelescope showing how rays coming from adistant object are received at the eyepiece.Write its two important advantages over arefracting telescope. Delhi 2016. Foreign 2013

14. (i) A giant refracting telescope has anobjective lens of focal length 15 m. Ifan eyepiece of focal length 1.0 err{ isused, what is the angularmagnification of the telescope?

294

(ii) If this telescope is used to view themoon, what is the diameter of theimage of the moon formed by theobjective lens? The diameter of themoon is 3.48x 106 m and the radius oflunar orbit is 3.8 x 108 m. AllIndia 2015

15. Which two of the following lenses ~, L2 and~ will you select as objective and eyepiecefor constructing best possible (i) telescope(ii) microscope? Give reason to support youranswer. 0.lh12015C~-~.~ri.iii-~::i~

__ LL_ lOD J 1 em

16. (i) Draw a labelled ray diagram showingthe formation of a final image by acompound microscope at least distanceof distinct vision.

(ii) The total magnification produced by acompound microscope is 20. Themagnification produced by the eyepieceis 5. The microscope is focused on acertain object. The distance betweenthe objective and eyepiece is observedto be 14 ern. If least distance of distinctvision is 20 cm. Calculate the focallength of the objective and theeyepiece. Oelhl 2014 C

17. Draw a labelled ray diagram of a refractingtelescope. Define its magnifying power andwrite the expression for it.Write two important limitations of arefracting telescope over a reflecting typetelescope. AllIndia 2013

18. Draw a ray diagram showing the imageformation by a compound microscope.Hence, obtain expression for totalmagnification when the image is formed atthe infinity. Oelhl2013

19. A compound microscope uses an objectivelens of focal length 4 em and eyepiece lensof focal length 10 em. An object is placed at6 emfrom the objective lens. Calculate themagnifying power of the compoundmicroscope. Also, calculate the length of themicroscope. AllIndia 2011

o Chopterwise C8SE Solved Papers PHYSICS

20. A giant reflecting telescope at anobservatory has an objective lens of focallength 15 m. If an eyepiece lens of focallength 1.0 ern is used, find the angularmagnification of the telescope.If this telescope is used to view themoon, what is the diameter of the imageof the moon formed by the objectivelens? The diameter of the moon is3.42 x 106 m and the radius of the lunarorbit is 3.8x 108 m. AllIndia 2011

21. Two convex lenses of focal length 20 emand 1em constitute a telescope. Thetelescope is focused on a point which is1 m away from the objective. Calculatethe magnification produced and thelength of the tube if the final image isformed at a distance 25 em from theeyepiece. Delhi 2011C

22. The objective of an astronomicaltelescope has a diameter of 150 mm anda focal length of 4.00 m. The eyepiecehas a focal length of 25.00 mm.Calculate the magnifying and resolvingpower of telescope. (A. = 6000 A foryellow colour).Delhl2011C

23. (i) Draw a neat labelled ray diagramof an astronomical telescope innormal adjustment. Explain brieflyits working.

(ii) An astronomical telescope uses twolenses ofpowers 10D and lD. Whatis its magnifying power in normaladjustment? AllIndia 2010

24. (i) Draw a neat labelled ray diagramof a compound microscope. Explainbriefly its working.

(ii) Why must both the objectiveand theeyepieceofa compoundmicroscopehave short focallengths? AllIndia 2010

25. Draw a schematic diagram of areflecting telescope (Cassegrain). Writetwo important advantages that thereflecting telescope has over a refractingtype.Farelgn 2010


26. Explain with the help of a ray diagram,the working of an astronomical telescope.The magnifying power of a telescope in itsnormal adjustment is 20. If the length ofthe telescope is 105 em in this adjustment,find the focal lengths of the two lenses. AllIndia 2010C

o 5 Marks Questions27. (i) Draw a labelled ray diagram to obtain

the real image formed by anastronomical telescope in normaladjustment position. Define itsmagnifying power.

(ii) You are given three lenses of power0.5 D, 4 D and 10 D to design atelescope.

