ISC Class 12 Physics Sample Paper 2022 with Solutions

SECTION A - 7 MARKS

Question 1

(i) What is meant by a wavefront? [1 Mark]

Answer:
Wavefront is defined as the imaginary surface in a medium such that all the medium particles lying on the surface are in the same phase of oscillation. It propagates along the direction of propagation of the wave with the same velocity as that of the wave.

Teacher's Note:
a) Ensure the definition highlights the locus of points vibrating in the same phase.
b) Students often miss mentioning that the wavefront moves along the direction of wave propagation.

(ii) Find the de Broglie wavelength of electrons moving with a speed of \( 7 \times 10^{6} \text{ ms}^{-1} \text{.} \) [1 Mark]

Answer:
De Broglie wavelength \( \lambda = \frac{h}{p} = \frac{h}{mv} \)
Here, \( h = 6.6 \times 10^{-34} \text{ Js} \), \( m = 9.1 \times 10^{-31} \text{ kg} \), \( v = 7 \times 10^{6} \text{ ms}^{-1} \)
\( \lambda = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 7 \times 10^{6}} = 1.03 \times 10^{-10} \text{ m} \).

Teacher's Note:
a) Use the standard formula \( \lambda = \frac{h}{mv} \) for matter waves.
b) Pay close attention to powers of 10 during simplification to avoid calculation errors.

(iii) State how a p-type semiconductor will be obtained from a pure crystal of a semiconductor. [1 Mark]

Answer:
A p-type semiconductor can be obtained by adding a trivalent impurity such as aluminium with a pure crystal of a semiconductor (silicon or germanium).

Teacher's Note:
a) Mentioning the specific type of impurity (trivalent) and giving an example (aluminium, boron, indium) is essential.
b) Do not confuse trivalent impurities with pentavalent impurities which form n-type semiconductors.

(iv) In case of a regular prism, in minimum deviation position, angle made by the refracted ray (inside the prism) with the normal drawn to the refracting surface is: [1 Mark]
(A) \( 90^{\circ} \)
(B) \( 60^{\circ} \)
(C) \( 45^{\circ} \)
(D) \( 30^{\circ} \)

Answer: (D) \( 30^{\circ} \)

In the position of minimum deviation for a regular prism (where refracting angle \( A = 60^{\circ} \)), \( r_{1} = r_{2} = r \). Since \( r_{1} + r_{2} = A \), we have \( 2r = 60^{\circ} \), which gives \( r = 30^{\circ} \).

Teacher's Note:
a) Recall that for a regular prism, the refracting angle is \( 60^{\circ} \).
b) At minimum deviation, the ray passes symmetrically, making equal angles with the refracting surfaces.

(v) In Young's double slit experiment, what is the effect on fringe pattern if the slits are brought closer to each other? [1 Mark]
(A) Fringes disappear.
(B) Fringe width increases.
(C) Fringe width decreases.
(D) Fringe width remains unaltered.

Answer: (B) Fringe width increases.

Fringe width \( \beta = \frac{\lambda D}{d} \). When the slit separation \( d \) decreases, the fringe width \( \beta \) increases.

Teacher's Note:
a) Relate the fringe width formula \( \beta = \frac{\lambda D}{d} \) to see the inverse proportionality between \( \beta \) and \( d \).
b) Students frequently confuse the effect of changing distance between slits versus distance between screen and slits.

(vi) First line of Balmer series (\( \text{H}_{\alpha} \)) in the spectrum of hydrogen is obtained when an electron of hydrogen atom goes from: [1 Mark]
(A) \( 2^{\text{nd}} \) orbit to \( 1^{\text{st}} \) orbit
(B) \( 2^{\text{nd}} \) orbit to \( 3^{\text{rd}} \) orbit
(C) \( 3^{\text{rd}} \) orbit to \( 2^{\text{nd}} \) orbit
(D) \( 3^{\text{rd}} \) orbit to \( 1^{\text{st}} \) orbit

Answer: (C) \( 3^{\text{rd}} \) orbit to \( 2^{\text{nd}} \) orbit

For the Balmer series, transitions terminate at \( n = 2 \). The first line corresponds to the lowest energy transition, which is from \( n = 3 \) to \( n = 2 \).

Teacher's Note:
a) Remember that spectral series are named after their lower energy level (Lyman: \( n=1 \), Balmer: \( n=2 \), Paschen: \( n=3 \)).
b) The first line of any series always involves an electron transition from the next higher consecutive orbit.

(vii) Which of the following graphs correctly represents the variation of maximum kinetic energy (\( E_{k} \)) of photoelectrons with the frequency (\( \nu \)) of the incident radiation? [1 Mark]
(A)
(B)
(C)
(D)

[Figure: Four graphs showing variation of \( E_{k} \) with frequency \( \nu \). Graph (a) passes through origin with positive slope. Graph (b) shows a non-linear rising curve starting from the origin. Graph (c) shows a decreasing exponential curve. Graph (d) is a straight line with a positive slope intersecting the frequency axis at threshold frequency and starting above zero frequency.]

Answer: (D)

According to Einstein's photoelectric equation \( E_{k} = h\nu - \phi \), \( E_{k} \) varies linearly with frequency \( \nu \) with a positive slope \( h \), and intercepts the frequency axis at the threshold frequency where \( E_{k} = 0 \).

Teacher's Note:
a) Einstein's photoelectric equation represents a straight line equation of the form \( y = mx + c \).
b) The graph starts only after the threshold frequency, making graph (d) correct instead of graph (a) which starts at the origin.

SECTION B - 10 MARKS

Question 2

(i) What is meant by Constructive interference? [1 Mark]

Answer:
When two waves meet at a point with phase difference \( \phi = 2m\pi \) (where \( m = 0, 1, 2, \dots \)) or a path difference \( x = m\lambda \) (where \( m = 0, 1, 2, \dots \)), they interfere constructively resulting in maximum intensity.

Teacher's Note:
a) Both phase difference and path difference conditions must be clearly stated for full credit.
b) Mention that constructive interference produces maximum intensity or bright fringes.

(ii) In Young's double slit experiment, what should be the phase difference between the two overlapping waves to obtain \( 5^{\text{th}} \) dark band/fringe on the screen? [1 Mark]

Answer:
For the \( n^{\text{th}} \) dark fringe, the phase difference is given by \( \phi = (2n - 1)\pi \).
For the \( 5^{\text{th}} \) dark fringe (\( n = 5 \)):
\( \phi = (2(5) - 1)\pi = 9\pi \).

Teacher's Note:
a) The condition for dark fringe phase difference is an odd multiple of \( \pi \).
b) Substitute \( n = 5 \) carefully into the formula \( (2n - 1)\pi \).

Question 3

(i) A thin converging lens of focal length \( 5\text{ cm} \) is used as a simple microscope. Calculate its magnifying power when image formed lies at: [2 Marks]
(a) Infinity.
(b) Least distance of distinct vision (\( D = 25\text{ cm} \)).

Answer:
Given focal length \( f = 5\text{ cm} \), \( D = 25\text{ cm} \).
(a) When image is formed at infinity:
\( M = \frac{D}{f} = \frac{25}{5} = 5 \)
(b) When image is formed at the least distance of distinct vision:
\( M = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 1 + 5 = 6 \).

Teacher's Note:
a) Distinguish clearly between the two standard formulae for a simple microscope.
b) Magnifying power is dimensionless, so no units should be attached to the final values.

OR

(ii) A thin converging lens of focal length \( 12\text{ cm} \) is kept in contact with a thin diverging lens of focal length \( 18\text{ cm} \). Calculate the effective/equivalent focal length of the combination. [2 Marks]

Answer:
Given \( f_{1} = +12\text{ cm} \), \( f_{2} = -18\text{ cm} \).
Using the combination formula for lenses in contact:
\( \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \)
\( \frac{1}{f} = \frac{1}{12} + \left(-\frac{1}{18}\right) = \frac{3 - 2}{36} = \frac{1}{36} \)
\( f = 36\text{ cm} \) (converging lens).

Teacher's Note:
a) Proper sign convention must be applied: positive for converging lens and negative for diverging lens.
b) Conclude whether the final combination behaves as a converging or diverging lens based on the sign of \( f \).

Question 4

(i) Define angular dispersion. [1 Mark]

Answer:
The angle between the emergent rays of any two extreme colours (such as violet and red) after passing through a prism is called angular dispersion between those colours.

Teacher's Note:
a) Angular dispersion is mathematically represented as \( \delta_{v} - \delta_{r} \) or \( (\mu_{v} - \mu_{r})A \).
b) It measures the spread of colours produced by a prism.

(ii) State any one difference between a primary rainbow and a secondary rainbow. [1 Mark]

Answer:
A primary rainbow is formed due to a single total internal reflection and two refractions inside water droplets, whereas a secondary rainbow is formed after double total internal reflections and two refractions.

Teacher's Note:
a) Mentioning the number of total internal reflections (one vs two) is the key differentiator.
b) Primary rainbows are brighter with red on the outer edge, while secondary rainbows are fainter with violet on the outer edge.

Question 5

Explain the following terms: [2 Marks]
(i) Intrinsic semiconductor.
(ii) Extrinsic semiconductor.

Answer:
(i) Intrinsic semiconductor: An intrinsic semiconductor is a pure semiconductor which is free from any chemical impurity. Pure germanium and pure silicon are examples of intrinsic semiconductors.
(ii) Extrinsic semiconductor: When a pure semiconductor crystal is doped with a small amount of suitable trivalent or pentavalent impurity to improve its electrical conductivity, the resulting material is known as an extrinsic semiconductor.

Teacher's Note:
a) Emphasize that intrinsic refers to pure form and extrinsic refers to doped form.
b) Examples like silicon and germanium add clarity to the definition.

Question 6

(i) Calculate maximum kinetic energy of photoelectrons emitted by a metal (work function = \( 1.5\text{ eV} \)) when it is illuminated with light of wavelength \( 198\text{ nm} \). [2 Marks]

Answer:
According to Einstein's photoelectric equation:
\( E_{k} = h\nu - \phi = \frac{hc}{\lambda} - \phi \)
Given: \( h = 6.6 \times 10^{-34}\text{ Js} \), \( c = 3.0 \times 10^{8}\text{ ms}^{-1} \), \( \lambda = 198 \times 10^{-9}\text{ m} \), \( \phi = 1.5\text{ eV} = 1.5 \times 1.6 \times 10^{-19}\text{ J} = 2.4 \times 10^{-19}\text{ J} \)
\( \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{198 \times 10^{-9}} = \frac{19.8 \times 10^{-26}}{198 \times 10^{-9}} = 10 \times 10^{-19}\text{ J} \)
\( E_{k} = 10 \times 10^{-19} - 2.4 \times 10^{-19} = 7.6 \times 10^{-19}\text{ J} \).

Teacher's Note:
a) Convert the work function from electron volts (eV) to Joules (J) before subtraction.
b) Keep track of powers of ten carefully during division and subtraction steps.

OR

(ii) Calculate the minimum amount of energy which a gamma ray photon should have for the production of an electron and a positron pair. [2 Marks]

Answer:
The rest mass energy of an electron or positron is:
\( E_{0} = m_{0}c^{2} = 9.1 \times 10^{-31}\text{ kg} \times (3.0 \times 10^{8}\text{ ms}^{-1})^{2} = 8.19 \times 10^{-14}\text{ J} \)
Converting to MeV:
\( E_{0} = \frac{8.19 \times 10^{-14}}{1.6 \times 10^{-13}} \approx 0.51\text{ MeV} \)
For pair production (electron and positron pair), the minimum energy of the gamma ray photon must be equal to the total rest mass energy of both particles:
\( E = 2 \times 0.51\text{ MeV} = 1.02\text{ MeV} \).

Teacher's Note:
a) Pair production requires energy at least equal to the combined rest mass energy of the particle-antiparticle pair.
b) Memorizing the rest mass energy of an electron as \( 0.51\text{ MeV} \) saves valuable time in examinations.

SECTION C - 18 MARKS

Question 7

(i) In a single slit diffraction experiment, how does the angular width of the central maxima change when: [3 Marks]
(a) screen is moved away from the plane of the slit?
(b) width of the slit is increased?
(c) light of larger wavelength is used?

Answer:
The angular width of the central maximum is given by \( \theta = \frac{2\lambda}{e} \), where \( \lambda \) is the wavelength and \( e \) is the slit width.
(a) Angular width remains unchanged when the screen is moved away, because it does not depend on the distance of the screen from the slit.
(b) Angular width decreases when the width of the slit (\( e \)) is increased, due to inverse proportionality.
(c) Angular width increases if light of larger wavelength (\( \lambda \)) is used, due to direct proportionality.

Teacher's Note:
a) Emphasize that angular width does not depend on the screen distance, whereas linear width does.
b) Clearly state the formula \( \theta = \frac{2\lambda}{e} \) to justify each answer.

