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Praganya Prakashan

CBSE Class-12th

Physics Model Paper 2

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General Instructions1. All questions are compulsory. There are 27 questions in all.2. This question paper has four sections: Section A, Section B, Section C and Section D.3. Section A contains five questions of one mark each, Section B contains seven questions two marks each, Section C contains

twelve questions of three marks each and Section D contains three questions of five marks each.4. 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 attempt only one of the choices in such questions.5. You may use the following values of physical constants wherever necessary. /m sc 3 108

#= , . Jsh 6 63 10 34#= − , . Ce 1 6 10 19

#= − , TmA4 1007 1

#μ π= − − , . C N m8 854 10012 2 1 2

#ε = − − − ,

Nm C41 9 10

0

9 2 2#πε = − , Mass of electron . kg9 1 10 31

#= − ,

Mass of neutron . kg1 675 10 27#= − ,

Mass of proton . kg1 673 10 27#= − , Avogardro’s number .6 023 1023

#= per gram mole, Boltzmann constant . JK1 38 10 23 1#= − −

.Time : 3 Hrs Max. Marks : 70

Section A

QUE 1.1

The lab instructor told a student that in a galvanometer a coil has been wrapped on a conducting frame. Why? Which value is shown by the lab instructor?

Ans :

Eddy currents in conducting frame help in stopping the coil soon, i.e. in making the galvanometer dead beat. Value shown by lab instructor is ‘Imparting knowledge to students’. (1)

QUE 1.2

The horizontal component of earth’s magnetic field at a place is B and angle of dip is 60c. What is the value of vertical component of earth’s magnetic field at this place?

Ans :

Bv sinB θ = sinB 60= c

B23= (1)

QUE 1.3

The graph given below represents the variation of the opposition offered by the circuit element to the flow of alternating current with the frequency of the applied emf. Identify the circuit element.

Ans :

From graph, it is clear that resistance (opposition to current) is not changing with frequency, i.e. resistance does not depend on frequency of applied voltage, so the circuit element here is pure resistance.

QUE 1.4

A converging lens is kept coaxially in contact with a diverging lens both the lenses being of equal focal lengths. What is the focal length of the combination?

Ans :

Let f and f− be focal lengths of the converging and the diverging lenses, respectively.Then, focal length of the combination is given by

F1 f f

1 1= + −

or F1 0=

F 3=The focal length of the combination is infinite. (1)

QUE 1.5

Name the types of communication system according to the type of transmission channel.

Ans :

According to the type of transmission channel, the type of communication system are1. Line communication2. Space communication (1)

Section B

QUE 1.6

Calculate the distance of an object of height ( )h from a concave mirror of focal length 10 cm, so as to obtain a real image of magnification 2.

Ans :

Given,

f cm10=− , m 2=−

m uv=−

2− uv=−

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v u2= (1)Using mirror formula,

v u1 1+ f

1=

u u21 1+ 10

1= −

u23 10

1=−

u cm15=−

Thus, the object is placed in front of concave mirror at a distance of 15 cm (1)

OR QUE

Sky waves are not used in transmitting TV signals, Why? State two factors by which the range of transmission of TV signals can be increased.

Ans :

TV signals have a frequency of 100 to 220 MHz, which cannot be reflected by the ionosphere.So, transmission of TV signals via sky wave is not possible. (1)Range of TV transmission can be increased by using1. Tall antenna2. Geostationary satellites. (1)

QUE 1.7

Guess a possible reason, why water has a much greater dielectric constant than mica?

Ans :

Dielectric constant of water is much greater than that of mica because of the following reasons 1. Water has a symmetrical shape as compared to mica2. Water has permanent dipole moment. (1+1)

QUE 1.8

Apply Kirchhoff’s laws to the loops PRSP and PRQP to write the expressions for the currents I1, I2 and I3 in the given circuit.

