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CBSE Class 12 Physics Previous Year Question Paper 2022

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CBSE Class 12 Previous Year’s Question Paper for Physics is a valuable resource for students By using the solution provided, students can also learn the appropriate method to answer the questions in the board examination.

Using the CBSE previous year’s paper for Physics, students can gain more confidence in attempting the board examination, as they get to know the areas where they are lacking in their preparation. Also, the mistakes made while attempting this question paper as a mock examination can be rectified to avoid the repetition of such errors in the board examinations.

CBSE Class 12 2022 Physics Paper Solutions

PHYSICS (Subject Code–042)  


Q1. Draw Energy Band diagrams of p-type and n-type semiconductors at temperature T > 0 K, depicting the donor and acceptor energy levels. Mention the significance of these levels. (2 marks)

Answer 1:

The Energy Band diagrams of p-type and n-type semiconductors at temperature T > 0 K are given below,

Energy Band diagrams of p-type and n-type semiconductors at temperature T greater than 0 K

Significance of n-type and p-type semiconductor are,

The n-type semiconductors have small energy gap between donor level and conduction band so that they can be covered by thermally excited electrons easily.

The p- type semiconductors have small energy gap between acceptor level and valence  band so that they can be covered by thermally excited electrons easily. 

In p-type semiconductors, there is a small amount of energy that can cause an electron from its valence band to jump to an acceptor energy level. This creates “holes” in the electrons’ outer shells, making them the majority charge carriers in these particular types of semiconductors.

Electrons are the majority charge carriers in the conduction band. These electrons come from donor impurity atoms and require very little energy to move into the conduction band.

Learn more about Semiconductors

Q2. (a) Draw the graph showing the variation of the number (N) of scattered alpha particles with scattering angle (θ) in Geiger- Marsden experiment. Infer two conclusions from the graph. (2 marks)

(b) Plot suitable graphs to show the variation of photoelectric current with the collector plate potential for the incident radiation of

(i) Same intensity but different frequencies v1,v2,v (v1<v2<v3)

(ii) Same frequency but different intensities I1,I2,I3 ( I1<I2<I3)

Answer 2. (a)

The graph showing the variation of the number (N) of scattered alpha particles with scattering angle (θ) in the Geiger- Marsden experiment is given below,

Geiger- Marsden Experiment


Conclusion are as follows

  • Most of the alpha particles pass through the gold foil without any deviation. 
  • Few alpha particles gets deflected by 90° or more. 
  • About 0.14% of the alpha particles gets reflected by larger angles.
  • A very few alpha particles goes back retracing their path.

NOTE: Write any two conclusions


Answer 2.(b)

i) Same Intensity and Different Frequency

The graph showing the variation of photoelectric current with the collector plate potential for the incident radiation of the same intensity but different frequencies v1, v2, v (v1<v2<v3) is given below,

Graph of Photoelectric current with the collector plate potential with same intensity and different frequency


ii) Same Frequency but Different Intensity

The graph showing the variation of photoelectric current with the collector plate potential for the incident radiation of the Same frequency but different intensities I1, I2, I3 ( I1<I2<I3) is given below,

Same Frequency but Different Intensity


Check Photoelectric Effect

Q3. Write the characteristics of a p-n junction that make it suitable for rectification. (2 marks)

Answer 3:

  • Unidirectional conductivity: A diode is an electronic component with unidirectional conductivity, meaning that it allows the current to flow in only one direction. Because of this property, diodes are often used in rectification circuits to convert alternating current (AC) into direct current (DC). This further means that the diode conducts only when it is forward biased and does not conduct when it is reverse biased.
  • Non-linear voltage-current relationship: A diode is an electrical component with two leads that has a low voltage drop when current passes through it in the forward direction. This makes diodes highly efficient components in circuits, as they suffer less power loss due to their low forward voltage drop.
  • High reverse breakdown voltage: When a voltage is applied in the reverse direction to a p-n junction, it can reach a point where the junction breaks down and allows current to flow in the reverse direction.


Q4. Define the term- Distance of closest approach. How will it be affected, for an α – particle, if the kinetic energy of the particle is doubled? (3 marks)

Answer 4:

Distance of closest approach is defined as the minimum distance up to which an alpha particle can travel along the central line of the nucleus before it rebounds or before it bounces back. To be specific, when two charged particles are traveling towards one another, the distance between them is distance of closest approach.        

