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® CBSE XII EXAMINATION-2018

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CLASS-XII / (CBSE)

PAGE # 1

Roll No.

bookanswertheofpagetitle

theoncodewritemustCandiates

Please check that this question paper contains 19 printed pages.

Code number given on the right hand side of the question paper should be written on the title page of theanswer- book by the candidate.

Please check that this question paper contains 26 questions.

Please write down the Serial Number of the question before attempting it.

15 minute time has been allotted to read this question paper. The question paper will be distributed at10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will notwrite any answer on the answer-book during this period.

PHYSICS (Theory) & SOLUTIONTime allowed : 3 hours Maximum Marks : 70

General Instructions :

(i) All questions are compulsory. There are 26 questions in all.

(ii) This question paper has five sections : Section A, Section B, Section C, Section D and Section E.

(iii) Section A contains five questions of one mark each, Section B contains five questions of two markseach, Section C contains twelve question of three marks each, Section D contains one value basedquestion of four marks and Section E contains three questions of five marks each.

(iv) There is no overall choice. However, an internal choice has been provided in one question of two marks,one question of three marks and all the three questions of five marks weightage. You have to attemptonly one of the choices in such questions.

(v) You may use the following values of physical constants wherever necessary :

c = 3×108 m/s

h = 6.63×10�34 Js

e = 1.6×10�19 C

0 =4 × 10�7 T m A�1

Series SGN SET-1

P.T.O

http://www.pspd.ersonance.ac.in

mailto:[email protected]





CLASS-XII / (CBSE)

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0 = 8.854 ×10�12 C2 N�1 m�2

041 = 9×109 Nm2 C�2

Mass of electron = 9.1 × 10�31 kg

Mass of neutron = 1.675 × 10�27 kg

Mass of proton = 1.673 × 10�27 kg

Avogadro�s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38×10�23 JK�1

SECTION A

Q 1. A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, actingperpendicular to their paths. Which of them will move in a circular path with higher frequency ? [1]

Ans. ProtonReason :Frequency of charge revolving in circular orbit is given by :

2 mf

qB

For proton

pp

2 mf

eB

For electron

ee

2 mf

eB

but mp > m

p

So, fP > f

e

Hence, proton will move in a circular path with higher frequency

Q 2. Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery. [1]Ans. (a) Water purification Ultraviolet rays

(b) eye-surgery Ultraviolet rays

Q 3. Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations ofequal frequancy and different intensities. Mark the graph for the radiation of higher intensity. [1]

Ans.

Here I1 and I2 are intensities of incident wave, V0 = stopping potentialP.T.O

http://www.pspd.resonance.ac.in






CLASS-XII / (CBSE)

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P.T.O

Q 4. Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of thetwo � the parent or the daughter nucleus � would have higher binding energy per nucleon ? [1]

Ans. Daughter nucleus have higher binding energy per nucleon because they are more stable and higherstability means higher binding energy per nucleon.

Q 5. Which mode of propagation is used by short wave broadcast services ? [1]Ans. Sky wave propagation.

SECTION B

Q 6. Two electric bubls P and Q have their resistances in the ratio of 1 : 2. They are connected in series acrossa battery. Find the ratio of the power dissipation in these bulbs. [2]

Ans. Power dissipation is given by (for series )P = 2RNow, For bulb PP

1 = 2 × R1

P1 = 2 R1

For bulb QP

2 = 2 × R2

P2 = 22 R1

2

1

P 2P 1

2 : 1 or 21

2

1 PP

Q 7. A 10V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V andinternal resistance 38 as shown in the figure. Find the value of current in the circuit. [2]

Ans.net emf

total resis tance

3810200

1905A

38







CLASS-XII / (CBSE)

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P.T.O

ORIn a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in opencircuit is 350 cm. When a resistance of 9 is used in the external circuit of the cell, the balance pointshifts to 300 cm. Determine the internal resistance of the cell. [2]

Ans. = 350 cmR = 9

2 300cm

Vr R

V

1 2

2

r R

350 3009

300

r = 1.5 Ans.

Q 8. (a) Why are infra-red waves often called heat waves ? Explain.(b) What do you understand by the statement, �electromagnetic waves transport momentum� ? [2]

Ans . (a) Infrared waves are prdocued by vibration of atoms of molecules which increases the temperaturedue which they are also the cause of heat energy

(b) Electromagnetic waves also show particle nature, i.e., they aslo behave as a matter according toPlank�s Quantum theory, So, they also possess some momentum and they can also change

momentum.

Q 9. If light of wavelength 4125 nm is incident on each of the metals given below, which ones will showphotoelectric emission and why ? [2]

Metal Work Function (eV)

Na 192

K 215

Ca 320

Mo 417

Ans. = 4.125 nm = 412.5 × 10�9 mEnergy possessed by the incident light :

34 8

9

hc 6.634 10 3 10E

412.5 10

= 4.82 × 10�19J

3 eVSo, Na, and K will show photoelectric emission because their work function is less than the energy ofincident light .

Q 10. A carrier wave of peak voltage 15V is used to transmit a message signal. Find the peak voltage of themodulating signal in order to have a modulation index of 60%. [2]

Ans. Given peak voltage V0 = 15V

Modulation index = 60%

60100







CLASS-XII / (CBSE)

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P.T.O

Modulation idedx

m

0

(v )Peak voltage of modulating signalPeak Voltage (v )

Vm = × V

0

Vm =

6015 9V

100

V09V

SECTION-CQ 11. Four point charges Q, q, Q and q are placed at the corners of a square of side �a� as shown in the figure.

[3]

Find the(a) resultant electric force on a charge Q, and(b) potential energy of this system.

Ans. (a) Force on Q due to Q :

(a) F1 =

2

2

KQ

2a

2

2

KQ2a

(direction is shown in figure)

Force on Q due to single �q� :

2

KQqF

a (direction is same as F1)

Resultant Force on Q due to both �q� :

2 2

2KQqF 2F

a

Resulatant electric force on Q :

F = F1 + F2 = 22

2

a

KQq2

2a

KQ

2

KQ QF 2q

2a

2

KQ Q 2 2qF

2a

Directed outwards along the line joing charges Q and Q. (Show in figure)

(b) Potential Energy of system :

U = aKQq

a2

Kqa

KQq

a2

KQa

KQqa

KQq 22







CLASS-XII / (CBSE)

PAGE # 6

P.T.O

= 2 24KQq KQ Kq

a 2a 2a

P.E.=

2

q

2

Q4Qq

aK 22

OR(a) Three point charges q, � 4q and 2q are placed at the vertices of an equilateral triangle ABC of side

�l� as shown in the figure. Obtain the expression for the magnitude of the resultant electric force

acting on the charge q.

(b) Find out the amount of the work done to separate the charges at infinite distance. [3]

Ans . (a) Force on q due to �4q

F1 =

2

K(4q) q

along A � B

2

2

4Kq

along A � B

Force on q due to 2q

2 2

K 2q qF

along C � AA

2

2 2

2KqF

along C � A

Angle between F1 of F2 ( = 120°):

Resultat Force :

2 21 2 1 2F F F 2FF cos120

2 2

2 2 2 2

1F 2F F 2 2F F

2

1 2F 2F

= 2 22 2 25F 2F 3 F

F = 2

2

2Kq3

Direction tan = 2

1 2

F sin

F F cos







CLASS-XII / (CBSE)

PAGE # 7

= 2

2 2

3F

21

2F F2

So, resultant forces is making angle

1 1tan

3

with the line AB

(b) Workdone to gather the charges from infinity = potential energy of system (U)

Kq 2q Kq 4q K 2q 4qU

2 2Kq kq

2 4 8 10

Workdone to separate the charges to infinity = l

210kq

Q 12. (a) Define the term �conductivity� of a metallic wire, Write its S unit.(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a

wire in terms of number density and relaxation time. Hence obtain the relation between currentdensity and the applied electric field E. [3]

Ans. (a) Conductivity of a metallic wire is defined as the conductance (reciprocal of resistance) of a metallicwire of unit side and unit cross sectional area perpendicular to the current flow

ORConductivity is defined as the reciprocal of resistvity of a metallic wire.

S.I. unit �1 m�1 or mho (m�1)

(b)A

Rl

l

RA

butA

R2ne

ml

So, l

l A

A

2ne

m

2ne

m or

mne

2

..............(1)

where

1

= conductivity

= relaxation timen = number density of electrone = charge on one electronm = mass of one elecrtronNow, current density is given by

J = neVd







CLASS-XII / (CBSE)

PAGE # 8

J = ne

m

eE

m

eEVd

Em

neJ

2

From equation (1)

J = E or EJ

Q 13. A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 044 T.Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magneticfield, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii). [3]

Ans. m = 6J/T

1 = 60°

B = 0.44T(a) W = � mB [cos

2 � cos

1]

(i) 2 = 90°

W = � 6 × 0.44 [cos90° � cos60°]

= � 2.64 1

02

W = 1.32 J Ans.