OR

(ii) Using Huygen's wave theory of light, prove Snell's law of refraction of light. [3 Marks]

Answer:
Consider a plane wavefront \( AB \) incident on a surface \( PQ \) separating two media 1 and 2. Medium 1 is rarer with refractive index \( n_{1} \) and speed \( c_{1} \), while medium 2 is denser with refractive index \( n_{2} \) and speed \( c_{2} \).
Let \( i \) be the angle of incidence and \( r \) be the angle of refraction.
From triangles \( \Delta BAD \) and \( \Delta ACD \):
\( \sin i = \frac{BD}{AD} = \frac{c_{1}t}{AD} \)
\( \sin r = \frac{AC}{AD} = \frac{c_{2}t}{AD} \)
Dividing \( \sin i \) by \( \sin r \):
\( \frac{\sin i}{\sin r} = \frac{c_{1}t / AD}{c_{2}t / AD} = \frac{c_{1}}{c_{2}} = \text{constant} \)
Since absolute refractive index \( n = \frac{c}{v} \), we get \( \frac{c_{1}}{c_{2}} = \frac{n_{2}}{n_{1}} = {}^{1}n_{2} \), which proves Snell's law.

Teacher's Note:
a) Draw a clean, well-labelled diagram showing wavefronts, incident rays, and refracted rays.
b) Clearly define the time \( t \) taken by the wavefront to travel from one point to another.

Question 8

Draw a ray diagram of a refracting astronomical telescope when final image is formed at infinity. Also write the expression for its angular magnification (magnifying power). [3 Marks]

Answer:
[Figure: Ray diagram of a refracting astronomical telescope with objective lens of focal length \( f_{o} \) and eye lens of focal length \( f_{e} \). Parallel rays from object at infinity enter the objective lens, form a real inverted image at the common principal focus (\( F_{o}, F_{e} \)), and the eye lens forms the final virtual image at infinity with parallel emergent rays subtending angle \( \beta \) at the eye.]
The expression for its angular magnification (magnifying power) when the final image is formed at infinity is:
\( M = -\frac{f_{o}}{f_{e}} \)
When the image is formed at the least distance of distinct vision \( D \):
\( M = -\frac{f_{o}}{f_{e}}\left(1 + \frac{f_{e}}{D}\right) \).

Teacher's Note:
a) Ensure arrows on light rays and proper focal point alignments (\( F_{o} \) coinciding with \( F_{e} \)) are shown in the diagram.
b) The negative sign in the magnification formula indicates that the final image formed is inverted with respect to the distant object.

Question 9

Radius of curvature of an equi - convex lens of glass (\( n = 1.5 \)) is \( 30\text{ cm} \). Find its focal length. An object of height \( 5.0\text{ cm} \) is placed at a distance of \( 60\text{ cm} \) from the optical centre of the lens. Find the position and the height of the image formed. [3 Marks]

Answer:
Given: \( n = 1.5 \), \( R_{1} = +30\text{ cm} \), \( R_{2} = -30\text{ cm} \), \( h_{o} = 5.0\text{ cm} \), \( u = -60\text{ cm} \).
Using lens maker's formula:
\( \frac{1}{f} = (n - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) \)
\( \frac{1}{f} = (1.5 - 1)\left(\frac{1}{30} - \frac{1}{-30}\right) = 0.5 \times \left(\frac{2}{30}\right) = \frac{1}{30} \)
\( f = +30\text{ cm} \).
Using lens formula to find image position \( v \):
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies \frac{1}{30} = \frac{1}{v} - \frac{1}{-60} \)
\( \frac{1}{v} = \frac{1}{30} - \frac{1}{60} = \frac{2 - 1}{60} = \frac{1}{60} \implies v = +60\text{ cm} \).
Magnification \( m = \frac{v}{u} = \frac{60}{-60} = -1 \).
Also, \( m = \frac{h_{i}}{h_{o}} \implies -1 = \frac{h_{i}}{5.0} \implies h_{i} = -5.0\text{ cm} \).
The image is formed at a distance of \( 60\text{ cm} \) on the other side of the lens, and its height is \( -5.0\text{ cm} \) (inverted and of the same size).

Teacher's Note:
a) Follow sign conventions strictly: \( R_{1} \) is positive and \( R_{2} \) is negative for an equi-convex lens.
b) The negative sign in height indicates that the image is real and inverted.

Question 10

On the basis of Bohr's theory, derive an expression for the radius of the \( n^{\text{th}} \) orbit of an electron of hydrogen atom. [3 Marks]

Answer:
Let \( e, m, v \) be the charge, mass, and velocity of the electron, and \( r \) be the radius of the orbit. The positive charge on the nucleus of atomic number \( Z \) is \( Ze \).
The centripetal force is provided by the electrostatic force of attraction:
\( \frac{mv^{2}}{r} = \frac{1}{4\pi\varepsilon_{0}}\frac{Ze^{2}}{r^{2}} \implies mv^{2} = \frac{Ze^{2}}{4\pi\varepsilon_{0}r} \quad \text{--- (i)} \)
By Bohr's quantization condition for angular momentum:
\( mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr} \quad \text{--- (ii)} \)
Squaring equation (ii) and substituting \( v^{2} \) into equation (i):
\( m\left(\frac{n^{2}h^{2}}{4\pi^{2}m^{2}r^{2}}\right) = \frac{Ze^{2}}{4\pi\varepsilon_{0}r} \implies \frac{n^{2}h^{2}}{4\pi^{2}mr} = \frac{Ze^{2}}{4\pi\varepsilon_{0}} \)
Solving for radius \( r \):
\( r = \frac{n^{2}h^{2}\varepsilon_{0}}{\pi Ze^{2}} \)
For a hydrogen atom (\( Z = 1 \)):
\( r = \frac{n^{2}h^{2}\varepsilon_{0}}{\pi me^{2}} \).

Teacher's Note:
a) Clearly state both electrostatic force balance and Bohr's angular momentum quantization postulate.
b) Show all algebraic steps clearly while substituting and isolating \( r \).

Question 11

Read the passage given below and answer the questions that follow.
Mr. Ravi had been living in a small town with his family for many years. He decided to shift to a city when he learnt that a nuclear power plant would be built in his town. His son Prakash, who was a science teacher, did not agree with his father's decision. Prakash explained to his father how mass defect during nuclear fission released energy. He assured his father that the authorities were taking all the possible safety measures to avoid any kind of nuclear mishap. Mr. Ravi understood the scientific explanation given by his son and decided not to shift to another place.

(i) What is the cause of energy generation of a nuclear reactor during nuclear fission? [1 Mark]

Answer:
The cause of energy generation during nuclear fission is the conversion of mass defect into energy, according to Einstein's mass-energy equivalence relation \( \Delta E = (\Delta m)c^{2} \).

Teacher's Note:
a) Mention mass defect and Einstein's equation explicitly.
b) This is a value-based application question derived from nuclear physics.

(ii) Write one nuclear reaction for nuclear fission that takes place in a nuclear reactor. [1 Mark]

Answer:
\( {}_{92}^{235}\text{U} + {}_{0}^{1}\text{n} \to {}_{56}^{144}\text{Ba} + {}_{36}^{89}\text{Kr} + 3{}_{0}^{1}\text{n} + \text{Energy} \).

Teacher's Note:
a) Ensure mass numbers and atomic numbers balance on both sides of the nuclear reaction equation.
b) Including three neutrons and the release of energy completes the equation.

(iii) Give the formula for the energy generation that takes place in the nuclear reactor. [1 Mark]

Answer:
\( \Delta E = (\Delta m)c^{2} \), where \( \Delta E \) is the energy produced, \( \Delta m \) is the mass defect, and \( c \) is the speed of light in vacuum.

Teacher's Note:
a) State the symbols clearly with their standard meanings.
b) This is the standard mathematical form of mass-energy equivalence.

Question 12

(i) Answer the following questions.
(a) Draw the circuit diagram of a full wave rectifier.
(b) Draw labelled graphs showing the input and output voltages. [3 Marks]

Answer:
(a) [Figure: Circuit diagram of a full wave rectifier showing a transformer with primary \( P_{1}P_{2} \) and secondary \( S_{1}S_{2} \) with center tap \( T \), two diodes \( D_{1} \) and \( D_{2} \), and a load resistance \( R_{L} \).]
(b) [Figure: Labelled graphs showing alternating sinusoidal input voltage wave with period \( T \) and the unidirectional pulsating output voltage wave showing successive positive half-cycles for both halves of input.]

Teacher's Note:
a) The center tap transformer connection and the polarity of diodes \( D_{1} \) and \( D_{2} \) must be clearly shown.
b) Input should be a complete sine wave while output should contain only positive half-cycles.

OR

(ii) With reference to Semiconductor Physics, [3 Marks]
(a) Name the diode that emits spontaneous radiation when forward biased.
(b) Draw a labelled energy band diagram for a semiconductor.
(c) Name the process that causes depletion region in a p-n junction.

Answer:
(a) Light Emitting Diode (LED).
(b) [Figure: Labelled energy band diagram for a semiconductor showing valence band filled with electrons, conduction band empty at absolute zero separated by a small forbidden energy gap of approximately \( 1\text{ eV} \).]
(c) The diffusion of electrons from the n-region to the p-region and holes from the p-region to the n-region (diffusion of charge carriers) causes the depletion region in a p-n junction.

Teacher's Note:
a) LED works on minority carrier injection under forward bias resulting in spontaneous emission.
b) Mentioning diffusion of charge carriers is vital for part (c).