​As the formula says,

r=2 Ze2/4πϵ0K

r is distance of closest  approach
Z is the atomic number of the nucleus
K is the kinetic energy

Now, from the above formula it is clear that,

Radius is inversely proportional to the kinetic energy i.e.

r ∝ 1/K


 K is doubled then r is halved 

Q5. A point source in the air is kept 24 cm in front of a concave spherical glass surface (aμg  = 1.5) and a radius of curvature of 60 cm. Find the nature of the image formed and its distance from the point source. (3 marks)

Answer 5:

To solve the following question, we will use mirror formula for spherical mirrors, which relates the object distance (u), image distance (v), and focal length (f) of the mirror.

Relation for concave spherical surface: is given below:

μ1/(-u) + μ2/v = (μ2 – μ1)/R

1/(-u) + μ/v = (μ-1)/R

1/-(-24) + 1.5/v = (1.5 – 1)/(-60)

1/24 + 3/2v = -1/120

3/2v = -1/120 – 1/24

        = -1/120 – 5/120

3/2v = -6/120

v = -30 cm


v = -30 cm

Distance of image from point source = -24 -(-30) = 6 cm 

Nature of image = Virtual image

Learn about, Image Formation by Spherical Mirror

Q6. Calculate the energy released in MeV in the following reaction: 

12H + 13H -> 24He + n + Energy


m(12H) = 2.014102 u

m(13H) = 3.016049 u

m(24H) = 4.002603 u  

mn = 1.008665 u  (3 marks)

Answer 6:

The given equation is:

12H + 13H -> 24He + n + Energy

We know by the formula:

Mass defect = Mass of Reactants – Mass of Products

Mass defect = (2.014102+3.016049)-(4.002603+1.008665)

                    = 5.030151-5.011268

                    = 0.018883u

Energy released = ∆m × 931.5 MeV
                         = 0.018883 × 931.5 MeV
                         =17.58 MeV 

Q7. Explain with the help of a suitable diagram, the phenomenon on which an optical fibre works. Mention any two uses of optical fibres. (3 marks)

Answer 7:

An optical fibre works on the principle of Total Internal Reflection (TIR).

The ray diagram explaining the TIR phenomenon is disussed in the image below,

Total Internal Reflection (TIR)


TIR is an optical phenomenon where a ray of light pass through a medium such as water or glass from the surrounding surfaces back into the medium. 

In TIR Angle of Incidence is greater than Critical Angle.

Uses of optical fibres are as follows:

  • Medical and Optical Examination: It is used in many areas due to its unique properties, such as telecommunication, bandwidth transmission, cable television and medical applications like endoscopy, laser surgeries and microscopy. Optical fibre has also been found to be essential for biomedical research.
  • Transmission and Reception of Signals: It is used to transmit digital data at high speeds. It has largely replaced copper wiring for the delivery of digital services such as HDTV, CATV and video-on-demand due to its lower cost and higher bandwidth. This revolutionary technology has enabled faster internet connection speeds for consumers around the world.
  • Industrial Use: It is used in a range of applications such as medical imaging and monitoring, automotive lighting systems, research and testing purposes, and transmitting data at lightning speed for communication networks. Its flexibility makes it ideal for use in environments where typical cables would not do the job, such as cars or hard to-reach locations.

NOTE: Write any two uses.

Q8. (a) A parallel beam of light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. If the resulting diffraction pattern is observed on a screen 1 m away, find the distance of  

(i) First minimum,

(ii) Second maximum, from the central maximum.  (3 marks)


b) A thin biconvex lens of the radius of curvature R made of material of refractive index μ1 is kept coaxially, in contact with a biconcave lens of the same radius of curvature and refractive index μ2(>μ1).  


(i) Ratio of their powers

(ii) Power of the combination and its nature. 

Answer 8:


(i) By formula of fringes, we know that first fringes is given by the formula,

The fringe width is the distance between two bright or dark fringes of light observed from Young’s double-slit experiment and is represented by the equation β = λD/d. 