Circuit diagram of loops

Ans :

Apply Kirchhoff’s Ist law,

I3 I I2 1= + (1/2)Applying Kirchhoff’s lInd law to loop PRSP ,

I I20 200 53 2− − + 0=

I I40 42 3+ 1= (1/2)Applying Kirchhoff’s lInd law to loop PRQP ,

I I20 60 43 1− − + 0= (1)

I I15 51 3+ 1=

QUE 1.9

Write Einstein’s photoelectric equation. Plot a graph showing the variation of stopping potential versus the frequency of incident radiation.

Ans :

Einstein’s photoelectric equation is,

( )h v v0− eV0= (1)Where, v0 is the threshold frequency, v is the frequency of incident radiation, h is Plank’s constant and V0 is the stopping potential v v0 − graph is shown below

(1)

QUE 1.10

An electron and a proton have the same kinetic energy. Which of the two has a greater wavelength? Explain.

Ans :

We know that de-Broglie wavelength is given by

λ mvh=

mKh

2= (1)

As, m m>p e , thus it is clear that for same kinetic energy,

p

e

λλ m

me

p=

>e pλ λ , i.e. de-Broglie wavelength of electron will be greater than that of a proton. (1)

QUE 1.11

Distinguish between point-to-point and broadcast communication modes. Give one example each.

Ans :

Point-to-point Communication Broadcast Communication

In such a mode, the communication link is between a single transmitter and receiver. e.g. Telephone.

In such a mode, large number of receivers are linked to a single transmitter. e.g. Radio.

The information is sent from a sender to a specific receiver i.e., from one point to the other point.

The information is broadcasted to all the receivers simultaneously.

(2)

QUE 1.12

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While studying a chapter ‘current electricity’, a student asked the teacher that, when is more power delivered to a light bulb, just after it is turned on and the glow of the filament is increasing or after it has been ON for a few seconds and the glow is steady. Write the appropriate explanation the teacher would have given about the students query. Which value is shown by the student?

Ans :

As, the volage is applied across the cold filament, when it is first turned on, the resistance of the filament is low, the current is high and a relatively large amount of power is delivered to the bulb. (1)As the filament warms, its resistance increases and the current decreases. As a result, power delivered to bulb decreases. (1)The value shown by stint is local affwoh.

Section C

QUE 1.13

In a single slit diffraction pattern, how does the angular width of central maximum changes when1. Slit width is decreased?2. Distance between the slit and screen is increased?3. Light of smaller visible wavelength is used? Justify your

answer in each case.

Ans :

We know that angular width of central maximum of diffraction pattern of a single slit is given by,

2θ a2λ =

1. If slit width a is decreased, the angular width will increases because 2 a

1?θ . (1)2. If the distance between the slit and the screen increases,

then it does not affect the angular width of diffraction maxima. (1)

3. If the light of smaller visible wavelength is used, the angular width is decreased because 2 ?θ λ. (1)

QUE 1.14

1. Steel is preferred for making permanent magnets, whereas soft iron is preferred for making electron magnets. Why?

2. A uniform magnetic field exists normal to the plane of the paper over a small region of space. A rectangular loop of wire is slowly moved with a uniform velocity across the field as shown in the figure.

Draw the graph showing the variation of (a) Magnetic flux linked with the loop and (b) The induced emf in the loop with time.

Ans :

1. Steel is preferred for making permanent magnets on account of its high retentivity and high coercivity. Soft iron is preferred for making electromagnets on account of low retentivity, low corecivity and low hysteresis loss. (1)

2. (a) Variation of magnetic flux linked with the loop

(1) (b) Variation of induced emf in the loop with time

(1)

OR QUE

A small compass needle of magnetic moment m and moment of inertia I is free to oscillate in a magnetic field B . It is slightly disturbed from its equilibrium position and then released. Show that it executes simple harmonic motion. Hence, write the expression for its time period.

Ans :

Let a small magnetic needle of magnetic moment m be freely suspended in a uniform magnetic field B, so that in equilibrium position, magnet comes to rest along the direction of B.

