cpp-11 class - xi batches - phonon lens - fiitjee jaipurjaipur.fiitjee.com/phononcpp/lens-d.pdf ·...

CPP-11 Class - XI Batches - PHONON

LENS1. A thin concavo-convex lens has two surfaces of radii of curvature R and 2R. The material of the lens has a refractive

index µ. When kept in air, the focal length of the lens :(A) Will depend on the direction from which light is incident on it(B*) Will be the same, irrespective of the direction from which light is incident on it

(C) Will be equal to 1µ R

(D*) Will be equal to µ 1R

Sol. From left to right 1

1

Lf = (µ – 1) 1 1–

– –2R R

= (µ = 1) 1 1–

2R R µ

concave surface

R

convex surface

2R

1

1

Lf = – –1

2µ

R

From right to left 2

1

Lf = (µ – 1)

1 1–2 ( )R R

= (µ – 1) 1 1–

2R R

= – –1

2µ

R

1

1

Lf =2

1

Lf = –

2( –1)

Rµ

Option (B) and (D) are correct.

2. An object is placed 10 cm away from a glass piece (n = 1.5) of length20 cm bound by spherical surfaces of radii of curvature 10 cm. Find theposition of the final image formed after twice refractions.

20 cmair

Bn = 1.5

ROC = 10cm10 cm

Aobject

ROC = 10cm

air

Ans. [50 cm]Sol. Refraction at first surface :

1

1.5V –

1(–10) =

1.5 –1( 10) V1 = – 30 cm from A

Refraction at second surface.

2

1V –

1.5(–50) =

1–1.5(–10) V2 = +50 cm from B

Hence final image will be 50 cm from B.

3. A concave mirror of radius R is kept on a horizontal table (figure). Water (refractiveindex = ) is poured into it upto a height h. What should be distance of a pointobject from surface along principal axis so that its final image is formed on itself.Consider two cases.(i) h 0(ii) in terms of h

Ans. [(i) R

; (ii) ( )R h

]

Sol. Object should appear to be at distance R from mirror. µ(d) + h = R

d =–R hµ

c

d

hif h << R

d =Rµ

4. A person's eye is at a height of 1.5 m. He stands infront of a 0.3 m long plane mirror which is 0.8 m above the ground,The length of the image he sees of himself is :(A) 1.5 m (B) 1.0 m (C) 0.8 m (D*) 0.6 m

Sol. 2 × height of mirror = 2 × 0.3

0.3 m

0.8 m

1.5m= 0.6 m

5. The values of d1 & d2 for final rays to be parallel to the principal axis are :(focal lengths of the lenses are written above the respective lenses)(A*) d1 = 10 cm, d2 = 15 cm(B*) d1 = 20 cm, d2 = 15 cm(C*) d1 = 30 cm, d2 = 15 cm(D) None of these

Sol.

O

f1=10cm f2=20cm f3= 5cm–

L1 L2 L3

d1 d210cm

Point object O in placed on focus of lines L1 as shown in figure so there is no condition for d1.After reflection from L3, ray become parallel so ray must comes from focus of L3.So,

d2 = 15 cmor d2 = 25 cm

6. A thin lens made of a material of refractive index 2 has medium with refractive index µ1 and –µ3 on either side. The lensis biconvex and the two radii of curvature has equal magnitude R. A beam of light travelling parallel to the principal axisis incident on the lens. Where will the image be formed it the beam is incident from (a) the medium 1 and (b) from themedium 3 ?

Ans. [(a) 3

2 1 32R

; (b) 1

2 1 32R

]

Sol.µ2

µ1 µ3

S2

S1

For surface S1

2

1

µV – 1µ

= 2 1–µ µ

R...(1)

For surface S2

3µV –

2

1

µV = 3 2–

–µ µ

R...(2)

From equation (1) and (2)

3µV = 2 1 32 – –µ µ µ

R

V =3

2 1 32 – –µ R

µ µ µSimilarls

2

1

µV – 3µ

= 2 3–µ µ

R...(1)

µ3µ2

µ1

1µV –

2

1

µV = 1 2–

–µ µ

R...(2)

V =1

2 1 32 – –µ R

µ µ µ

7. An equiconvex lens of refractive index n2 is placed such that the refractive :(A*) Must be diverging if n2 is less than the arithmetic mean of n1 and n3(B*) Must be converging if n2 is greater than the arithmetic mean of n1 and n3(C) May be diverging if n2 is less than the arithmetic mean of n1 and n3(D*) Will neither be diverging nor converging if n2 is equal to arithmetic mean of n1 and n3