 β = λD/d

Here, D=600 nm , a =0.2 mm

  = (600×10-9 ×1)/0.2 × 10-3

  = 3 × 10-3 m = 3 mm

(ii) The following formula is used for a particular condition that is “maxima” or “bright fringe”

y = (n + 1/2) λD/a

n = 2

y = (2 + 1/2) λD/a

y = (5/2) λD/a

y = (5/2) × (600×10-9 × 1)/(0.2 × 10-3)

   =7.5 × 10-3= 7.5 mm



(i) A thin biconvex lens of the radius of curvature R has power=P1 and another biconcave lens of the same radius of curvature has power=P2

Calculating Ratios,

The formula P = (μ – 1) (1/R1 – 1/R2) relates the power (P) of a lens to its refractive index (μ) and the radii of curvature (R1 and R2) of its two surfaces.

From P =( μ − 1) (1/R1 – 1/R2)

P1 =  PCONVEX = ( μ − 1) (1/R1 – (-1/R2))

                       =( μ − 1) (2/R)

P2 =  PCONCAVE = ( μ − 1) (-1/R1 – 1/R2)

                        =-( μ − 1) (2/R)

P1/P2 = ( μ1 − 1)/ -( μ2 − 1) 

          = ( μ1 − 1)/-( μ2 − 1)

          = ( μ1 − 1)/(1 – μ2)

(ii) Calculation for Combination of Powers,

The below equation is used for calculating the combined power of both the lenses when they are kept coaxially.

P = P1 + P2  

   =( μ1 − 1) (2/R) + (-( μ2 − 1))(2/R)

P  = 2(μ1 – μ2)/R

As  μ2 >  μ1, P is negative

So, Nature is diverging

Q9. Photoelectrons are emitted from a metal surface when illuminated with UV light of wavelength 330 nm. The minimum amount of energy required to emit electrons from the surface is 3.5 x 10-19 J. Calculate  (3 marks)

(i) Energy of the incident radiation

(ii) Kinetic energy of the photoelectron

Answer 9:

(i) Here, wavelength = 330 nm

Minimum amount of energy required =3.5 x 10-19J.

We know, the formula to calculate, Energy of incident radiation is given by:

E = hν =  hc/λ 

The formula E = hν = hc/λ relates the energy (E) of a photon of electromagnetic radiation to its frequency (ν) or wavelength (λ), where h is Planck’s constant and c is the speed of light.

=6.63 × 10-34  ×  3 × 108

=330  × 10-9

=6.027 × 10-19  Joules

(ii) To calculate kinetic energy we have:

Kinetic energy of photoelectron

K.E. = E – φ0

The formula K.E. = E – φ0 relates the kinetic energy (K.E.) of an electron in a material to its total energy (E) and the work function (φ0) of the material.

= (6.027 × 10-19 – 3.5 × 10-19) Joules

= 2.527 × 10-19 Joules

Q 10: State the working principle of an LED. Write any two important advantages and two disadvantages of LED. (3 marks)

Answer 10:

A diode is a two-terminal semiconductor device. It has two regions, the n-region and the p-region, and when forward biased, sends electrons from the n-region to the p-region, and holes from the p-region to the n-region. At this junction where the carriers recombine energy is released in the form of photons.

Advantages of LEDs

  • Low Operational Voltage
  • Less Power Consumption
  • Fast Action
  • Long Life and Ruggedness

Disadvantages of LEDs

  • High Cost
  • Can get damaged due to Overheating
  • Excess of Voltage or Current can damage LED

NOTE: Write any two advantages and disadvantages.

Q11. (a) (i) Monochromatic light is incident on a surface separating two media. The frequency of the light after refraction remains  
unaffected but its wavelength changes. Why? (3 marks)

(ii) The frequency of electromagnetic radiation is 1.0 x 1011 Hz  Identify the radiation and mention its two uses.  

(b) (i) Trace the path of a ray of light PQ which is incident at an angle i on one face of a glass prism of angle A. It then emerges out from the other face at an angle e. Use the ray diagram to prove that the angle through which the ray deviates is given by ∠δ = ∠i + ∠e – ∠A.

(ii) What will be the minimum value of δ if the ray passes symmetrically through the prism?  

Answer 11:


(i) Refraction is a phenomenon that occurs when light is bent or ‘refracted’ due to its interaction with the atoms of matter. This bending of light occurs because the atoms act as oscillators and take up the frequency of the light. When this takes place, the wavelength of the light changes even if its frequency remains constant.

As a result, when light passes through a material it is bent at an angle, creating refraction and changing the direction of travel – this also ensures that the frequency of refracted light is equal to that of incident light.

(ii) The waves can be Infrared/ Microwaves/ Radio waves

Uses of Infrared rays

  • Remote Control Devices
  • Green House Effect
  • Photography in foggy condition
  • To reveal secret writings
  • Infrared lamps

Uses of Microwaves

  • Radar System
  • Geostationary satellite
  • Microwave ovens

Uses of Radiowaves

  • TV transmission
  • Radio broadcast
  • Mobile transmission

NOTE: Write two uses of each wave.



Q12. The principle of superposition is used to understand the phenomenon of interference of light waves. The principle states that at a particular point, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each wave. Light waves from two coherent sources produce an interference pattern. Thomas Young devised a way to obtain two coherent sources using two identical pinholes (S1 and  S2) illuminated by a single monochromatic pinhole source S. Using these sources in his experiment known as Young’s double slit experiment, Young studied the interference pattern. The pattern consists of alternate bright and dark fringes. The distance between two successive bright or dark fringes depends on the distance between S1 and S2, the distance of the screen from the plane of S1S2 and the wavelength of light used.  (5 marks)

I. Consider the following waves  

(i) y1 = a Sinωt

(ii) y2 = a Sin2ωt

(iii) y3 = a Sin (2ωt + ϕ)

(iv) y4 = a Sin (4ωt + π/2) 

Which pair of waves coming from two sources S1 and S2 will produce interference?  

(A) (i) and (ii)      (B) (ii) and (iii)  
(C) (iii) and (iv)    (D) (iv) and (i)  

Answer 12:

The answer to the question is 12 I is an option (B)

y2 and y3 both waves have similar amplitude and therefore these waves produce interference.

II. Two light waves of the same intensity I0 each, having a path difference of λ/4, emanating from two coherent sources, meet at a  point. The resultant intensity at the point will be  

(A) Zero      (B) I0

(C) 2 I0        (D) 4 I0

Answer 12:

The answer to the question is 12 II is an option (C)

We know by the formula, φ=2π/λ × Δx. Here, Δx= λ/4

So, φ=2π/λ × λ/4= π/2.


Iresultant = 4 I0 cos2 φ/2

Iresultant = 4 I0 cos2 π/4 ( φ=π/2)

            =4 I0 × 1/2 =2 I0

III. Vandana performs Young’s double slit experiment by using orange,  green and red lights successively. If the fringe widths measured in the three cases are ω1, ω2 and ω3 respectively, then which of the following is correct?  

(A) ω2 > ω1 >  ω3

(B) ω1 > ω2  >  ω3

(C) ω2 > ω3  >  ω1

(D) ω3 > ω1 >  ω2

Answer 12: 

The answer to question 12 III is option (D)

We know, (fringe width)β=λD/d


λred > λorange > λgreen

So, ω3 > ω1 >  ω2        (as β is directly proportional to λ)

IV. In Young’s double slit experiment, the slit separation is 0.8 mm and the interference pattern is obtained on a screen kept 50 cm from the plane of the slits S1 and S2. If the first bright fringe is formed 0.4  mm from the central maximum, the wavelength of light used is  

(A) 480 nm   (C) 640 nm  
(B) 560 nm   (D) 680 nm 

Answer 12:

The answer to the question is 12 IV is option (C)

y= λD/d

Hence, λ= dy/D 

λ = (0.08 × 0.04)/500

   = 640 nm

V. Consider the effect on the angular separation of the fringes in Young’s double slit experiment due to the following operations: 

(i) The screen is moved away from the plane of the slits

(ii) The separation between the two slits is increased till fringes are observed.  

Which of the following options is correct?  

(A) It remains constant in both cases. 
(B) It decreases in both cases. 
(C) It remains constant in (i) but decreases in (ii). 
(D) It decreases in (i) but remains constant in (ii). 

Answer 12:

The answer to the question is 12 V is an option (C)



So, it remains constant in condition (i) but as the distance increases, angular separation decreases as they are inversely propotional.

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Last Updated : 24 Mar, 2023
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