SECTION B

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

where 
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

Thus,

 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 (_aμ_g  = 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

Hence, 

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

Given: 

m(₁²H) = 2.014102 u

m(₁³H) = 3.016049 u

m(₂⁴H) = 4.002603 u  

m_n = 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,

 

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)

OR

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).  

Find  

(i) Ratio of their powers

(ii) Power of the combination and its nature. 

Answer 8:

(a) 

(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

                                                                                                                   OR

(b) 

(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 =  P_CONVEX = ( μ − 1) (1/R1 – (-1/R2))

                       =( μ − 1) (2/R)

P₂ =  P_CONCAVE = ( 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 >  μ_1, 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.  
                                                                                                       OR

(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:

(a) 

(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.

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.