Hence, Restoring torque,

τ m B#= sinmB θ =− (1/2)If I is the moment of inertia of magnetic needle about the axis of suspension, then

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τ Iα = Idtd

2

2θ = (1/2)

Hence, in equilibrium state, we have

Idtd

2

2θ sinmB θ =−

If θ is small, then sinθ θ = , we get

Idtd

2

2θ mBθ =−

dtd

2

2θ ImB θ =− (1/2)

But angular acceleration is directly proportional to angular displacement and directed towards the equilibrium position, motion of the magnetic needle is simple harmonic motion.Angular frequency of SHM,

ω ImB= (1)

Time period of oscillation,

T 2ωπ= mB

I2π = (1/2)

QUE 1.15

A circular coil of N -turns and radius R is kept normal to a magnetic field given by cosB B t0 ω = . Deduce an expression for the emf induced in this coil. State the rule which helps to detect the direction of induced current.

Ans :

Induced emf in the coil,

e N dtdφ =− ( )cosNdt

d BA θ =− [ ]cosBAφ θ=

( )cosNdtd BA 0=− c

NA dtdB=− [ ]cos0 1=c (1)

( )cosN R dtd B t2

0π ω=− [ ]A R2π =

sinN R B t20π ω ω= ( )cos sindt

d t tω ω ω=−: D

(1)The direction of induced current is given by Lenz’a law which sates the direction of induced current is always in such a way that it opposes the cause due to which it is produced. (1)

QUE 1.16

A convex lens made of a material of refractive index n1 is kept in a medium of refractive index n2. A parallel beam of light is incident on the lens. Complete the path of rays of light emerging from the convex lens, if1. n n>1 2

2. n n1 2=3. n n<1 2.

Ans :

The path of rays of light emerging from the convex lens in different cases is shown below.1. n n>1 2

(1)2. n n1 2=

(1)3. n n<1 2

(1)

QUE 1.17

1. What will be the effect on the fringe width, if the entire Young’s double slit experiment’s apparatus is immersed in water?

2. Draw a diagram showing the formation of primary rainbow and explain at what angles the primary rainbow is visible.

Ans :

1. We have, wλ w

a

μλ=

i.e. the wavelength of light decreases in water.

Therefore, the fringe width β dDλ = , also decreases in water

as ?β λ. (1)2. Formation of primary rainbow

(a) Primary rainbow

The primary rainbow is formed by those rays which suffer one internal reflection and two refractions and come out of the raindrop at angle of minimum deviation.

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(b) Angles of primary rainbow

The violet and red light colours emerge at 40c and 42c, respectively and can be viewed by observer. (2)

QUE 1.18

Find the half-life period of a radioactive material, if its activity drops to (1/16)th of its initial value in 30 yr.

Ans :

Activity ? Number of atoms present

Hence, N N16

0= , if t yr30= (1)

Let half-life period of sample be T .Number of atoms left after n half-lives is given by,

N N 21 n

0= b l

NN

0 2

1 n= b l

161 2

1 n= b l

& 2n 16 24= =

& n 4=

Hence, Half-life period, T nt= 4

30= . yr7 5= (2)

QUE 1.19

The ground state energy of hydrogen atom is -13.6 eV.1. What is the kinetic energy of an electron in the second

excited state?2. If the electron jumps to the ground state from the second

excited state. Calculate the wavelength of the spectral line emitted.

Ans :

1. E1 . eV13 6=− , if n 1= Energy of electron in the second excited state n 3= is

E3 .

n13 6

2=− .9

13 6=− . eV1 51=− (1)

Hence, Kinetic energy of electron in the second excited state is given by

KE3 E3=− . eV1 51= (1)

2. Energy of the photon emitted is given by,

E E E3 1= − . ( . )1 51 13 6=− − − . eV12 09=

. . J12 09 1 6 10 19# #= −

Wavelength of the spectral line emitted,

λ Ehc=

. ..12 09 1 6 10

6 63 10 3 1019

34 8

# ## # #= −

−

. m1 028 10 7#= − (1)

QUE 1.20

Write the truth table for circuit given in figure below consist-ing of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

Ans :

A and B are the inputs of the given circuit. The output of the first NOR gate is A B+ . It can be observed from the figure, that the outputs of the first NOR gate becomes the input of the second one.