Sol. for surface S1

2

1

nV = 1n

u = 2 1–( )

n nfR ...(1)

for surface S2

3nV –

2

1

nV =

3 2–(– )

n nR ...(2) n2

n1 n3

S2

S1

From (1) and (2)

3 1–n nV u = 2 1–n n

R – 3 2–n n

R =

= 2 1 32 – –n n nR

= 1 32

2 –2

n nnR

8. In the figure given below, there are two convex lens L1 and L2 having focal length off1 and f2 respectively. The distance between L1 and L2 will be :(A) f1

L1 L1

(B) f2(C*) f1 + f2(D) f1 – f2

Sol.

| |f2| |f1

| |+f1 | |f2

9. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQ andRS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm.An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens.If A B is the image after refraction from the lens and reflection from the mirror, find the distance A B (in cm) fromthe pole of the mirror and obtain its magnification. Also locate positions of A' and B' with respect to the optic axis RS.

0.6cm

PR

A

B Q

S

20 cm30 cm

Ans. [A' B' at 15 cm to the right of mirror. B' is 0.3 cm above RS & A' is 1.5 cm below RS. Magnification is – 1.5]Sol. Reflection from lens

1 1–V u =

1f

1 120V

=1

15V = 60 cm

m = Vu =

60–20 =

0

Ihh

hI = –3 h0 = –3 × 1.2 = – 3.6 cmFor mirror

A1

C1

B1

A'

C'

B'

1.5cm

0.3cmQ

5

30 cm 15 cm

0.6cm

3cm

(Image after reflectionfrom lens)

(Final image after reflection from lens and mirror)

1 1V u =

1f

1 130V

= – 130 V = – 15 cm

m = – Vu = –

(–15)30 =

12

10. A convexo-concave diverging lens is made of glass of refractive index 1.5 and focal length 24 cm. Radius of curvaturefor one surface is double that of the other. Then radii of curvature for the two surfaces are (in cm) :(A*) 6, 12 (B) 12, 24 (C) 3, 6 (D) 18, 36

Sol.1f =

2 1

1

–µ µµ

1 2

1 1–R R

–124

=1.5 –1

1

1 1–2R R

= 12

3–2R

124

=1

2RR = 6 cm, 2R = 12 cm

1. When a lens of power P (in air) made of material of refractive index µ is immersed in liquid of refractive index µ0. Thenthe power of lens is :

(A) 0

1

P (B) 01

P (C*) 0

1

. 0

P

(D) None of these

Sol. P =1f =

–11µ

1 2

1 1–R R

P' =0

0

–µ µµ

1 2

1 1–R R

'PP

=0

0

–µ µµ

×1

( –1)µ = 0––1

µ µµ

0

1µ

P' = 0––1

µ µµ

0

Pµ

2. What will the paths of the ray be after refraction in the lenses.[F1 – First focus, F2 – Second focus]

(a) (b)

Ans. [(a) ; (b) ]

Sol. (a)

f1 f2

(b)

f1 f2

3. A thin symmetrical double convex lens of power P is cut into three parts, as shown in the figure. Power of Ais :

(A) 2 P (B) 2P

(C) 3P (D*) P


LENS

Sol. If we cut lens along principal axis, power of lens remain unchanged.

4. Lenses are constructed by a material of refractive index 1.50. The magnitude of the radii of curvature are 20 cm and30 cm. Find the focal lengths of the possible lenses with the above specifications.Ans. [± 24 cm, ± 120 cm]

Sol.1f = (µ – 1)

1 2

1 1–R R

Case I - convex lens

1f = ± (1.5 – 1)

1 120 30

= ± 12

112

f = ± 24 cmCase II

1f = ± (1.5 – 1)

1 1–20 30

f = ± 120 cm

5. A quarter cylinder of radius R and refractive index 1.5 is placed on atable. A point object P is kept at a distance of mR from it. Find the valueof m for which a ray from P will emerge parallel to the table as shown inthe figure. mR R

P

Ans. [m = 4/3]Sol. for plane surface

1

1.5V –

1mR =

1.5 –1 A B

mR RV1 = 1.5 mRfor curved surface

1

–1.5

(1.5 )mR R =1–1.5

–R

1.51.5mR R

= 0.5R

3 = 1.5 m + 1 m = 4/3

6. A meniscus lens is made of a material of refractive index µ2. Both its surfaces have radiiof curvature R. It has two different media of refractive indices µ1 and µ3 respectively, onits two sides (shown in the figure). Calculate its focal length for µ1 < µ2 < µ3, when lightis incident on it as shown.

