NEET Previous Year Questions (2014-2024): Wave Optics | Physics Class 12 PDF Download

2024

Q1: If the monochromatic source in Young's double slit experiment is replaced by white light, then   (NEET 2024)
(a) Interference pattern will disappear
(b) There will be a central dark fringe surrounded by a few coloured fringes
(c) There will be a central bright white fringe surrounded by a few coloured fringes
(d) All bright fringes will be of equal width 
Ans: (c)
In Young’s double-slit experiment, if a monochromatic source is replaced by white light, which contains multiple wavelengths, the patterns observed on the screen will differ significantly.
First, remember that the pattern formed in the double-slit experiment consists of both bright and dark fringes due to constructive and destructive interference, respectively. The position and intensity of these fringes depend on the wavelength of the light used. For monochromatic light (light of a single wavelength), the interference pattern is stable, with bright and dark fringes evenly spaced. Each fringe is uniformly bright or dark.
When using white light, which is a combination of various wavelengths of light, each color, or each wavelength, forms its own interference pattern with slightly different fringe spacing. This occurs because the separation between fringes Δy is given by the formula:
Δy = λL / d
Where:
λ  is the wavelength of light
L is the distance from the slits to the screen
d is the separation between the slits
Since different wavelengths have different values for λ, each color’s fringes will be at slightly different positions. The result is that near the center of the pattern, where there is the least path difference, all wavelengths constructively interfere to form a bright white central fringe. However, moving away from the center, the fringes start to show different colors as the path difference between the light from the two slits increases. This causes a dispersion of colors with different orders of fringes dominated by different colors. The constructive and destructive interference patterns of different wavelengths slightly offset one another.
Thus:

• Option A - "Interference pattern will disappear" is incorrect because the interference pattern does not disappear but changes due to the dispersion of colors.
• Option B - "There will be a central dark fringe surrounded by a few coloured fringes" is incorrect as the central fringe in the presence of white light is bright, not dark.
• Option C - "There will be a central bright white fringe surrounded by a few coloured fringes" is correct. This is because all the wavelengths interfere constructively at the center, creating a bright white fringe, succeeded by colored fringes due to the varying interference conditions for each wavelength.
• Option D - "All bright fringes will be of equal width" is incorrect. The width and spacing of the fringes vary by wavelength, so the interference pattern will not have uniform fringe widths.

Therefore, the correct answer is Option C: "There will be a central bright white fringe surrounded by a few coloured fringes."

Q2: An unpolarised light beam strikes a glass surface at Brewster's angle. Then   (NEET 2024)
(a) The reflected light will be partially polarised.
(b) The refracted light will be completely polarised.
(c) Both the reflected and refracted light will be completely polarised.
(d) The reflected light will be completely polarised but the refracted light will be partially polarised.
Ans: (d)According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.

Q3: In Young's double slit experiment, if the wavelength of light used is increased (say from violet to red), then the:   (NEET 2024)
(a) fringe width decreases.
(b) fringe width increases.
(c) central bright fringe becomes dark.
(d) fringe width remains unaltered.
Ans: (b)
In Young's double slit experiment, the fringe width (β) is given by the formula:
β = λL / d
Where:

• λ is the wavelength of light,
• L is the distance between the slits and the screen,
• d is the distance between the slits.

If the wavelength of light is increased (say from violet to red), then λ increases. Since the fringe width is directly proportional to the wavelength, the fringe width will increase as well.
Correct Answer: (b) fringe width increases.

Q4: If an unpolarised light is incident on a plane surface of refractive index √3 at Brewster's angle, then the angle of refraction is:   (NEET 2024)
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Ans: (b)
When unpolarised light is incident on a plane surface at Brewster's angle, the angle of refraction can be determined using Brewster's Law, which states that at Brewster's angle (θᵦ), the angle of incidence (θᵦ) and the angle of refraction (θᵣ) are related by the refractive index (n) of the medium.
Brewster's angle is given by the formula:
tan(θᵦ) = n₂ / n₁
Where:

• n₁ is the refractive index of the medium from which the light is coming (usually air, where n₁ ≈ 1),
• n₂ is the refractive index of the medium the light is entering (given as √3).