(1)Hence, the output of the combination is given as,

Y ( ) ( )A B A B= + + +

( ) ( )A B A B$= + +

( ) ( )A B A B$= + + A B= + (1)The truth table for this operation is given as,

A B Y A B= +

0 0 0

0 1 1

1 0 1

1 1 1This is the truth table of an OR gate. Hence, this circuit functions as an OR gate. (1)

QUE 1.21

A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of magnetic field inside the solenoid?

Ans :

Given, N 500=

l . m0 5= , I A5= , r cm m1 10 2= = −

As, rl .

100 5

2= − 50= (1)

l r>>Hence, we can use the formula of magnetic field induction at a point inside the solenoid due to its length. (1)

Hence, B lNI0μ = .4 10 0 5

500 57# # #π = −

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. T6 28 10 3#= − (1)

QUE 1.22

A little boy while crossing the footpath saw an overhead street light transmitting current in the day time. Thinking it to be wastage of current and powerloss, he switched OFF the street light over when it was at some height on the pole and it was not easy for him to switch OFF.Read the above passage and answer the following questions:1. What words would you frame for the little boy and his

thinking?2. A small town with a demand of 800 kW and 200 V of

electric power is situated 15 km away from the electric plant generating power of V440 0. A wire (electric power line of .0 5 Ω per km) connects the town with the plant. The town receives the power from the line through a 4000-200 V step-down transformer of a sub-station located in the town. Estimate the power loss in the form of heat.

Ans :

1. The little boy was having the sense of nationality as he considered himself as a responsible citizen of the country and did a tough task for saving the electricity. His action and sense of belongingness in favour of all proved him to be a member of the family, who loves his nation. (1)

2. Transformer voltage ( )V4000 200−= Length of two conductor 15 2#= km30= Resistance of the conductor .0 5 30#= 15Ω = and power kW800= W800 103

#=

As, VI Power=

I4000# W800 103#=

I A200= Now, power loss in the form of heat

I R2= W200 200 15# #=

600000= kW600= (2)

QUE 1.23

What is a photodiode? Explain its working principle. Why is a photodiode operated in reverse bias?

Ans :

A p n− junction diode which is made up of photo-sensitive semiconductors designed to operate in reverse bias and its reverse current increases, when light of suitable frequency and intensity is incident on it, is known as photodiode . (1)

(a) Symbol of Photodiode

(b) Reverse Biased Photodiode

When light of suitable frequency ( )v is incident on the junction such that energy of photon is greater than the band gap of the semiconductor, then more electron-hole pairs are generated and hence, reverse current increases. The saturation value of reverse current increases with the increase in the intensity of incident light. (1)The photodiode is operated in reverse bias because1. Fractional charge of minority charge carrier is much higher

than fractional charge in majority charge carrier, when light of suitable frequency and intensity is incident on it.

(1/2)2. Reverse bias current due to minority charge carrier is easily

measurable than majority charge carrier dominated forward current. (1/2)

QUE 1.24

Give reason of the followings1. D-lazer and E-layer disappear at night in the earth’s

atmosphere.2. Radioengineers always considers microwaves better carriers

of singnals than radio waves.3. In AM broadcast, range of frequencies are limited to 30

MHz.

Ans :

1. It is due to the de-ionisation of the ions into molecules in these layers, in the absence of radiations coming fro the sun. (1)

2. On account of smaller wavelengths of microwaves, they can be transmitted as beam signals in particular direction, much better than radio waves, as microwaves do not bend around the corners of the obstacles coming in their way. (1)

3. In standard AM broadcast, ground wave propagation is used for transmitting a signal. Attenuation of surface wave increases very rapidly with increase in frequency i.e. why it is limited to frequencies upto a few MHz Thus in AM broadcast, range of frequencies are limited to 30 MHz (1)

Section D

QUE 1.25

Find an expression for the capacitance of a parallel plate capacitor.