(a) Which lenses should be used asobjective and eyepiece? Justify youranswer.

(b) Why is the aperture of the objectivepreferred to be large? All India 2013

28. (i) Draw a labelled ray diagram showingthe image formation of a distantobject by refracting telescope.Deduce the expression for itsmagnifying power when the finalimage is formed at infinity.

(ii) The sum of focal lengths of the twolenses of a refracting telescope is105 cm. The focal length of one lens is20 times that of the other. Determinethe total magnification of thetelescope when the final image isformed at infinity. All India 2014

29. Define magnifying power of a telescope.Write its expression.A small telescope has an objective lens offocal length 150 cm and an eyepiece offocal length 5 cm. If this telescope is usedto view a 100 m high tower 3 km away,find the height of the final image when itis formed 25 cm away from the eyepiece.Oelhl2012

295

30. How is the working of a telescopedifferent from that of a microscope?The focal lengths of the objective andeyepiece of a microscope are 1.25 cm and5 cm, respectively. Find the position ofthe object relative to the objective inorder to obtain an angular magnificationof 30 in normal adjustment. Deihl 2012

31. Draw a ray diagram to show the workingof a compound microscope. Deduce anexpression for the total magnificationwhen the final image is formed at thenear point.In a compound microscope, an object isplaced at a distance of 1.5 em from theobjective of focal length 1.25 cm. If theeyepiece has a focal length of 5 em andthe final image is formed at the nearpoint. Estimate the magnifying power ofthe microscope. Delhi 2010

32. (i) (a)Draw a labelled ray diagram toshow the formation of image in anastronomical telescope for adistant object.

(b)Write the three distinctadvantages of a reflecting typetelescope over a refracting typetelescope.

(ii) A convex lens of focal length 10 ern isplaced coaxially 5 em away from aconcave lens of focal length 10 cm. Ifan object is placed 30 ern in front ofthe convex lens. Find the position ofthe final image formed by thecombined system. All India 2009

33. Draw the labelled ray diagram for theformation of image by an astronomicaltelescope.Derive the expression for its magnifyingpower in normal adjustment. Write twobasic features which can distinguishbetween a telescope and a compoundmicroscope. Foreign 2009

o Explanations1. Angular magnification or magnifying power of

compound microscope is defined as ratio of anglemade at eye by image formed at infinity to theangle made by object, if placed at distance ofdistinct vision from an unaided eye. (1)

ificati LDMagm icanon = --fo·1.where, L is length of the tube of microscope.

As, m oc 2.. and m oc 2..fo I.

:.Both eyepiece and objective must be of smallerfocal lengths, so that magnification is higher. (1)

2. When final image is at D,then magnifying power,

M= -fo(l+ f.)I. D

In normal adjustment,

M = -foI.

For telescope,Focal length of objective lens, fo = 150 cmFocal length of eye lens, I. = 5emWhen final image forms at D = 25cm

.. Magnification, M = -fa (1 + f.)I. D

=-150(1+2)=-150x~ ~ M=-365 25 5 5 (1)

Let height of final image be h cm.

~ tanp=~25

P = visual angle formed by final image at eye.a = visual angle subtended by object at objective.

100m 1tanu = --- = -3000m 30

B M = tanput,tan«

~ -36 = (:5) ~ -36 = ~ x 30(;0) 25

~ -36 = 6h ~ h = -36 x 55 6

h = - 30cm

3. Maximum magnification of a compoundmicroscope is

m= :: [1+ ~]

So, for m to be 30,

30 = Vo [1 + 25] or 30 = Vo [6]Uo 5 Uo

Vo = 5uo ... (i) (1)

For objective of focal length 1.25 ern,1 1 1---=-Vo Uo fo

fa =1.25 em

~

(1)

1---=-5uo -Uo 1.25

1+ 5 1----5 Uo 1.25

5 Uo = +7.5cm or Uo = 1.5cm So, Vo = +7.5cmNow, ue for required magnification

1 1 1 1 1 1---=-or---=-ve u, I. -25 -ue 5

1 1 1 5+ 1 25- = - + - = -- or ue = - cm», 5 25 25 6

Hence, separation between two lenses should be25Vo + ue = 7.5cm+ -cm = 11.67cm6 rn