Now, applying Brewster's law:
tan(θᵦ) = √3
So,
θᵦ = tan⁻¹(√3) = 60°
At Brewster's angle, the light is partially reflected and refracted. The angle of refraction (θᵣ) is related to the angle of incidence (θᵦ) using Snell's law:
n₁ * sin(θᵦ) = n₂ * sin(θᵣ)
Substituting the values:
sin(60°) = √3 * sin(θᵣ)
Since sin(60°) = √3 / 2, we solve for θᵣ:
(1) * (√3 / 2) = (√3) * sin(θᵣ)
sin(θᵣ) = 1 / 2
θᵣ = 30°
Correct Answer: (b) 30°

Q5: Two slits in Young's double slit experiment are 1.5 mm apart, and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is 600 × 10⁻⁹ m, then the fringe separation is:        (NEET 2024)
(a) 4 × 10⁻⁵ m
(b) 9 × 10⁻⁸ m
(c) 4 × 10⁻⁷ m
(d) 4 × 10⁻⁴ m
Ans: (d)
The formula for fringe separation in Young's double-slit experiment is:
β = λL / d
Where:

• λ is the wavelength of light,
• L is the distance from the slits to the screen,
• d is the distance between the slits.

Given:

• d = 1.5 mm = 1.5 × 10⁻³ m (distance between the slits),
• L = 1 m (distance from the slits to the screen),
• λ = 600 × 10⁻⁹ m (wavelength of the light).

Substituting the values into the formula:

β = (600 × 10⁻⁹ m) × (1 m) / (1.5 × 10⁻³ m)
β = (600 × 10⁻⁹) / (1.5 × 10⁻³)
β = 4 × 10⁻⁴ m
Correct Answer:  (d) 4 × 10⁻⁴ m

Q6: An interference pattern can be observed due to the superposition of more than one of the following waves:     (NEET 2024)
(A) y = a sin(ωt)
(B) y = a sin(2ωt)
(C) y = a sin(ωt − ϕ)
(D) y = a sin(3ωt)
Identify the waves from the options given below:
(a) (B) and (C) only
(b) (B) and (D) only
(c) (A) and (C) only
(d) (A) and (B) only
Ans: (c)
For interference to occur, the waves must have a constant phase relationship.
(A) y = a sin(ωt): A simple harmonic wave with a sinusoidal form.
(C) y = a sin(ωt − ϕ): This represents a wave with a phase shift, and interference is possible between waves that have a phase difference.
In the case of (B) y = a sin(2ωt) and (D) y = a sin(3ωt), these are waves of different frequencies (ω and 2ω, or 3ω), which typically do not produce a simple interference pattern.
Thus, interference is possible between waves (A) and (C).
Correct Answer: (c) (A) and (C) only

Q7: A beam of unpolarised light of intensity I₀ is passed through a polaroid A, through another polaroid B, oriented at 60°, and finally through another polaroid C, oriented at 45° relative to B as shown in the figure. The intensity of the emergent light is:     (NEET 2024)

(a) I₀ / 16
(b) I₀ / 4
(c) I₀ / 2
(d) I₀ / 32
Ans: (a)
First Polaroid (A): When unpolarised light passes through the first polaroid, the intensity is reduced by half:
I₁ = I₀ / 2

Second Polaroid (B):
The light passing through polaroid B, which is oriented at 60° to the initial polarisation, will have its intensity reduced according to Malus' Law:
I₂ = I₁ × cos²(60°) = (I₀ / 2) × (1/2)² = I₀ / 8

Third Polaroid (C):
The light passing through polaroid C, which is oriented at 45° to polaroid B, will again be reduced by Malus' Law:
I₃ = I₂ × cos²(45°) = (I₀ / 8) × (√2/2)² = (I₀ / 8) × (1/2) = I₀ / 16
Correct Answer: (a) I₀ / 16

