CBSE Class 12 Marking Scheme 2022 for Physics Term 2

MARKING SCHEME
CLASS XII
PHYSICS THEORY
TERM II
SESSION 2021 - 22
MM:35 TIME: 2 Hours

ANS 1
As given in the statement antimony is added to pure Si crystal,
then a n -type extrinsic semiconductor would be so obtained, 1
Since antimony(Sb) is a pentavalent impurity. Mark

Energy level diagram of n-type semiconductor

Donar Energy level

1
Mark

ANS 2 No 1/2
mark
Because according to Bohr's model,
. 1/2
En = − and electrons having different energies belong to
mark
different levels having different values of n.

So, their angular momenta will be different, as 1
mark
L = mvr =
OR
(i) The increase in the frequency of incident radiation has no effect on
photoelectric current. This is because of incident photon of 1/2

increased energy cannot eject more than one electron from the mark
metal surface.

1/2
mark

(ii) The kinetic energy of the photoelectron becomes more than the
double of its original energy. As the work function of the metal is 1/2
fixed, so incident photon of higher frequency and hence higher mark
energy will impart more energy to the photoelectrons.

1/2
mark

ANS 3 Photodiodes are used to detect optical signals of different 1/2 mark
intensities by changing current flowing through them.

1/2 mark

I-V Characteristics of a photodiode

Applications of photodiodes:
1. In detection of optical signals.
2. In demodulation of optical signals.
3. In light operated switches.
4. In speed reading of computer punched cards. (1/2) X 2= 1
5. In electronic counters mark

(any two out of these or any other relevant application)

SECTION B
ANS 4 From Bohr’s theory, the frequency f of the radiation emitted when
an electron de – excites from level n to level n is given as
2π mk z e 1 1
f= −
h n n
2 marks
Given n = n − 1 , n = n , derivation of it
2π mk z e (2n − 1)
f=
h (n − 1) n
For large n, 2n − 1 = 2n , n − 1 = n and z = 1
1 mark
Thus, f =
which is same as orbital frequency of electron in n orbit.
v 4π mk e
f= =
2πr n h
ANS 5
A junction diode allows current to pass only when it is forward
biased. So, if an alternating voltage is applied across a diode the 1 mark
current flows only in that part of the cycle when the diode is
forward biased. This property is used to rectify alternating voltages
and the circuit used for this purpose is called a rectifier.

1 mark

Circuit Diagram

1 mark

Working with input and output waveforms
ANS 6 Number of atoms present in 2 g of deuterium = 6 × 10
Number of atoms present in 2.0 Kg of deuterium = 6 × 10 1 mark
Energy released in fusion of 2 deuterium atoms
= 3.27 MeV
Energy released in fusion of 2.0 Kg of deuterium atoms
.
= × 6 × 10 MeV
= 9.81 × 10 MeV 1 mark
= 15.696 × 10 J
Energy consumed by bulb per sec = 100 J
. ×
Time for which bulb will glow = s = 4.97 × 10 𝑦𝑒𝑎𝑟
1 mark
ANS 7 A locus of points, which oscillate in phase is called a wavefront.
OR
A wavefront is defined as a surface of constant phase. 1 mark

1 mark

Diagram
Proof n sin i = n sin r (Derivation) 1 mark
This is the Snell’s law of refraction.
ANS 8 Diagram of Compound Microscope for the final image formed at D:
(a)

𝟏
𝟏
𝟐
marks

(b) m = 30, f = 1.25 cm, f = 5 cm
when image is formed at least distance of distinct vision,
D = 25cm
Angular magnification of eyepiece
D 25 1/2 mark
m = 1+ = 1+ =6
f 5
Total Angular magnification, m = m m ⇒ m = = =5
As the objective lens forms the real image,
m = = −5 ⇒ v = −5u
1/2 mark
using lens equation, u = −1.5 cm, v = −5 × (−1.5)cm = +7.5 cm
Given v = −D = −25 cm, f = +5 cm, u = ?
using again lens equation u =
Thus, object is to be placed at 1.5 cm from the objective and
separation between the two lenses should be 1/2 mark
L = v + Iu l = 11.67 cm

OR
ANS 8
(a) Ray diagram of astronomical telescope when image is formed at 𝟏
𝟏 marks
𝟐
infinity.