4. Ray diagram of a reflecting type telescopeRefer figure on page 291. (1)

Advantages(i) Reflecting telescopes have high resolving

power due to a large aperture of mirrors.(ii) Due to availability of paraboloidal mirror, the

image is free from chromatic and sphericalaberration. (1'12)

5. Refer figure on page 292. (1'12)

Limitations of refracting telescope over areflecting type telescope.

(i) Refracting due to telescope suffers fromchromatic aberration, due to uses of large sizedlenses. (1)

(ii) It is difficult and expensive to make such largesized lenses. (1)


6. A compound microscope consists of two convexlenses coaxially separated by some distance. Thelens nearer to the object is called the objective.The lens through which the final image is viewedis called the eyepiece. The focal length ofobjective lens is smaller than eyepiece.Refer figure on page 290. (2)

7. Diagram of a reflecting telescope (Cassegrain) isshown as below:Cassegrain reflecting telescope It consists of alarge concave paraboloidal (primary) mirrorhaving a hole at its centre. There is a smallconvex (secondary) mirror near the focus of theprimary mirror. The eyepiece is placed on the axisof the telescope near the hole of the primarymirror.Advantages of reflecting telescope over arefracting telescope are given as below:

(i) A reflecting telescope reflects all wavelengthsof light at the same angle, so there are nocolour halos.

Secondarymirror">,

(ii) A mirror has only one surface to be figured, soit is easier to control the shape.

(ill) (a) A mirror reflects the light, so the materialthat is made from does not have to betransparent.

(b) Infrared and ultraviolet light reflectsequally well. (2)

8. Resolving power of telescope,A

R" = 1.221..

where, A = aperture or diameter of the objectivetelescopeand A = wavelength of the objective.~ RocA

.. Ratio of resolving powers of two telescopesRI = AlR2 A2

A2> Al.. R2 > RI (1)

The larger the aperture of objective, higher theresolving power of telescope. As well moregathering of light to form the image and hence,brighter image would be obtained. (1)

297

9. Resolving power of a microscope is defined asthe reciprocal of its limit of resolution (d) i.e.RP of microscope = lidwhere, limit of resolution is equal to the smallestdistance between two closest objects whose vividor clean image can be seen through the

microscope and given by d = __ 1..__211 sin 9

:. Resolving power of microscope = 211 sin 9A

where, A = wavelength of light used, (1)

9 = semivertical angle of the cone formed by objectat objective and 11 = refractive index of moleculebetween object and lens.(a) Resolving power increases with the increase

of'u. .(b) Resolving power decreases as resolving power

oc III... (1/2 x 2 = 1)

Magnifying power The magnifying power of atelescope is equal to the ratio of the visual anglesub tended at the eye by final image formed atleast distance of distance vision to the visualangle sub tended at naked eye by the object atinfinity.Telescope has objective of large aperture and largefocal length whereas microscope have objective ofsmall aperture and focal length.The relative distance between objective and eyelens may change in telescope whereas theseparation between objective and eye lens incompound microscope remains fixed. (2)

The book be placed at a distance of 25 ern fromthe lens to get its image at a distance 50 cm.i.e u = - 25 crn v = - 50 em

1 1

10.

11.

As,f v u1 1 1 1-=--+-=+-f - 50 25 50

f= 50 cm

p =100 =100 =2Df 50

p= 2D (1)

~

~~

~12. (i) An astronomical telescope is an optical

instrument which is used for observingdistinct images of heavenly bodies likeplanets, stars, etc. It has two convex lens(objective and eye lens) placed coaxially andseparated by some distance in normal

298 o Chopterwise CBSE Solved Popers PHYSICS

adjustment. Final image is formed atinfinity as depicted below.