µ1

µ2

µ3

Ans. [ f = 3

3 1( )R

]

Sol. For Ist surface µ1

µ2

µ3

2µu – 1µ

= 2 1–µ µ

R

u =2

2 1–µ Rµ µ

[ u = v2]

For IInd surface

3 2

2

–µ rv v = 3 2–µ r

R

3µv –

2

2

µµ R (µ2–µ1) = 3 2–µ r

R

3µv = 3µ

R – 1µ

R

v =3

3 1–µ Rµ µ

7. A thin concave-concave lens is surrounded by two different liquids A and B as shown in figure. The system issupported by a plane mirror at the bottom. Refractive index of A, lens and B are 9/5, 3/2 and 4/3 respectively. The radiusof curvature of the surfaces of the lens are same and equal to 10 cm. Where should an object be placed infront of thissystem so that final image is formed on the object itself.

Ans. [75 cm]Sol. For image to form on object itself rays should fall perpendicularly on plane mirror/

Focal length of combination will be :P = P1 + P2 + P3

1

qf=

1

1f +

2

1f +

3

1f

f1 = 12.5 cmf2 = –10 cmf3 = 30 cm

Thus we getfeq = 75 cm

Hence object should be placed at 75 cm.So, that light rays becomes parallel to principal axis.

8. The radius of curvature of the left & right surface of the concave lens are 10 cm & 15 cm respectively. The radius ofcurvature of the mirror is 15 cm :(A*) Equivalent focal length of the combination is – 18 cm(B) Equivalent focal length of the combination is + 36 cm(C*) The system behaves like a concave mirror(D) The system behaves like a convex mirror

Sol. HerePeq = 2PL1 + 2PL2 + PM air water

µ= 4

__3

Glass

µ= 4

__2

=1

1

Lf

+ 2 2

1

Lf

– 1

Mf

Peq = – 2 1

12

+ 445 –

2–15

[ Mirror is converging so in power is +ve.]

Peq =1

18

–1f =

118

f = – 18 cmHere system acts as concave mirror.

9. A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The spacebetween the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object isplaced 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with anotherliquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of theliquid.

Ans. [n = 8/5 = 1.6]Sol. For image to form on object itself, ray should strike the mirror perpendicularly.

HerePeq = P1 + P2

=1

1f +

2

1f

1

1f =

3 –12

DR

1

1f = R

2

1f =

4 –13

1 1– –R

2

1f = –

13

1R

= –13R

1

eqf =1R

– 1

3R = 3 –13R =

23R

feq =32

R = 15 cm [ object is at focus]

R = 10 cmSame experiment is repeated using some other liquid.

feq =1

1f +

2

1f

2

1f = (µ – 1)

1–R

=1R

–( –1)µ

R

1

eqf =1– 1µ

R

= 2 – µR R =

125

210 – 10

µ = 0.04

10µ

= 0.2 – 0.04

µ = (0.2 – 0.04) 10= 2 – .4 = 1.6

10. An object O is kept infront of a converging lens of focal length 30 cm behindwhich there is a plane mirror at 15 cm from the lens :(A) The final image is formed at 60 cm from the lens towards right of it(B*) The final image is at 60 cm from lens towards left of it(C*) The final image is real(D) The final image is virtual

Sol. Image formed after refraction form lens.30 cm

15cm 15cm

O

1v –

1u =

1f

1v –

1(–15) =

130

1u =

130 –

230

v =–130

v = – 30 cmfor this virtual image, image formed by plane mirror will be at 45 cm on light of mirror this image will be real.And for this image thus is at distance 2f i.e. at 60 cm. Hence its real image again will be formed on 2f of lens in its leftside.

1. A thin linear object of size 1 mm is kept along the principal axis of a convex lens of focal length 10 cm. The object is at15 cm from the lens. The length of the image is :(A) 1 mm (B*) 4 mm (C) 2 mm (D) 8 mm

Sol.1v –

1u =

1f

1v –

1–15 =

110

2

– 1dvdu v

+ 2

1v = 0

1v =

110 –

115

dvdu =

2

2

vu

=3

30 –2

30

dv =2

2

vu

du1v =

130 v = 30

dv =2

2

(30)(15) × 1

= 4 × 1 = 4 mn.2. A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal

axes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0 mm, is incidenton the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of theemergent beam. Also find out the ratio of emergent and incident intensities.Ans. [1.0 cm if the light is incident from the side of concave lens and 2.5 mm if it is incident from the side of the convexlens and the corresponding ratio of intensities are 1/4 and 4]

Sol. Case I:- 1st convex t hen concave Image from convex long is at focus of concave so emergent light ray becomesparallel to principal axis.

A

BCD

tF

0.5cm

10cm 10cm

f = 20 f = 10

From geometry :ABBF

=CDDF

[CD — width of emergent beam]

0.520 = 10

CD

CD =10 0.5

20

CD = 2.5 mm

Intensity area beam croos section.


LENS

I emergentI incident =

2

2

(2.5)(5) =

14

= 1 : 4 ans

Case II :- 1st concave, then convex

A

B D

C

10cm 10cm

0.5cmF

Since image of concave lens is at focus of convex lens so final emergent ray becomes parallel to principal axis.

From geometry Þ CDDF

= ABBF

Þ 20CD

cm = 0.510

cmcm

CD = 1 cm

I emergentI incident =

210.5

cmcm

= 4 Ans.

3. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5has a radius of curvature 20 cm. The concave surface has a radius of curvature 60cm. The convex side is silvered and placed on a horizontal surface as shown infigure. (a) Where should a pin be placed on the axis so that its image is formed atthe same place ? (b) If the concave part is filled with water ( = 4/3), find thedistance through which the pin should be moved so that the image of the pinagain coincides with the pin.Ans. [(a) 15 cm from the lens on the axis; (b) 1.14 cm towards the lens]

Sol. (a) Net focal length.

–1

eqf = feq =1

Lf +

1

Lf –

1

Mf

R=60cm

µ=3/2R=20cm

=2

Lf –

1

Mf = 2

1 1 1 2–2 60 20 20

–1

eqf =4

30 cm

feq = 7.5 cmfeq = 7.5 × 2 = 15 cm

Hence object should be placed at centre of curvature of equivalent mirror i.e. at 15 cm from mirror.

(b) –1

eqf= Peq =

2

wf+

2

Lf–

2

Mf

= 213

1160

+ 2

12

–1 160 20

+ 2

20

=23

160

+ 1 1–20 60

+ 2

20

=190 +

60 – 2060 20

+ 2

20

=190 +

4060 20

+ 220

=190 +

4120 +

110

=190 +

130 +

110 =

190 +

390 +

990 =

1390

feq =1390

Peq =18013 = 13.85 cm

Shift of object 15 – 13.85 = 1.15 cm

4. An insect at point 'P' sees its two images in the water-mirror system as shownin the figure. One image is formed due to direct reflection from water surfaceand the other image is formed due to refraction, reflection & again refractionby water mirror system in order. Find the separation between the two images.M has focal length 60 cm. (nw = 4/3).Ans. [Distance P' P" = 36 – 12 = 24 cm]

Sol. Image (1) By direct reflection from top water surface. 12 cm below water furpace.Image (2) 1st refraction from top water surface.

1

4 / 3v –

1(–12) =

(4 / 3 –1)

v1 – 16cm

reflection from convex mirror

2

1 1(–40)v

=1

( 60) v2 = +24 cm

2nd refraction from top trasfer surface.

3

1 4 / 3–(–48)v =

1– 4 / 3

v3 = 36 cm

Image (2) is 36 cm below top water surface Separation of two image = 36 – 12 = 24 cm

5. A symmetrical converging convex lens of focal length 10 cm & diverging concavesymmetrical lens of focal length – 20 cm are cut from the middle & perpendicularlyand symmetrically to their principal axis. The parts thus obtained are arranged asshown in the figure. The focal length of this arrangement will be :(A) (B) 20 cm(C) 40 cm (D*) 80 cm

Sol. For L1 P1 = (µ – 1) 1

1 1–R

L1

L2L3

R1R2 R2

2c

= +1

( –1)µR

For L2 P2 = (µ – 1) 2

1 1–( )R

= – 2

( –1)µR

For L3 P3 = (µ – 1) 2

1 1–(– )R

= – 2

( –1)µR

For converging 10 =1

2( –1)Rµ R1 = 20 (µ – 1)

For diverging 20 =2

2( –1)Rµ R2 = 40 (µ – 1)

P1 = +1

( –1)µR = +

120

P2 = –2

( –1)µR = –

140 = P3

P12 = P1 + P2 = +120 –

140 = +

140

Combined P123 = P12 + P3 – dP12 P3

=140

+ 1–40

–20140

1–40

= +140 –

140 +

180

P123 = + 1

80 F123 = +80 cm

6. A hollow sphere of glass of R.I. n has a small mark M on its interiorsurface which is observed by an observer O from a point outside thesphere. C is centre of the sphere. The inner cavity (air) is concentricwith the external surface and thickness of the glass is everywhere equalto the radius of the inner surface. Find the distance by which the markwill appear nearer than it really is, in terms of n and R assuming paraxialrays.

glass

airCM

O

2R

4RAns. [(n – 1)R/(3n – 1)]Sol. Refraction from surface CD:

1

nv –

1(–2 )R =

( –1)(– )n

R

v1 =2

(1– 2 )nR

n from surface (1) CMO

(1)(2)

Refraction from surface (2)

2

1v – (4 –1)–

(2 –1)

nR n

n

=1–(–2 )

nR

2

1v +

(2 –1)(4 –1)

n nR n =

( –1)2

nR

v2 = – 2 (4 –1)

(3 –1)R n

n

Shift = 3R – 2 (4 –1)

(3 –1)R n

n

= (3 –1)R

n [3(3n – 1) – 2(4n – 1)]

Shift =( –1)

(3 –1)R n

n

7. Two media each of refractive index 1.5 with plane parallel boundaries areseparated by 100 cm. A convex lens of focal length 60 cm is placed midwaybetween them with its principal axis normal to the boundaries. A luminouspoint object O is placed in one medium on the axis of the lens at a distance125 cm from it. Find the position of its image formed as a result of refractionthrough the system.Ans. [200 cm, w.r.t. lens]

Sol. Distance of object as seen by lens

= 50 + 751.5

4=1

50

75= 100 cm

Now1 1–v u =

1f

1 1–

(–100)v =1

60 1 1–

100v = 160

v =6000

40 = 150

= 50 +100Apparent distnace

50 + 100µ = 200 cm

8. A point object is placed at distance of 20 cm from a thin planoconvex lens of focal length15 cm. The plane surface of the lens is now silvered. The image created by the system is at :(A) 60 cm to the left of the system

20(B) 60 cm to the right of the system(C*) 12 cm to the left of the system (D) 12 cm to the right of the system

Sol. Image after 1st refraction from lens

1v –

1(–20) =

115

1v =

520 15

v = 60 cm

After reflation from mirrorv' = – 60 cm.

again after refraction from lens

1v –

1( 60) =

115

1v =

115 +

160

v =45

60 15

v = 12 cm, to the left.

9. An object O is kept in air and a lens of focal length 10 cm (in air) is kept at the bottom ofa container which is filled upto a height 44 cm by water. The refractive index of water is4/3 and that of glass is 3/2. The bottom of the container is closed by a thin glass slab ofrefractive index 3/2. Find the position of the final image formed by the system.Ans. [90 cm]

Sol. Focal length of lens is water =

3 1 410 –2 8 33 9–2 3

=

1 4102 326

= (30) × 43 = 40 cm

Object will appear to lens at 44 + 12 × 43

= 60 cmLens image will be at

1 1–(–60)v =

140 v =

40 6020

= 120 cm

After refraction from water, image will be at

120 × 34

= 90 cm

10. A stationary observer O looking at a fish F (in water of, µ = 4/3) through a converginglens of focal length 90 cm, The lens is allowed to fall freely from a height 62.0 cm withits axis vertical. The fish and the observer are on the principal axis of the lens. Thefish moves up with constant velocity 100 cm/s. Initially it was at a depth of 44.0 cm.Find the velocity with which the fish appears to move to the observer att = 0.2 sec. (g = 10 m/s2)

Ans. [ 914 m/s = 2275 cm/s (upwards)]

Sol.

20cm

200cm/sec42cm

24cmF '

F

u=75cm/sec

After .2 sec [ lens fall by 20 cm in 0.2 sec]

Image of fish F1 after 0.2 sec will be at 18 cm from water air surface so distance of fish from lens will be(42 + 18) = 60 cm

Image of this fish due to lens is

1v –

1u =

1f

v = – 180 cmWe know that

dvdt =

2

2

vu

dudt [Here u is –ve and u decrease with time as fish is coming near lens]

dvdt =

2

2

vu

(200 + 75)

dvdt =

218060

(275)

dvdt = 2475 cm/sec

So this is the speed of image wrto lensvIL= vI – vL = 2475vI = 2475 + vL [ vL = – 200]

vI = 2275 cm/sec.

cpp-11 class - xi batches - phonon lens - fiitjee jaipurjaipur.fiitjee.com/phononcpp/lens-d.pdf ·...

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