An air capacitor has a capacitance of F2 μ , which becomes F12 μ , when a dielectric medium is filled in the space between

the plates. Find dielectric constant of that material.

Ans :

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Suppose a parallel plate capacitor consists of two conducting parallel plates X and Y , each of cross-sectional area A and separated by a distance d consisting of material having dielectric constant K .

q+ charge is given to plate X while plate Y is connected to the earth. (1)Charge density on plates,

σ Aq=

Electric field intensity between the plates,

E K 0εσ= (1)

We know that potential difference between plates,

V E d= K d0ε

σ= (1)

Substituting the value of σ , we get

V AKq d

oε =

The capacitance of a parallel plate capacitor,

C Vq=

KAqdq

0ε

=c m

dKA 0ε = b l (1)

For air, the capacitance of the capacitor,

C0 dA 0ε = F2 μ = ...(i)

When a dielectric medium is placed between the plates,

then C dKA 0ε = 1 F2 μ = ...(ii)

Dividing Eq. (ii) by Eq. (i), we get

dAd

KA

0

0

ε

ε

b

b

l

l 2

12=

& K 6= (1)

OR QUE

Find an expression for the electric field intensity at a point on equatorial line due to an electric dipole.

Ans :

Consider an electric dipole AB , consists of two charges q+ and q− separated by a distance a2 . We have to find electric field

at a point P on equatorial line separated by a distance r from centre O .

(2)Electric field at point P due to charge q+ .

EA r a

q4

10 2 2 2$πε=

+^ h

( )

, ( )along APr a

q4

10

2 2$πε=+

Electric field at point P due to charge q− ,

EB , ( )along PBr a

q4

10

2 2$πε=+^ h

(2)

On resolving EA and EB into rectangular components, sinEA θ and sinEB θ cancel each other. [ ]E EA B=Hence, Resultant electric field at point P ,

E cos cosE EA Bθ θ= +

( )

cosr a

q2 41

02 2# $πε θ=+

E EA B=8 B

( ) ( )r a

qr aa

41 2

02 2 2 2

$ #πε=+ +

cos BPOB

r aa

2 2θ = =

+; E

But, q a2$ p= , electric dipole moment.

Hence, E ( )r a

p4

1/

02 2 3 2$πε=+

(1/2)

If r a> , then r a>>2 2

Therefore, neglecting a2 in comparison to r2, we get

E rp

41

03$πε=

In opposite direction of electric dipole moment. (1/2)

QUE 1.26

Explain with the help of diagram, the principle and working of an AC generator. Write the expression for the emf generated in the coil in terms of its speed of rotation.

Ans :

The labelled diagram of an AC generator is shown below

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(1)PrincipleAn AC generator is based on the principle of electromagnetic induction. If a rectangular armature rotates about its axis in a uniform magnetic field, then the magnetic flux linked with the coil changes and an emf is induced in the coil. The direction of induced current is given by Fleming’s right hand rule. (1)WorkingWhen an armature coil of N turns and each turn enclosing area A is placed in a uniform magnetic field of strength ( )B making an angle θ with normal to the direction of magnetic field.Magnetic flux linked with the coil is cosB A NBA$φ θ= = .As, the coil is rotated about its own axis with an angular speed ( )ω , then value of angle θ tω = and hence, magnetic flux changes and an induced emf is developed across the ends of coil. (2)Hence, Induced emf,

ε dtdφ =− sinNBA tω ω= sin t0ε ω=

Where, 0ε NBAω = = maximum (peak) value of induced emf. Induced emf is sinusoidal in nature. (1)

OR QUE

The primary coil of an ideal step-up transformer has 100 turns and the transformation ratio is also 100. The input voltage and the power are 220 V and 1100 W. Find:1. Number of turns in secondary2. The current in the primary3. Voltage across the secondary4. The current in the secondary5. Power in the secondary.

Ans :

Given,

NP 100= , k 100= ,

VP V220= , Pin W1100=

1. NS K NP$= 100 100#= 10000= (1)

2. IP VPin

P= 220

1100= A5= (1)

3. Vs K VP$= 100 220#= V22000= (1)

4. Is KIP= 100

5= . A0 05= (1)

5. Output power V Is s#= .22000 0 05#= W1100= (1)

QUE 1.27

1. What do you mean by the polarisation of light? Define law of Malus and then show that the intensity of light becomes half, when ordinary light is incident on a polariser.

2. Two polarising sheets have their polarising direction parallel, so that the intensity of the transmitted light is maximum. Through what angle must the either sheet be turned, if the intensity is to drop by one half?

Ans :

1. Polarisation of light : The ordinary light have electric vec-tors in all possible directions in a plane perpendicular to the direction of propagation of light waves. When it is pass through a tourmaline crystal, then in transmitted light, only those electric vectors are present which are parallel to the axis of crystal.

Such light is called plane polarised light. The phenomena of restricting the electric vectors of light

into particular direction is called polarisation of light. The tourmaline crystal acts as a polariser. ( )11

2

Law of Malus when completely plane polarised light is incident on an analyser, the intensity of transmitted light is proportional to the square of the cosine of the angle between the plane of polariser and analyser.

cosI 2? θ cosI I02& θ = (1)

Where, I0 is the intensity of incident light. Let the intensity of the ordinary light be I0 and it is incident

on a polariser. In ordinary light, electric vectors are in all possible directions

and therefore, intensity of transmitted light is

I ( )cosI av02θ = [average value of cos2θ ]

Here, ( )cos av2θ cos d2

1 2

0

2

π θ θ=π

#

cos d21

21 2

0

2

πθ θ= +

π

: D#

sin41

22

0

2

π θ θ= +π

: D sin41 2 2

4 0π π π= + −: D

41 2#π π= 2

1=

Hence, I I20= ( )11

2

2. We know that, I I20=

Using Malus’s law,

I cosI02θ =

I20 cosI0

2θ =

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cos2θ 21=

cosθ 2

1!= & θ ,45 135! != c c (1)

OR QUE

1. Define Brewster’s law. Show that the sum of angle of polarisation and angle of refraction is 90c.

2. Discuss the intensity of transmitted light, when polaroid sheet is rotated between two crossed polaroid.

Ans :

1. Brewster’s Law : When unpolarised light is incident on a transparent medium at polarising angle , the reflected light becomes completely polarised. (1)

Transparent Medium of Polarising Angle

(1) If refractive index of the transparent medium is μ , then

tan ipμ = This relation is called Brewster’s law, which gives relation

angle of polarisation ( )iP and angle of refraction. According to the Brewster’s law,

μ tan ip=

or μ cossin

ii

p

p= ...(i)

According to Snell’s law, Refractive index,

μ sinsin

rip

p= ...(ii) (1)

From Eqs. (1) and (2), we get

cossin

ii

p

p sinsin

rip

p=

& cos ip sin rp=

or ( )sin i90 p−c sin rp= ( )cos sin 90θ θ= −c6 @

( )i90 p−c rp=

& i rp p+ 90= c (1)2. Let I0 be the intensity of polarised light transmitted by

first polariser P1. Then, the intensity of light transmitted by second polariser P2 will be,

I ( )cosI02θ =

As, P1 and P3 are crossed, the angle between P2 and P3 will be ( / )2π θ− .

The intensity of light transmitted by P3 will be,

I ( / )cos cosI 202 2θ π θ= −

cos sin sinI I4 20

2 2 0 2θ θ θ= −

The transmitted intensity will be maximum When,

2θ /2π =

& θ /4π = (1)

F R E E B O A R D E X A M S O LV E D PA P E R S P D FBY

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