Objective lens--- '0----+-

FEye

The final Image is magnified and Inverted(1'12)

(ii) In the astronomical telescope, aperture ofobjective must be less than eyepiece.Therefore, possible combinations are (Lland L 3) or (Ll and L 2)' Also, focal length ofthe objective (10) must be greater than thatof eyepiece (f.).

1 110>1. ~-<- ~Po<p.10 I.., Power of objective (Po)must be less thanpower of eyepiece (P.).

Now, for (L1 and L 3) combination,

(10) = p. = 10I. 1 Po 3

For (Ll and L 2) combination,

(10) =P'=~=2«fo)1.2 Po 3 1.1

Thus, the best combination of the lenses is (Lland L 3)' (l'12J

13. Reflecting telescope consists of concave mirrorof large aperture and large focal length(objective). A planemirror is placed betweenthe concave mirror and its focus. A smallconvex lens works as eyepjece.

EyePiece(~

Rays arriveparallel fromvery distantobject------------~~----------~

M2 = plane mirror (1)

Advantages of Reflecting Telescope overRefracting Telescope

14.

For a reflecting telescope, the mirror affords severaladvantages over the objective lens in astronomicaltelescope(i) A mirror is easier to produce with a larger

diameter, so thln it can intercept rays crossing alarger area and direct them to the eyepiece. (1)

(ii) The mirror can be made parabolic to reduce'spherical aberration. Aberration is furtherreduced because passage through one layer ofglass (the objective lens) is eliminated.

(i) Let 10 = focal length of the objective lens= 15m = 1500 cm

I. = focal length of the eye lens = 1em.,' Angular magnification of giant refractingtelescope is given by

nlo = 17.1=1151

001= 1500

(1)

(ii)(1'12)

Diameter of the image of the moon formed by theobjective lens, d = ex 10~ d = Diameter of the moon

Diameter of the lunar orbit6

148 x 10 x 15= 0135m = 13.5cm18 x 108 (1'12)

15. An astronomical telescope should have an objectiveof larger aperture and longer focal length while aneyepiece of small aperture and small focal length.Therefore, we will use L2 as an objective and L3 asan eyepiece. For constructing microscope, L3 shouldbe used as objective and L, as eyepiece because boththe lenses of microscope should have short focallengths and the focal length of objective should besmaller than the eyepiece.

(i) Refer to Ans. 6, refer figure on page 290. (1)(ii) Given, magnification, M = 20

Magnification of eyepiece, m. = 5Least distance vision, D = 20 cmDistance between objective and eyepiece,L = 14 cmWe know that,Magnification, M = me x mo

m 20 Dmo =-=-=4~ m, =1 +-me 5 I.

where, I. is focal length of eyepiece.20~ 5 = 1 + - ~ f, = 5 cmf.' (1)

Using lens formula for eyepiece,1 -1 1 -5 -1-=---=-=-ue 20 5 20 4

16.


U, =-4 cm(objective distance for eyepiece)

L = Vo + Iu, I=Ho = L -I u, 1= 14 - 4 = 10 cmMagnification produced by objective,

mo =-~Uo

Object distance for objective,-v -10

Uo =_0 = - = -25cmmo 4

Using lens formula for objective,IIII I I I-=---=---=-+-=> fa =2cm

fa Vo Uo 10 -2.5 10 2.5 (1)

17. Refer to Ans. 5. (3)

18. Refe+ ,« Ans. 6. (3)

19. F~r compound microscope,fa = 4 ern, .r. = 10 em

Uo = - 6 cm, v, = - D = - 25 cmFor objective lens,

~ = ~ - ~ => ~ = ~ + (~) => ~ = ~ - ~ = ~fa Vo Uo 4 Vo 6 Vo 4 6 12

Vo =12cm:. Magnifying power M

= _ ( Vo ') (I + E.)\.u. J F,

=-(¥)(I+ ~~)=-2G)=-7

Magnifying power, M = - 7 (1)

Length of microscope = Ivo I+ Iu, Iwhere, Vo = 12 emFor eye lens,

v, =-25cm..r. =IOcm,u, =7~=~ It: v, u,I I I I I=> -=---=----

U, v, .r. - 25 10- 2- 5 7

-=--=v, 50 50 (1)

U = - 50 cm = -714cm, 7. . Length of microscope

=lvol+luel=12+7.14=19.l4cm (1)

NOTE1. The separation between objective and eye lens is

known as length of microscope.2. The image formed by objective is an object for

eye lens.