(ii) 2 = 180°

W = � 6 × 0.44 [cos180° � cos60°]

= � 2.64 1

12

W = 3

2.642

W = 3.965 J Ans.

(b) When = 180°

Torque will be = mBsin= 6 × 0.44 × sin180°

= 0 Nm Ans.

Q 14. (a) An iron ring of relative permeability r has windings of insulated copper wire of n turns per metre.

When the current in the windings is , find the expression for the magnetic field in the ring.(b) The susceptibility of a magnetic material is 09853. Identify the type of magnetic material. Draw the

modification of the field pattern on keeping a piece of this material in a uniform magnetic field. [3]Ans. (a) Relative permeability =

r

Number of turns/ metre = nCurrent = IBy Ampere�s Circuital law

0B.d × total current

0B d N Here N = Total number of atoms

B = 0

N2 r







CLASS-XII / (CBSE)

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B = 0 n

Nn

2 r

For ring = 0

r

B = 0

r n

(b) m = 0.9853

The given magnetic material is paramagnetic in nature.

Q 15. (a) Show using a proper diagram how unpolarised light can be linearly polarised by reflection from atransparent glass surface.

(b) The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having

refractive index 23

, placed in water of refractive index 34

. Will this ray suffer total internal reflection

on striking the face AC ? Justify your answer.A

B CAns. (a)







CLASS-XII / (CBSE)

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When unpolarised light ray is incident at an angle such that the angle between reflected ofrefracted rays is 90°, then reflected ray is linearly polarised. In that case incident angle is called

polarising angle or Brewster angle(iP or iB).

(b) For Total internal reflection

DR

Ci

sin1

sin ic =

DR

Ci

sin1

sin ic =

wg

sin ic =

4 33 2

4 23 3

sin ic =

80.88

9 ................ (i)

Now, in this case

sin i = sin 60° = 3

0.8672

............... (ii)

sin i < sin ic

So, i < ic

So, ray will not suffer Total internal reflection

Q 16. (a) If one of two identical slits producing interference in Young�s experiment is covered with glass, so

that the light intensity passing through it is reduced to 50%, find the ratio of the maximum andminimum intensity of the fringe in the interference pattern.

(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light?

Ans. (a) 1 =

2

2 =

2

1 2max

min 1 2

2

2

2

II

II

2

1 2

1 2

= 2 1 2 2 3 2 2

2 1 2 2 3 2 2

(b) If we use white light, central fringe will be white, then there will be colourful fringes on both sides,starting from violet to red (fringe width wavelength) and then there will be general illumination ofscreen.







CLASS-XII / (CBSE)

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Q 17. A symmetric biconvex lens of radius of curvature R and made of glass of refrative index 1.5, is placed ona layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on theprincipal axis of the lens is moved along the axis until its real, inverted image coincides with the needleitself. The distance of the needle from the lens is measured to be x. On removing the liquid layer andrepeating the experiment, the distance is found to be y. Obtain the expression for the refractive index ofthe liquid in terms of x and y.

Ans. Let f1 = focal length of biconvex lens

f2 = focal length of plano concave liquid

f = focal length of combination of above two lenses.f = x

f1 = y .........(i)

From combination of thin lenses

fff111

21

12

111fff

= yx11

xyxy

f

2

1........(ii)

For bicovex lens

RRf11

)1(1

1

Rf2

)151(1

1

f1 = R = y .......(iii)

For planoconcave liquid

11)1(

1

2 Rf l

From equation (ii) & (iii)







CLASS-XII / (CBSE)

PAGE # 12

yxyxy

l1

)1(

lxxy

1

xxy

l

1

xxyx

l

1

xyx

l

2

Q 18. (a) State Bohr�s postulate to define stable orbits in hydrogen atom. How does de-Broglie�s hypothesis

explain the stability of these orbits ?(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level.

Estimate the frequency of the photon.Ans. (a) Quantum condition �

The electrons are permitted to circulate only in those orbits in which the angular momentum of an

electron is an integral multiple of hh

;2

being Planck�s constant. Therefore, for any permitted orbit.

,2nh

mvrL n = 1, 2, 3.........

where L, m and v are the angular momentum, mass and speed of the electron, r is the radius of thepermitted orbit and n is positive integer called principal quantum number, The above equation isBohr�s famous quantum condition.