(b) (i) In normal adjustment :
Magnifying power. ½ mark
m = f /f = (140/5) = 28

(ii) When the final image is formed at the least distance of 1 mark
distinct vision (25 cm) :

m= 1+ = (28 x 1.2) = 33.6

ANS 9
λ = 2000 Å = (2000 × 10 )m
W = 4.2eV
h = 6.63 × 10 JS
(a) Using Einstein's photoelectric equation 1 mark
K. E. = (6.2 – 4.2) eV = 2.0 eV

The energy of the emitted electrons does not depend upon 1 mark
(b)
intensity of incident light; hence the energy remains unchanged.

(c) For this surface, electrons will not be emitted as the energy of 1 mark
incident light (6.2 eV) is less than the work function (6.5 eV) of the
surface.

ANS 10 Given a = 1.5
Focal length of the given convex lens when it is placed in air is
f = + 20 cm
Refractive index of the given medium with respect to air is
a = 1.25
New focal length of the given convex lens when placed in a
medium is f
= a −1 + ------(A) 1/2 mark
1/2 mark
= m −1 + ------(B)
Dividing (A) by (B), we get
f a −1 (1.5 − 1) 0.5 5
= = = = = 2.5
f m −1 (1.2 − 1) 0.2 2
. 1 mark
f = 2.5f = (2.5 × 20)cm = +50cm as m = = = 1.2
.
New focal length is positive.
The significance of the positive sign of the focal length is that given 1/2 mark
convex lens is still converging in the given medium.
1/2 mark
ANS 11.
(a) Microwaves are suitable for the radar system used in aircraft

navigation. 1 mark
Range of frequency of microwaves is 108 Hz to 1011 Hz.
(b)
If the Earth did not have atmosphere, then there would be absence
of greenhouse effect of the atmosphere. Due to this reason, the 1 mark
temperature of the earth would be lower than what it is now.
(c)
An e.m. wave carries momentum with itself and given by
P = Energy of wave(U)/ Speed of the wave(c) 1 mark
= U/c
when it is incident upon a surface it exerts pressure on it.
OR
ANS. 11 The total intensity at a point where the phase difference is ∅, is
(a)
given by 𝐼 = 𝐼 + 𝐼 + 2 𝐼 𝐼 𝐶𝑂𝑆 ∅. Here 𝐼 and 𝐼 are the
intensities of two individual sources which are equal. 2 marks
When ∅ is 0, I = 4𝐼 .
When ∅ is 90o, I = 0
Thus intensity on the screen varies between 4𝐼 and 0.

Intensity distribution as function of phase angle, when diffraction of
ANS. 11
(b)
light takes place through coherently illuminated single slit

1 mark

ANS 12.
(a) Ans (i) Refraction, Total internal reflection 1 mark

(b) Ans iii) sin-1( )
1 mark
a =
⇒ sin C = ⇒ C = sin

(c) Ans (iv) n1 > n2 1 mark

The refractive index of the core should be greater than the
refractive index of the cladding.

(d)
1 mark
Ans (iv) increases
l = = ,ω = =

μ >μ
(e) Thus C > C
1 mark

Ans (ii) 0 < i < 60° , 1 =
.
Sin C = = 0.8571
.
⇒ C = 59°
Total internal reflection will occur if the angle 𝑖 > 𝑖 ,
i.e., if 𝑖 > 59o or when r < rmax’ where rmax = 90o – 59o = 31o.
Using Snell’s law,
= 1.68
or 𝑆𝑖𝑛 𝑖 = 1.68 × 𝑆𝑖𝑛 𝑟
= 1.68 x sin 31o = 1.68 x 0.5150 = 0.8662

∴ 𝑖 = 60o
Thus all incident rays which make angles in the range 0 < 𝑖 <
60 with the axis of the pipe will suffer total internal
reflections in the pipe.