20. For astronomical telescope,fa=15m=1500cm, .r.=lcm

Angular magnification, m = _ fat:= 15 x 100 cm = _ 1500

I cm

(112)

(1)

NOTE As clear from the figure of astronomical telescope,angle subtended by moon at the objective must beequal to the angle subtended by image formed byobjective on the objective lens.

.. The angle subtended by moon at objectiveDa=-------

Radius of lunar orbit3.42 x 106 ma = ---,,---3.8x 108 m

Also, then angle sublended by image formed byobjective on itself

da =-fa

where, d = diameter of imageFrom Eqs. (i) and (ii), w.e get

3.42 x 106 _ d3.8 X 108 - 1500

... (i)

... (ii)

(1'12)

6d=1500x142XIO =13.5cm

3.8 x 108

21. Given, fo = 20cm,.r. = I ern. ve = - 25cm

For objectiveUo = -100cm, fa = 20cmI I I I I I-=--- => -=----fa Vo Uo 20 Vo (- 100)I 1 1 5-1 4-=---=--=-Vo 20 100 100 100

Vo = 25cm (112)

For eye lens.r. = 1 ern u, = 7, ve = - 25I I 1 I I I-=----- => -=----.r. ve u, 1 - 25 ue

I + ~ = _ ~ => 26 = 125 ». 25 »,

25u =--e 26

lu, I = 0.96 ern (1/2)Magnification

m = - Vo (1+ E.) = _ (~) (I + 25)Uo I; 100 I

300

m = - ~ x 26 = - 6.54

Length of telescope, L = Vo + ue = 25 + 0.96L = 25.96 cm

22. The diameter of objective of the telescope= 150 x 10-3 m

10 =4mIe = 25 X 10-3 m and D = 0.25 m

Magnifying power,

M = - to (1 + E.)t: t: (1)

J.1 = - 4 (1 + 0.25 J25xl03 25xl03

M = _ 4000 (l + }'O) = __4000 x 1125 25

M = -1760 (1)

Resolving power = ..!..de

de= 1.2lAD

1.22 x 6 x 10-7 2.9 X 10-6 rad0.25 (1)

:. Resolving power = 1 = 0.34 X 106

29 x 10 6

23. (i) Refer to Ans ..5.The image formed by objective lens must fallon the focus of eye lens in order to form finalimage at infinity. (1)

1(ii) As, f. = - = 0.1 m=10 cme 10

1f. = - = 1 m = 100 cme 1 (1)

Magnifying power in normal adjustment,M = _ to = _100

Ie 10

.. M = -10 (1)

24. (i) Refer to Ans. 6.The magnification by compound microscope istwo step process.Firstly, the objective gives a magnified imageof the object and after that the eyepieceproduces the angular magnification. (1'!.)

(ii) to and Ie of compound microscope must besmall so as to have large magnifying power as

M=-~(l+ DJto Ie (1'!.)


(1)25. When rays from infinity, i.e. parallel rays are

reflected by a concave mirror they willtend to meetat focus after reflection.

(1) Diagram of a reflecting telescope~~~~~~~~~ Paraboloidal

objectivemirrorSecondary

mirror ->

Parallel rays from F <:distant object '

26.

(1'1.)

Refer to Ans. 4 (for advantages). (.1'1.)