Stationary orbits �While revolving in the permissible orbits, an electron does not radiate energy. These non-radiatingorbits are called stationary orbits.

According to de-Broglie, the circumference of nth circular orbit in which electron is revolving is given

by : 2rn = n ...........(i)

where = wavelength of wave emitted by electronalso, by de-Broglie�s hypothesis

mvh

............(ii)

From (i) and (ii)

2rn = mvnh

mvrn =

2nh

(b) 2 21 2

1 1 1R

n n

Here

1n 1

2n 4







CLASS-XII / (CBSE)

PAGE # 13

1 1R 1

4

1 3R

4

43R

Now c = f

f = Rcc

34

Where c = speed of light in vacuum = 3 × 108 m/s.

R = Rhydberg constant = 1.09 × 107 m�1.

f = 8

7

4 3 103 1.09 10

f = 36.69 Hz

Q 19. (a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy pernucleon (BE/A) versus the mass number A.

(b) A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to3.125% ?

Ans. (a)

Nuclear fission � Binding energy per nucleon is smaller for heavier nuclei (greater than 200) i.e.,heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, they becomestable, this happens in nuclear fission.Nuclear fusion � The binding energy per nucleon is small for light nuclei (between 2 to 20), i.e.,they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher bindingenergy per nucleon of the latter results in the release of energy. This is what happens in a nuclearfusion.

(b)

n

0 0

R N 1R N 2

5

0

R 3.125 1 13.125%

R 100 32 2

= n 5

1 1n 5

2 2







CLASS-XII / (CBSE)

PAGE # 14

t = nT1/2

= 5 × 10 = 50 year

Q 20. (a) A student wants to use two p-n junction diodes to convert alternating current into direct current.Draw the labelled circuit diagram she would use and explain how it works.

(b) Give the truth table and circuit symbol for NAND gate.

Ans. (a)

Working :� For +ve half cycle of AC source, let A be at +ve potential and B and �ve potential, So, D

1 will be

forward biased and D2 will be reversed biased. It means current will only flow through D

1 and current

will not flow through D2. Output is obtained across R

L.

� For �ve half cylce, B will be at +ve potential and point A at �ve potential, So, D1 will be reversed

biased and D2 will be forward biased It means that current will only flow through D

2 and not through

D1, output is obtained across R

L.

This process is called full wave rectification and the device is called full wave rectifier.

(b) Truth Table

A B A.B Y A.B

0 1 0 1

1 0 0 1

0 0 0 1

1 1 1 0

Logic Symbol :

Q 21. Draw the typical input and output characteristics of an n-p-n transistor is CE configuration. Show howthese characteristics can be used to determine (a) the input resistance (ri), and (b) current amplificationfactor ()

Ans 21.

V =6VCE

V =8VCE

V =10VCE

Input CharacteristicsVBE







CLASS-XII / (CBSE)

PAGE # 15

IC

Output CharacteristicsVCE

The input resistance ri of the transistor in CE configuration is defined as the ratio of the small

change in base-emitter voltage to the corresponding small change in the base current, when thecollector-emitter voltage is kept fixed. Thus

constantVCE

B

BEi

Vr

I

As input characteristic is non-linear, so ri varies. At any point of the curve, ri is equal to the slope of

the tangent to the curve.Current amplification factor () : It is defined as the ratio of the change in collector current to the

small change in base current at constant collector-emitter voltage (VCE) when the transistor is in the

active state.

constantVCE

B

Cac I

I

This is also known as small signal current gain and its value is very large.

Q 22. (a) Give three reasons why modulation of a message signal is necessary for long distancetransmission.

(b) Show graphically an audio signal, a carrier wave and an amplitude modulated wave.

Ans. (a) (1) Practical antenna lengthTo transmit a signal effectively, the height of the antenna should be comparable to the wavelengthof the signal (at least /4 in length), wavelength of signal are very large & such large antenna arepractically impossible to construct, so modulation becomes necessary

(2) Effective power radiated by an antennaFor linear antenna of length l, it is seen that

Power radiated

21

Thus for the same antenna length, the power radiated by short wavelength or high frequencysignals would be large. To transmit them over large distances, they are superimposed on highfrequency carrier waves.

(3) Frequency multiplexingIn order to transmit different audio signals (which fall in the same spectral range) simultaneouslythrough one antenna, so modulation becomes necessary







CLASS-XII / (CBSE)

PAGE # 16

(b)

SECTION D

Q 23. The teachers of Geeta�s school took the students on a study trip to a power generating station, located

nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over sucha long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receivingend in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced.Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lowervoltage.(a) Name the device used to change the alternating voltage to a higher or lower value. State one cause

for power dissipation in this device.(b) Explain with an example, how power loss is reduced if the energy is transmitted over long distances

as an alternating current rather than a direct current.(c) Write two values each shown by the teachers and Geeta.

Ans. (a) Trans formerCopper Loss : Energy is lost in the form of heat energy due to flow of current in the copper windingsof transformer because of its resistance. This loss is called copper loss.

(b) For long distances, alternating current of high voltage is used because current reduces siginificatlyand power loss due to heating (2R) also reduces significantly.Example : Electric power from power station is transferred to substation in 44 kV a.c., and then itis reduced to 22 kV a.c.

(c) Values shown by teachers good demonstrator, caringValues shown by Geeta Good listener, curious.

SECTION EQ 24. (a) Define electric flux. Is it a scalar or a vector quantity ?

A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown in







CLASS-XII / (CBSE)

PAGE # 17

the figure. Use Gauss� law to obtain the expression for the electric flux through the square.

q

d/2

d

d

(b) If the point charge is now moved to a distance �d� from the centre of the square and the side of the

square is doubled, explain how the electric flux will be affected.OR

(a) Use Gauss� law to derive the expression for the electric field )E( due to a straight uniformly chargedinfinite line of charge density C/m.

(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.(c) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2 > r1).

Ans. (a) Electric Flux : It is the total number of electric field lines passing normally through a surface held inelectric field.

orThe electric flux linked with a surface is the surface integral of the electric field over the surface.It is a scalar quantity.

We will draw a closed gaussian surface in the form of cube of side �d�By Gauss�s Law

0

q

For any surface, flux will be

0

q6

(b) Electric flux will not be affected because electric flux does not depend on dimensions or area ofsurface.







CLASS-XII / (CBSE)

PAGE # 18

0

q6

OR(a)

Consider a line charge of density coulomb per metre. Choose a Gaussian as a right circular

cylinder of unit length and of radius r with its length as axis. On the top and bottom,

E and

sare perpendicular to each other, so there is no electric flux through the faces. On the sides theyare parallel if is positive, anti parallel if is negative. Therefore, the total outward flux throughS is

)2(0cos lrEsESE

surfacecurved

Also, By Gauss�s Law )(00

ll

q

q

From (i) and (ii)

E(2rl) = 0

l

02

rE =

rk2

(b)







CLASS-XII / (CBSE)

PAGE # 19

(c)

Work done to move charge q by small distance dxdw = F dx cos180° = �qE dx

= 0

qdx

2 x

( E =

02 x

)

So, work done to move from r1 to r

2

w =

2

1

r

r

dw

2

1

r

0r

qdx

2 x

w = 2

1

r

0 r

q dx2 x

e 2 e 10

qw log r log r

2

1e

0 2

rqw log

2 r

Q 25. (a) State the principle of an ac generator and explain its working with the help of a labelled diagram.Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A,rotating with a constant angular speed �� in a magnetic field B , directed perpendicular to the axisof rotation.

(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate thepotential difference developed between the ends of its wings having a span of 20 m. The horizontalcomponent of the Earth�s magnetic field is 5 × 10�4 T and the angle of dip is 30°

ORA device X is connected across an ac source voltage v = v0 sin t the current through X is given

as I = I0 sin

2t

(a) Identify the device X and write the expression for its reactance.(b) Draw graphs showing variation of voltage and current with time over one cycle of ac, for X.(c) How does the reactance of the device X vary with frequency of the ac ? Show this variation graphi

cally.(d) Draw the phasor diagram for the device X.







CLASS-XII / (CBSE)

PAGE # 20

Ans 25. (a) It works on the phenomenon of electro magnetic induction i.e., whenever there is a change inmagnetic flux linked with a coil an induced emf is induced in it. A coil made to rotate in magnetic fieldabout an axis results in a change in effective area normal to the magnetic flux causing varying fluxand hence producing an alternating emf in the coil.

