700002231_topper_8_101_4_4_physics_2017_solution_up201703301955_1490883947_3274.pdf

CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 CBSE Class XII Physics – Set 1 Board Paper – All India – 2017 Solution SECTION A 1. Nichrome wire is heated more Heat dissipate in a wire is given by  H  I2Rt  I2 t A  H    I, and A remains same  But nichrome  copper  Hnichrome  Hcopper

2. Yes. As an electromagnetic wave is a combination of electric and magnetic fields, it carries both energy and momentum. It is given by hC E  p  mC where c = Speed of EM wave in vacuum  = Wavelength of EM wave 3. If the incident violet light is replaced with red light, then the angle of minimum deviation of a glass prism decreases. 4. Photoelectric effect shows the quantum nature of electromagnetic radiation. 5. According to Lenz law, Plate A of the capacitor is at a higher potential than Plate B, i.e. polarity of Plate A will be positive with respect to Plate B.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 SECTION B 6.

Intensity distribution graph for single slit diffraction

Difference between interference pattern and diffraction pattern Interference It is due to the superposition of two waves coming from two coherent sources. Width of the interference bands is equal.

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Diffraction It is due to the superposition of secondary wavelets originating from different parts of the same wavefront. Width of diffraction bands is not the same.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 OR According to law of Malus, when a beam of completely plane polarised light is incident on an analyser , resultant intensity of light (I) transmitted from the analyser varies directly as the square of cosine of angle between the plane of the analyser and polariser.

I  cos2   I  Io cos2  ........ Equation 1

polarizer

When   0 or   180

and analyzer and parallel

 cos   1  I  Io ........ Equation 2  When   90  cos   cos 90  0  I  0 ........ Equation 3 In unpolarised light, vibrations are probable in all the direction in a plane perpendicular to the direction of propagation.   can have any value from 0 and 2.  cos   av 2

1  2

2

1  2

 cos 0

2



2

 d

1  cos 2 

0

2

d

 sin2   1   0   2  2  2 

2

0

1 2 Usin g the law of malus 

I  Io cos2   I  Io  Graph:

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1 Io  2 2

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 7. (a) X-rays: Used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. (b) Microwaves: Used in radar systems in aircraft navigation. 8. Consider a charge q moving with velocity v in the presence of electric and magnetic fields. The force on an electric charge q due to both of them is F  q E r   v  B r    F  Felectric  Fmagnetic ........ Equation 1 where v  velocity of the charge r  location of the charge at a given time t E r  = Electric field B r   Magnetic field

Let us consider a simple case in which electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle.

FE  qE  qEj.FB  q v B



ˆ  q vi  Bk



 qBj  F  q E  vB  j

Thus, electric and magnetic forces are in opposite directions. Suppose we adjust the values of E and B such that magnitudes of the two forces are equal, then the total force on the charge is zero and the charge will move in the fields undeflected. This happens when

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 qE = qvB

E B This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass). The crossed E and B fields therefore serve as a velocity selector. or or v 

9. It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV = −1.1 eV. Orbital energy related to orbit level (n) is 13.6 E eV 2 n 

For n  3, 13.6 13.6 E eV  eV  1.5 eV 2 9 3 This energy is approximately equal to the energy of gaseous hydrogen. This implies that the electron has jumped from n = 1 to n = 3 level. During its de-excitation, electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum. Relation for wave number for the Lyman series is 1 1 1 R 2  2  n  1 For first member n  3 1 1 1  1 1   R   R 2  2 1 1 1 9  3   1 9  1 7 1   1.097  107   ......... where Rydberg constant R  1.097  10 m 1 9   







1 8  1.097  107  1 9

 1  1.025  107 m

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 For n  3 1 1 1  1 1   R   R 2  2 2 1 1 4  2   1  4  1   1.097  107  ......... where Rydberg constant R  1.097  107 m1  2  4  







1 3  1.097  107  2 4

 2  1.215  107 m

Relation for wave number for the Balmer series is 1 1 1 R 2  2  n  2 For first member n  3 1 1 1  1 1    1.097  107     R 2  2 3 2 4 9 3    3  6.56  107 m



10. Two properties of material used for making permanent magnets are (a) high coercivity (b) high retentivity (c) high hysteresis loss Two properties of material used for making electromagnets are (a) high permeability (b) low coercivity (c) low retentivity

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 SECTION C 11. (a)