For figure refer to Ans. 7. (1)

The parallel beam of light from distant object getfocused in focal plane of objective at angle u, Thisimage A' B' acts as an object for eye lens which liebetween optical centre of eye lens and its focus.Eyepiece form virtual, erect, magnified imageA" B" at least distant of distinct vision. (1)

In normal adjustment, M = 17.1 = 20

to = 20leAlso, length of telescope,

to + Ie = 10520le + t; = 105 => 21 Ie = 105

=> Ie = 5 em10 = 201e = 20 x 5= 100 em (1)

(i) In astronomical telescope for normaladjustment final image is formed at infinityand it is virtual.The labelled ray diagram to obtain one of thereal image formed by the astronomicaltelescope is as follows.

t+---- fa ' I , fe-+<

27.

Magnifying power is defined as the ratio of theangle sub tended at the eye by the focal imageas seen through the telescope to the anglesubtended at the eye by the object seendirectly, when both the image and the objectlies at infinity. (3)


28.

(ii) (a) We know objective lens of a telescopeshould have larger focal length andeyepiece lens should have smaller focallength. And focal length is inverse of

power, so lens of power (p = -7)10D can be used as eyepiece and lens ofpower 0.5 D can be used as objective lens.

(b) The objective lens of a telescope shouldhave larger aperture, in order to formbright image of an distant objects, so thatit can gather sufficient light rays from thedistant objects. 121

(i) For figure refer to Ans. 5.When the final image is formed at infinity,angular magnification is given by M = ~a.However, 13 and a. are very small.• • 13 '" tan 13 or a. '" tan a.==> M = tan 13

tan a.I is the image formed by the objective, J;, and Ieare the focal length of objective and eyepiece,respectively.

tan a. = ~ orJ;,I

M = ~ or M = - J;,~ IeJ;,

(ii) Given, 10 + J;, = 105,J;, = 20 L105f. = - = 5 ==> J: = 20 x 5= 100cm

e 21 0

M = J;, = 100 = 20Ie 5

Itan 13 =---Ie

29. The magnifying power of a telescope is equal tothe ratio of the visual angle sub tended at the eyeby final image formed at least distance of distinctvision to the visual angle sub tended at naked eyeby the object at infinity. IIIWhen final image is at D,

magnifying power, M = J;, (1 + Ie)t. D

In normal adjustment, M = _ J;,t: IIIFor telescopeFocal length of objective lens, J;, = 150 cmFocal length of eye lens, Ie = 5 em

301

When final image forms at D = 25 cm

:. Magnification, M = - J;, (1 + Ie)t: D

= _150 (1+ 2) = _150 x ~5 25 5 5

M = - 36 11%1

Let height of final image is h ernhtan 13 =-25

13 = visual angle formed by final image at eyea. = visual angle sub tended by object at objective

tan a. = 100 m = ~3000 m 30

M= tan 13tan a.

But,

III_ 36= (:5)

(;0)==> -36 = ~ x 30= 6h

25 5

h = - 36 x 5 = _ 30 cm6 11%1

Negative sign indicates inverted image.30. Differences between telescope and microscope

are given as below:

III

Characteristics Telescope MicroscopeH· • __ H •• H •• ••• _

1. Position of At infinity Near objective at aobject distance lying

between 10 and 2102. Position of

imageBeyond 210 when10 is the focallength ofobjective.

Focalplane ofobjective

1112 x 2 = 1)III For microscope

J;, = 1.25 cm, Ie = 5 ernWhen final image forms at infinity, thenmagnification produced by eye lens is given by

M = -.£. ~ ==> - 30 = - ~ x 25J;, I; 1.25 5

L = 30 x 1.25 ==> L = 7.50 ern5 nl

For objective lensVo = L = 7.5 cmJ;, = 1.25 em, Uo =?

Applying lens formula1 1 1 1 1-=--- ==>-J;, Vo Uo 1.25 7.5 Uo

302 o ehapterwise eSSE Solved Papers PHYSICS

1 1.25 - 7.5._-=---», 7.5 1.25 7.5 x 1.25

=> U = _ 7.5 x 1.25 _ 1.5 cmo 6.25

6.257.5 x 1.25

31.

The object must be at a distance of 1.5 cm fromobjective lens.For figure Refer to Page 290. (1)

The objective lens forms real. inverted magnifiedimage A' B' of object AB in such a way that AB' fallsome where between pole and focus of eye lens.So, A' B' acts as an object for eye lens and its virtualmagnified image A" BIf formed by the lens. (1)

The magnifying power of a compound microscopeis defined as the ratio of the visual anglesubtended by final image at eye (J3) and the visualangle subtended by object at naked eye when bothare at the least distance of distinct vision from theeye.

Visual angle with instrument (J3)m=>---,---,-----=-"'----::-----,---:--"-:'--Visual angle when object is placed at least

distance of distinct vision (a)

B'A'

=> m= ~ = tan~ = ~ = (B' A')x!!...= mom,u,a tan « BA BA ue

D

m = mo me' where mo and me are magnificationproduced by objective and eye lens, respectively.

B'A' vNow, mo = __ = _0_

BA - Uo

D Dme = - = 1 + - [By lens formula]v, !"

m = - (:: J (1 + ~ JThis is the required expression.

Also, Uo = + 1.5 cm =>10 = 1.25 em, !" = 5 emve = - D = - 25 cm

For objective lens,

1

10 Vo Uo

1 1 1=>-=---=Vo 1.25 1.5

1 1 1=> -=-+-1.25 Vo 1.51.5-1.25 0.251.5x 1.25 1.5x 1.25

1

7.50

Vo = 7.5 cm (1)

.. Magnifying power,

m=-(::J(l+ ~J=-G:)(I+ 2;)=-5x6

m= - 30. (1)

(3)

(1)

32. (i) (a) For figure Page 291. (1)(b) For advantages Refer to Ans. 9. (1)

(ii) J; = 10 ern, 12= -10 ern, U = - 30 cm fromconvex lens.

For lIens,

=>

U = - 30 ern 1= + 10 ern v =?1 1 1 1 1 I-=--- => -=----1 v U 10 v (- 30)

~ = ~ _ ~ = 3-1 => v = 15ernv 10 30 30 (1)

This image would be a virtual object for II lens.For II lens,

U2 = + 10 cm[·:concave lens is at a distance

of 5 em from convex lens]12=-lOcm,v=?~=~ 11 v U

_1_=~-~=> ~=O =>V=OO-10 v 10 v

=>

33.So, final image forms at infinity. (2)

For figure Refer to page 291. (2)Magnifying power of astronomical telescope innormal adjustment.

Magnifying power = ~a

= Visual angle formed by final image at eye lensVisual angle formed by object at naked eye

A'B'_ ~ _ tan ~ _ B' E _ OB' _ + fam-------------a tan o A'B' B'E - ue

OB'When final image is formed at D

m=-/O(I+ !,,)!" D (1)

For final image at infinity, B' point must lie onfocus of eye lens, i.e. u, = !"... Magnifying power in normal adjustment,

m=-~ (1)!"

(i) Telescope has objective of a large aperture andlarge focal length whereas microscope haveobjective of small aperture and focal length.

(ii) The relative distance between objective and eyelens - may change in telescope whereas theseparation between objective and eye lens incompound microscope remain fixed. (1)

Value Based Questions (From Complete Chapter)o 4 Marks Questions

1. Kanchan while driving her scooty sees awoman behind driving a moped throughhis rear view mirror. She sees that hersaree is almost touching the wheels of thevehicle. She stops her and alert-that itmay cause a severe accident.Read the above passage and answer thefollowing questions.

(i) What values you observe in Kanchan?(ii) Name the mirror used in rear view in

scooty?Draw a ray diagram for thesame.

Ans. (i) The values we observe in Kanchan are(a) concern about others, awareness(b) observation of traffic rules, applicative

mind(c) responsible citizen (3)

(ii) Convex mirror is used in rear view in scooties.The ray diagram for convex mirror is

I_U (1)

2. Kritika's mom is finding difficult to cook inthe kitchen as there was power-cut, andshe told the same to Kritika. Sheimmediately took a plane mirror from hershelf made it stand against a wall suchthat sun rays were focussed into thekitchen due to the reflection of the mirror.There was some lighting and her motherwas able to finish her work.Read the above passage and answer thefollowing questions.