Magnetic flux B = NBA cos t

B = NBA cos t

and induced emf by Lenz low E =dtd B

= NBA sin t

(b) V = 900 km/h = 900 ×5

18m/s

Plane is moving horizontally, So it will cut the vertical component of earth�s magnetic field

normally.V = 250 m/s

= 20 mB

H = 5 × 10�4 T

tan 30° = H

V

B

B

BV =

4HB5 10 3

tan30

BV = 45 3 10

Induced emf = B

V V

= 45 3 10 20 250

= 125 3 10

= 2.5 3 Volts or 4.3Volts

OR(a) V = V

0 sin t

= I0 sin t

2







CLASS-XII / (CBSE)

PAGE # 21

(a) X is capacitor.

It�s capacitive reactance is given by XC = fCC 211

(b)

(c)

(d)

Q 26. (a) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted andmagnified image of the object.

(b) Obtain the mirror formula and write the expression for the linear magnification.(c) Explain two advantages of a reflecting telescope over a refracting telescope.

OR(a) Define a wavefront Using huygens principle verify the laws of reflection at a plane surface(b) In a single slit diffraction experiment the width of the slit is made double the original width, How does

this affect the size and intensity of the central diffraction band ? Explain.(c) When a tiny circular obstacle is placed in the path of the light from a distant source, a bright spot is

seen at the centre of the obstacle Explain why.

Ans. (a) Object between F and C. The image is Beyond CNature � Real , Inverted and Larger than object.

PF

A

OCI

B







CLASS-XII / (CBSE)

PAGE # 22

(b) Consider an object AB placed on the principal axis beyond the centre of curvature C of a concavemirror of small aperture, as shown in Fig. A ray AM from the object travels parallel to the principalaxis and after reflection from the mirror it passes through focus F. Another ray AP is incident on thepole P of the mirror and is reflected along P A� in accordance with the laws of reflection so that

APB = BPA�. The two reflected rays meet at point A�. Thus A� is the real image of A. The image

of any point on AB will lie on a corresponding point of A� B. Hence A� B is the real image of AB formed

by reflection. from the concave mirror.Using cartesian sign convention, we find

A

B

B'

M

A'

Object distance, BP = � u

Image distance, B� P = �v

Focal length, FP = �f

Radius of curvature, CP = �R = �2 f

Now A� B� C ~ ABC

RuvR

CPBPPB'CP

BCCB'

ABB'A'

.....(1)

As A� PB� = APB, therefore,A� B� P ~ ABP.

Consequently,

uv

uv

BPPB'

ABB'A'

........(2)

From equations (1) and (2), we get

uv

RuvR

or �uR + uv = �uv + vR

or vR + uR = 2 uv

Dividing both sides by uvR, we get

R2

v1

u1

But R = 2 f

f1

v1

u1

This proves the mirror formula for a concave mirror, when it forms a real image.

Linear magnification is given by : m = uv

hh

1

2

here h2, h1 are heights of image and object respectively

v, u are distances of image and object from the pole of mirror respectively

(c) (i) No chromatic aberration(ii) Spherical aberration reduces significantly(iii) Low set up cost, lighter, forms brighter image.

OR(a) A wavefront is defined as the continuous locus of all such particles of the medium which are vibrating

in the same phase at any instant.







CLASS-XII / (CBSE)

PAGE # 23

Laws of reflection on the basis of Huygens� wave theory�

As shown in Fig. consider a plane wavefront AB incident on the plane reflecting surface XY, both thewavefront and the reflecting surface being perpendicular to the plane of paper.

Wavefronts and corresponding rays for reflection from a plane surface.

First the wavefront touches the reflecting surface at B and then at the successive points towards C.In accordance with Huygens� principle, from each point on BC, secondary wavelets start growing

with the speed c. During the time the disturbance from A reaches the point C, the secondarywavelets from B must have spread over a hemisphere of radius BD = AC = ct, where t is the timetaken by the disturbance to travel from A to C. The tangent plane CD drawn from the point C over thishemisphere of radius ct will be the new reflected wavefront.Let angles of incidence and reflection be i and r respectively. In ABC and DCB, we have

BAC = CDB [Each is 90°]

BC = BC [Common]AC = BD [Each is equal to ct]

ABC DCBHence ABC = DCBor i = ri.e., the angle of incidence is equal to the angle of reflection. This proves the first law of reflection.

(b) If slit width (d) is doubled

(i) Width of central maximum

dD

2

will become half.

(ii) Intensity of central maxima will become double ( d)

(c) This is due to diffraction, central maximum is obtained as all the waves reaching at the centresuperimpose at same phase.



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