Heat generated H  I2Rt 

V2 t R

Given that H'  9H V '2 t V2 t  9 R R 2 2  V'  9  V  V '  3V 

(b) Given : emf E  12 V Internal resis tan ce r  2  External resis tan ce r  4  Current I 

E 12 12   R r 42 6

 I  2A We know V  E  Ir 

 12   2  2 

V  8 V

Thus, reading of the ammeter will be 2 A and of the voltmeter will be 8 V. 12. (a) In amplitude modulation, the amplitude of the carrier varies in accordance with the information signal. At the input of the transistor as CE, the low frequency modulating signals are superimposed on high frequency carrier wave. The output signal is carrier signal varying in amplitude in accordance with biasing modulation voltage. Thus, AM wave is produced.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 (b) Given fc  fm  660 kHz ........ Equation 1 fc  fm  640 kHz ........ Equation 2  Adding equation 1 and 2, we get 2fc  660 kHz+640 kHz  fc  650 kHz Now fc  fm  660 kHz  fm  660 kHz  650 kHz  fm  10 kHz

Band width required for amplitude modulation = Upper side  lower side = (fc +fm)  (fc fm) = 2 fm ∴ Band width required for amplitude modulation = 2  10 kHz = 20 kHz 13. (a) Reverse biased. (b) This circuit uses two diodes and a centre-tap transformer whose secondary is wound into two equal parts. The voltages at any instant at S1 and S2 are out of phase with respect to each other. If the voltage at any instant at S1 is positive, then diode D1 conducts and because the voltage at S2 is negative, D2 does not conduct.

Input and output wavefront

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017

Working:  During this positive half-cycle, we get an output current, and hence, voltage across load resistance.  If the voltage at S1 becomes negative, only D2 conducts, and we get an output in the negative half-cycle as well, which is however identical to the output in the positive half-cycle. Thus, we find that during both halves, current flows in the same direction. 14. When a photon of energy 'h’ falls on a metal surface, the energy of the photon is absorbed by the electrons and is used in the following two ways: (a) A part of energy is used to overcome the surface barrier and come out of the metal surface. This part of energy is known as a work function and is expressed as o = ho (b) The remaining part of energy is used in giving a velocity ‘v’ to the emitted photoelectron which is equal to the maximum kinetic energy of photo 1  electrons  mv2max  2  (c) According to the law of conservation of energy, 1 2 h  o  mvmax 2 1 2 h  h o  mvmax 2 1  mv2max  KEmax  h  h o 2  KEmax  h     o  or KEmax  h  o

This equation is called Einstein photoelectric equation.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 Features of photoelectric effect which cannot be explained by wave theory: (i) Maximum kinetic energy of emitted electrons is independent of the intensity of incident light. (ii) There exists a ‘threshold frequency’ for each photosensitive material. (iii) ‘Photoelectric effect’ is instantaneous in nature. 15. (a) Given  air  589 nm  589  109m Speed of light c  3  108m/s a

w  1.33

Frequency of the refracted light, f  f 

c  air

3  108  5.09  1014 Hz 9 589  10

Wavelength of refracted light  water 

 air w

a

589  109  4.42  107 m 1.33 c Speed of refracted light, v water  a w   water 

 vwater 

3  108  2.26  108 ms1 1.33

(b) Given f  20 cm and   1.55 Let the radius of the curvature of each of the two surfaces of the lens be R. If R1  R  R 2  R 1 1 1     1    f  R1 R 2  1 1 1    1.55  1    20 R R  1 1.1  20 R  R  20  1.1  R  22 cm 

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 16. Mutual inductance is the property of two coils by the virtue of which each opposes any change in the strength of current flowing through the other by developing an induced emf.

Consider two long co-axial solenoids, each of length l. Let n1 be the number of turns per unit length of the inner solenoid S1 of radius r1, n2 be the number of turns per unit length of the outer solenoid S2 of radius r2. Imagine a time varying current I2 through S2 which sets up a time varying magnetic flux 1 through S1. ∴1 = M12 (I2) ……… (Equation 1) where M12 = Coefficient of mutual inductance of solenoid S1 wrt solenoid S2 Magnetic field due to the current I2 in S2 is B2 = on2I2 ∴Magnetic flux through S1 is ∴ 1 = B2A1N1 where N1 = n1 and l = total number of turns in S1 1 = (on2I2) (r12) (n1l)  1 = o n1n2 r12 l I2 ……… (Equation 2) From Equations 1 and 2, we get  M12 = o n1n2 r12 l ……… (Equation 3) Let us consider the reverse case. A time varying current I1 through S1 develops a flux 2 through S2. ∴2 = M21 (I1) ……… (Equation 4) where M21 = Coefficient of mutual inductance of solenoid S2 wrt solenoid S1 Magnetic flux due to S1 is confined solely in S1 as the solenoids are assumed to be very long. There is no magnetic field outside S1 due to current I S1. The magnetic flux linked with S2 is ∴ 2 = B1A1N2 = = (on1I1) (r12) (n2l)  2 = o n1n2 r12 l I1 ……… (Equation 5) From Equations 4 and 5, we get  M21 = o n1n2 r12 l ……… (Equation 6) From Equations 3 and 6, we get www.topperlearning.com

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 M12 = M21 = M = = o n1n2 r12 l We can write the above equation as N N  M  o  1   2  r2  l  l  l 

M

 



oN1N2 r 2 l

OR Self-inductance is a property of a coil by virtue of which it opposes any change in self-flux linked with the coil. When a current is passed through a coil, it produces a magnetic field. A flux is linked with the magnetic field produced by the coil. This flux is called selfflux. Self-flux is directly proportional to the current flowing through the coil. If I is current flowing through a coil and ‘’ is the magnetic flux linked with its own magnetic field, then

I Or

  LI

where L is the proportionality constant and is known as self-inductance.

Consider a simple circuit having a coil, a battery and a key. The coil has a selfinductance L. On pressing the key K, current flows through the circuit. However, self-inductance gives rise to induced current which opposes the growth of current in the circuit. Thus, to increase the current from zero to its maximum value Io, some work has to be done. This work done is stored as the magnetic field of the inductor. Similarly, when the key is opened, the induced emf tends to maintain current in the circuit.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 Magnitude of emf is given by dI dt eI dt  LI dI ......... Equation 1 eL

dq dt I dt  dq

But I 

Also, work done  dW   Voltage  e   Ch arg e  dq  or dW  e  dq  eI dt Substituting in equation 1, we get dW  LI dI Integrating the above equation to find the total work done in increasing the current from zero to Io , we get



W

0

dW 

Io

 LI dI 0

1 2 LIo 2 This work done in increasing the current flowing through the inductor is stored as potential energy (U) in the magnetic field of the conductor. 1 U  LI2o 2 or W 

17. (a) Principle of working of a meter bridge: Working of a meter bridge is based on the principle of Wheatstone bridge. According to the principle, the balancing condition is P R  Q S For balancing lengths in a meter bridge, P R l R    Q S 100  l S 100  l S  R l (b) For balancing length l1, the condition is l1 R  ...... (1) S 100  l1 When a resistance X is connected in parallel with S, the net resistance becomes XS Snew  XS For balancing length l2, the condition is www.topperlearning.com

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 l2 R  Snew 100  l2 

R  X  S XS



l2 100  l2

...... (2)

From (1) and (2), l1 l2 XS   100  l1 X 100  l2 

l2 100  l1 XS   X 100  l2 l1



X  S l2 100  l1   X l1 100  l2 



l 100  l1  S 1  2 X l1 100  l2 



S l2 100  l1   1 X l1 100  l2 



S l2 100  l1   l1 100  l2   X l1 100  l2 



l1 100  l2  X  S l2 100  l1   l1 100  l2 

X  S

18.

l1 100  l2 

l2 100  l1   l1 100  l2 

Generalised communication system:

Transmitter: A transmitter is an arrangement which processes the incoming message signal to a form suitable for transmission through a channel and subsequent reception. Channel: Channel is the medium through which the signal is transmitted from the transmitter to the receiver. Receiver: A receiver extracts the desired message signals from the received signals at the channel output.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 19. (a) Three segments of a transmitter are Emitter (E), Base (B) and Collector (C). Emitter: It is of moderate size and heavily doped. It supplies a large number of majority carriers for the current flow through the transistor. Base: It is the control segment and is very thin and lightly doped. Collector: It is the segment which collects a major portion of the majority carries supplied by the emitter. It is moderately doped and large in size as compared to the emitter. (b) Output waveform for AND gate is

Truth table for the same is A 0 0 1 1

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B 0 1 0 1

Y  A B 0 0 0 1

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 20. (a) Telescope:

(b) Lens L1 has higher aperture of 8 cm. So, it can gather more light and will have high resolving power. Hence, L1 should be used as the objective lens. Lens L3 should be used as an eyepiece because it has high power. So, it will give higher magnification. 21. (a) Biot–Savart law states that the magnetic field dB due to the current element dl at any point P is (i) directly proportional to the current I, dB  I (ii) directly proportional to the length dl of the element, dB  dl (iii) directly proportional to sin, where  is the angle between dl and r, dB  sin (iv) inversely proportional to the square of the distance r from the current 1 element AB, dB  2 r Therefore, we have Idl sin  dB  r2  Idl sin   dB  0 4 r2 In vector notation,  Idl  r dB  0 4 r3 (b) Two coils are placed as shown in the figure below:

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 I1  1 A I2 

3 A

The magnetic field due to coil P at its centre is I BP  0 1 along z-axis 2R The magnetic field due to coil Q at its centre is I BQ  0 2 along x-axis 2R Hence, the field structure is

Therefore, the resultant field is B  BP2  B2Q 2

B 

2

 0I1   0I2        2R   2R 

2

 0   0     3   2R   2R 

2

   B  2 0   0  2R  R

The direction of this field is in the x–z plane. 22. Two capacitors are connected in parallel. Hence, the potential on each of them remains the same. So, the charge on each is QA  CV  QB Hence, the energy stored in the system is 1 1 Uinitial  CV2  CV2  CV2 2 2 When a dielectric slab is introduced, the capacitance changes to KC. As the switch is open, only voltage across A remains the same. The voltage across B changes to V’ = Q/C’ = Q/KC = V/K. Hence, new energy stored in the system is Ufinal   Ufinal 

1 1 V2 KCV2  KC 2 2 2 K 1 1 CV2 1 1   KCV2   CV2  K   2 2 K 2 K 

Uinitial  Ufinal

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1 1 K K



K K2  1

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 23. (a) The installation at Chernobyl was a nuclear power plant. The disaster took place because a fire broke out in the reactor and released harmful radiations in the atmosphere. (b) Inside a reactor, nuclear energy is first converted to heat energy. This heat energy is then converted to mechanical energy of the turbine which is finally converted to electrical energy. (c) Values displayed by Asha: Awareness about real-life scenarios, helping nature towards her mother Values displayed by Asha’s mother: Curiosity towards an incident SECTION E 24. (a) Consider an electric dipole consisting of two point charges +q and q separated by a small distance 2a.

Field intensity at point P due to charge q is 1 q 1 q E1   2 40 AP 40 r  a2 It is along PA. The field intensity at point P due to charge +q is 1 q 1 q E2   40 BP2 40 r  a2 It is along BP. Hence, the resultant field is 1 q 1 q E  E2  E1   2 40 r  a 40 r  a2 q  1 1  q   E   2 2 40  r  a r  a  40  E 

  4ar  r 2  a2 



  2  



q 2a  2r 40 r2  a2 2





Now, the dipole moment is q  2a = p p 2r E  40 r2  a2 2

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 (b) For r ≫ a p 2r 2p E   40 r 4 40r 3 E 

1 r3

Hence, the graph is

(c) For stable equilibrium:

  pEsin   pEsin0

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 For unstable equilibrium:

  pEsin   pEsin180

OR (a) Consider a thin infinite plane uniformly charged sheet having a surface charge density .

We take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z coordinates and its direction. We have to find the electric field intensity because of this sheet at any point which is at distance r from the sheet. Let the Gaussian surface be a rectangular parallelepiped of cross-sectional area A. From the figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces, and they therefore do not contribute to the total flux. No field lines cross the side walls of the rectangular parallelepiped, i.e. component of E normal to the walls of the parallelepiped is zero.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 The unit vector normal to surface 1 is in the –x direction, while the unit vector normal to surface 2 is in the +x direction. Therefore, flux (E.ΔS) through both surfaces is equal and adds up. Therefore, the net flux through the Gaussian surface is 2 EA. The charge enclosed by the closed surface is σ A. Therefore by Gauss’s law, A 2 EA  0

E 

 20

(b) Let V0 be the potential at the point in front of the large thin plane sheet. This point is at a distance r from its surface. dV  E  dr

 V0  V 

 rW 20

25. (a) The device ‘X’ is a capacitor. (b) Curve A: Power consumed in the circuit Curve B: Voltage Curve C: Current This is because current leads voltage in a capacitive circuit. (c) Impedance: 1 XC  2fC Therefore, impedance is directly proportional to frequency.

(d) Voltage applied to the circuit is V  V0 sin t Due to this voltage, a charge will be produced which will charge the plates of the capacitor with positive and negative charges. If the potential difference across the plates of the capacitor is V, then q V C  q  CV Therefore, the instantaneous value of the current in the circuit is

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017

dq d  CV  d   CV0 sin t  dt dt dt V    i  CV0  cos t  0 sin  t   1 2  C Thus, the peak value of current is V i0  0 1 C    i  i0 sin  t   2  i

Hence, current leads voltage by

 . 2 OR

(a) AC Generator: An alternating current (AC) generator/electrical generator/alternator is a device which produces an electromotive force (EMF) by changing the number of magnetic flux lines (lines of force), Φ, passing through a wire coil. The basic elements of an AC generator are shown in the figure below. The AC generator consists of a coil mounted on a rotor shaft.

When the coil is rotated with a constant angular speed ω, the angle θ between the magnetic field vector B and the area vector A of the coil at any instant t is θ = ωt (assuming θ = 0º at t = 0) As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and the flux at any time t is B  BA cos   BA cos t

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 From Faraday’s law, the induced EMF for the rotating coil of N turns is d d   N  NBA (cos t) dt dt Thus, the instantaneous value of the EMF is   NBA sin t where NBA is the maximum value of the EMF If we denote NBA as 0, then we have   0 sin t (b) The emf induced in the rod is   Blv    0.3  104  10  5  1.5  103 V  1.5 mV

26. (a) A wavefront is the locus of points which oscillate in phase. Let PP represent the surface separating medium 1 and medium 2, as shown in the figure below. Let v1 and v2 represent the speed of light in medium 1 and medium 2, respectively. We assume a plane wave front AB propagating in the direction A. A ray is incident on the interface at an angle i as shown in the figure. Let  be the time taken by the wave front to travel the distance BC. Thus, BC = v Determine the shape of the refracted wave front by drawing a sphere of radius V2 from the point A in the second medium.

The figure shows a plane wave AB is incident at an angle i on the surface PP’ separating medium 1 and medium 2.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 The plane wave undergoes refraction and CE represents the refracted wave front. The figure corresponds to v2 < v1 so that the refracted waves bend towards the normal. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2 Consider triangles ABC and AEC, BC v1 sini   AC AC AE v2  sinr   AC AC where i and r are the angles of incidence and refraction, respectively. Therefore, v sini  1 sinr v2 From the above equation, we get the result that if r < i (i.e. if the ray bends towards the normal), the speed (v1) of the light wave in the second medium will be less than the speed (v2) of the light wave in the first medium. If c is the speed of light in vacuum, then c n1  v1 n2 

c v2

where c is the speed of light and n1 and n2 are the refractive indices of the first medium and the second medium, respectively. In terms of refractive index, n1 sini  n2 sinr This is known as Snell’s law of refraction. (b) If an incident, unpolarised wave falls on an atom, it makes the electrons in the atom oscillate along the random direction of the electric field. If light emitted by these oscillating electrons is viewed along a direction perpendicular to the incident direction, the light is seen to be polarised partially. This is because along this line, radiation is emitted with the electric field parallel to the direction of oscillation. Because it also has to be perpendicular to the line of sight, only one direction of oscillation is favoured. For example, sunlight scattered due to the sky observed perpendicular to the direction of the Sun is seen to be polarised.

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017

Brewster angle: It is related to refractive index as   tanip  ip  tan1   tan1 1.5  56.3

OR (a) Image formation by a combination of lens:

Let a point object O be placed on the common principal axis at a distance OC1 = u. Lens L1 will form an image at I2 with C1I2 = v2 Therefore, from the lens formula, 1 1 1   ...... (1) v2 u f1 Now, this image will act as an object for second lens L2 and it will form an image at I with C2I = v As the lenses are thin, u = C2I2 = C1I = v2 1 1 1   ...... (2) v v2 f2 Adding,

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CBSE XII | PHYSICS Board Paper – All India – Set 1 – Solution 2017 1 1 1 1 1     v u f1 f2 f 

1 1 1   f f1 f2

1 P f 1 1 P   f1 f2

(b) Angle of prism is A = 60 as the prism is an equilateral prism. 3 Given : i  A 4  i  45 It is also given that the ray undergoes minimum deviation. i  e A and r1  r2   30 2 1 sini sin 45 2     2   2  1.414 1 sinr sin30 2 2  Speed of light in the prism is v

c 3  108   2.12  108 m/s  1.414

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