(i) What are the values shown byKritika?

(ii) Name the types of mirror used inperiscope.

(iii) Give the nature of image in case ofplane mirror.

Ans. (i) Kritika's is intelligent and has goodknowledge of Physics. She is helpful and canapply her mind at required time. (1)

(ii) Plane mirror is used in periscope, plane mirrorfocusses all rays at a point (1)

(ill) Nature of image in plane mirror is(a) virtual(b) object distance = image distance(c) same size as that of object (2)

3. Ravi wanted to buy a gift for his sisterand so he entered inside a gift shop. Thegift shop had many glass items. Onlooking closely, he found many of thebeverage glasses used for cold drinks hadbig thick glass walls. He decided not tobuy these glasses because he knew thatthis gives a false impression that there ismore amount of liquid inside the glass.Read the above passage and answer thefollowing questions.

(i) Why are the beverage glasses madewith very thick glass walls?

(ii) What values can you associate withRavi decision?

Ans. (i) As a Physics student, he knows that ligh raysfrom inside the glass bends away from thenormal and appear to diverge. So, it gives falseimpression that there is more amount ofliquid in bottle. (2)

(ii) Affection, patience, knowledge aboutrefraction. (2)

4. One day Chetan's mother developed asevere stomach ache all of a sudden. Sheimmediately rushed to the doctor, whosuggested for an immediate endoscopytest and gave an estimate of expenditurefor the same. Chetan's immediatelycontacted his class teacher and shared

304

the information with her. The classteacher arranged the money and reachedto the hospital. On realising that Chetan'sbelonged to a below average income groupof family, even the doctor offeredconcession for the test fee. The test wasconducted successfully.Read the above passage and answer thefollowing questions.

(i) Which principle of optics is used inendoscopy?

(ii) Briefly explain the values reflected inthe action taken by the teacher.

(iii) In what way do you appreciate theresponse of the doctor on the givensituations? All India 2013

Ans. (i) Endoscopy is based on Total Internal Reflection(TIR) principle. It has tubes which are made upof optical fibres which are used for transmittingand receiving electrical signals which areconverted into light by suitable transducer. (2)

(ii) Humanity and charity. Teacher as a socialreformer to support humanity. (1)

(ill) Doctor gave monetary help to Chetan byunderstanding his poor financial condition. (1)

5. Mr. Viswanathan, a retired professor ofPhysics was walking with his grandson. Itwas last week of December and so it wasdark around 5.30 pm. The streetlightswere on and the yellow light flooded thearea around. The boy asked professor whyyellow lights were used when white lightwere brighter. The professor answeredthat during foggy days the tiny dropletsact as prisms splitting white light into itsconstituent colours and thus reducing theclarity.

o Chopterwise cast Solved Papers PHYSICS

Read the above passage and give theanswer of the following questions.

(i) What phenomena was the professorreferring to? Why does it happen?

(ii) Give one application of prism.(iii) What values of the boy reflect from

the conversation?Ans. (i) He refers to the process of dispersion of light.

Dispersion of light is due to the differentvelocity of light rays in a medium. (1)

(ii) Studying and analysing the spectrum ofdistant light sources. (2)

(iii) Curiosity research mindedness and awareness.(1)

6. A teacher has given three lenses of power0.5 D, 4 D and 10 D to a student. He is notsure as to which lenses would he use forconstructing a good astronomicaltelescope. So, he consults his seniors andthe teacher and then constructs atelescope. Later, he shows this telescopeto the junior classes and explain about thechoice oflenses.Read the above passage and give theanswer of the following questions.

(i) What values has he shown by doingthese?

(ii) Which lenses are used as objectiveand which one as eyepiece?

Ans. (i) The values shown by him are(a) consulting others in case of need(b) curiosity(c) sharing knowledge (2)

(ii) From these three lenses, he will use a lens ofpower 0.5 D for objective and lens of power10 D as eyepiece. (2)

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