2023

Q1: For Young's double-slit experiment, two statements are given below: (NEET 2023)
Statement I: If the screen is moved away from the plane of slits, angular separation of the fringes remains constant.
Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.
Which of the following is correct?
(a) Statement I is False but Statement II is True.
(b) Both Statement I and Statement II are True.
(c) Both Statement I and Statement II are False.
(d) Statement I is True but Statement II is False.
Ans: (d)
Statement I: If the screen is moved away from the slits, the fringe linear separation increases because the distance between the slits and the screen increases. However, the angular separation (θ = λ / d) of the fringes remains constant. So, Statement I is True.
Statement II: The angular separation of the fringes depends on the wavelength of the light. A larger wavelength (λ) leads to a larger angular separation because θ = λ / d. Therefore, Statement II is False, as the angular separation increases with a larger wavelength, not decreases.
Correct Answer: (d) Statement I is True but Statement II is False.

2022

Q.1. For Young’s double slit experiment, two statements are given below:
Statement I: If screen is moved away from the plane of slits, angular separation of the fringes remains constant.
Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.
In the light of the above statements, choose the correct answer from the options given below:    (2022)
A: Both Statement I and Statement II are true.
B: Both Statement I and Statement II are false.
C: Statement I is true but Statement II is false.
D: Statement I is false but Statement II is true.
Ans: C
Solution: For YDSE, angular fringe width is given by α = λ/d
It does not depend on the distance of screen from the slit, so statement I is correct.
Angular fringe width ∝ λ
angular separation of fringes increases
So, statement I is true and statement II is false.

Q.2. In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is                   (2022)
A: 8 
B: 9 
C: 12 
D: 6 
Ans: C
Solution:

2020

Q.3. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2m is:      (2020)
A: 7.32 × 10^–7 rad 
B: 6.00 × 10^–7 rad 
C: 3.66 × 10^–7 rad 
D: 1.83 × 10^–7 rad
Ans: C
Solution:


Q.4. The Brewster’s angle ib for an interface should be :       (2020)
A: 45° < i_b < 90° 
B: i_b = 90° 
C: 0° < i_b < 30° 
D: 30° < i_b < 45°
Ans: A
Solution:

m = tan (i_b)
i_b =  tan^-1 (μ)
(μ > 1)
i_b > 45
45 < i_b < 90°

Q.5. In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :       (2020)
A: four times 
B: one-fourth 
C: double 
D: half
Ans: A
Solution:
Fringewidth in YDSE is given by

2019

Q.6. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μ_water = 4/3) (2019)
A: 0.266°
B: 0.15°
C: 0.05°
D: 0.1°
Ans: B
Solution:
In air angular fringe width

Angular fringe width in water

2018

Q.7. Unpolarised light is incident from air on a plane surface of a material of refractive index 'μ'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation ?    (2018)
A: Reflected light is polarised with its electric vector parallel to the plane of incidence
B: Reflected light is polarised with its electric vector perpendicular to the plane of incidence
C:
D:
Ans: (b)
Solution:

Q.8. In Young's double slit experiment the separation between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to :-    (2018)
A: 1.8 mm
B: 1.9 mm
C: 2.1 mm
D: 1.7 mm
Ans: B
Solution:

2017

Q.9. Young's double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5^th dark fringe lies in air. The refractive index of the medium is nearly :- (2017)
A: 1.59
B: 1.69
C: 1.78
D: 1.25
Ans: C
Solution:
Fringe Width in a medium -

- where in
Since wavelength in a medium become

Fringe width in air
Fringe width in any other medium
According to question


Q.10. Two Polaroids P₁ and P₂ are placed with their axis perpendicular to each other. Unpolarised light I₀ is incident on P₁ . A third polaroid P₃ is kept in between P₁ and P₂ such that its axis makes an angle 45° with that of P₁. The intensity of transmitted light through P₂ is :- (2017)
A: