Wave Optics – Interference and Young's Double Slit Experiment: NEET MCQs with Answers

15 free questions on Wave Optics – Interference and Young's Double Slit Experiment for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Interference and Young's Double Slit Experiment: Interference, Young's double-slit experiment and expression for fringe width, coherent sources, and sustained interference of light.

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Question 1

Previous-year question
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 [NEET 2020]
  1. half
  2. four times
  3. one-fourth
  4. double
Show answer
Answer: (B) four times
Explanation: β = λD/d. Halving d doubles β and doubling D doubles it again, so the fringe width becomes four times its original value.

Question 2

Previous-year question
In a Young's double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and the distance D between the screen and the 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 the same λ and D) the separation between the slits needs to be changed to [NEET 2018]
  1. 1.8 mm
  2. 1.9 mm
  3. 1.7 mm
  4. 2.1 mm
Show answer
Answer: (B) 1.9 mm
Explanation: The angular width λ/d varies inversely with d, so d′ = d × 0.20/0.21 = 2 × 0.952 ≈ 1.9 mm.

Question 3

Previous-year question
Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the 8th bright fringe in the medium lies where the 5th dark fringe lies in air. The refractive index of the medium is nearly [NEET 2017]
  1. 1.59
  2. 1.69
  3. 1.78
  4. 1.25
Show answer
Answer: (C) 1.78
Explanation: The 5th dark fringe in air is at 4.5β and the 8th bright fringe in the medium is at 8β/µ. Equating, µ = 8/4.5 ≈ 1.78.

Question 4

Previous-year question
The intensity at the maximum in a Young's double slit experiment is I₀. The distance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d? [NEET 2016]
  1. I₀
  2. I₀/4
  3. 3I₀/4
  4. I₀/2
Show answer
Answer: (D) I₀/2
Explanation: In front of one slit y = d/2, so the path difference is yd/D = d²/2D = d/20 = λ/4. Then φ = π/2 and I = I₀cos²(π/4) = I₀/2.

Question 5

Previous-year question
In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K (λ being the wavelength of light used). The intensity at a point where the path difference is λ/4 will be [AIPMT 2014]
  1. K
  2. K/4
  3. K/2
  4. Zero
Show answer
Answer: (C) K/2
Explanation: A path difference λ gives φ = 2π, so K = 4I₀. A path difference λ/4 gives φ = π/2, so I = 4I₀cos²(π/4) = 2I₀ = K/2.

Question 6

Previous-year question
Interference is possible in [AIPMT 1989]
  1. light waves only
  2. sound waves only
  3. both light and sound waves
  4. neither light nor sound waves
Show answer
Answer: (C) both light and sound waves
Explanation: Interference results from the superposition of two coherent waves and requires no particular wave type. Both light (transverse electromagnetic) and sound (longitudinal mechanical) waves exhibit it. Polarisation, by contrast, is restricted to transverse waves.

Question 7

Previous-year question
In Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, without disturbing the geometrical arrangement, the new fringe width will be [AIPMT 1990]
  1. 0.30 mm
  2. 0.40 mm
  3. 0.53 mm
  4. 450 micron
Show answer
Answer: (A) 0.30 mm
Explanation: In water the wavelength becomes λ/μ, so the fringe width becomes β′ = β/μ = 0.4/(4/3) = 0.3 mm.
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Question 8

Previous-year question
In Young's double slit experiment performed with blue and with green light of wavelengths 4360 Å and 5460 Å respectively, if x is the distance of the 4th maximum from the central one, then [AIPMT 1990]
  1. x(blue) = x(green)
  2. x(blue) > x(green)
  3. x(blue) < x(green)
  4. x(blue)/x(green) = 5460/4360
Show answer
Answer: (C) x(blue) < x(green)
Explanation: The distance of the nth maximum is x = nλD/d, so x ∝ λ. Since the blue wavelength (4360 Å) is smaller than the green (5460 Å), x(blue) < x(green).

Question 9

Previous-year question
The ratio of intensities of two waves is given by 4 : 1. Then the ratio of the amplitudes of the two waves is [AIPMT 1991]
  1. 2 : 1
  2. 1 : 2
  3. 4 : 1
  4. 1 : 4
Show answer
Answer: (A) 2 : 1
Explanation: Intensity is proportional to the square of the amplitude, so I₁/I₂ = a₁²/a₂² = 4/1. Taking the square root gives a₁/a₂ = 2/1.

Question 10

Previous-year question
In Young's experiment, two coherent sources are placed 0.90 mm apart and fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be [AIPMT 1991]
  1. 60 × 10⁻⁴ cm
  2. 10 × 10⁻⁴ cm
  3. 10 × 10⁻⁵ cm
  4. 6 × 10⁻⁵ cm
Show answer
Answer: (D) 6 × 10⁻⁵ cm
Explanation: For a dark fringe x = (2n − 1)λD/2d; with n = 2 this is 3λD/2d. Hence λ = 2xd/(3D) = (2 × 10⁻³ × 0.9 × 10⁻³)/3 = 0.6 × 10⁻⁶ m = 6 × 10⁻⁵ cm.

Question 11

Previous-year question
If yellow light emitted by a sodium lamp in Young's double slit experiment is replaced by monochromatic blue light of the same intensity, then [AIPMT 1992]
  1. fringe width will decrease
  2. fringe width will increase
  3. fringe width will remain unchanged
  4. fringes will become less intense
Show answer
Answer: (A) fringe width will decrease
Explanation: The fringe width β = λD/d is directly proportional to wavelength. Blue light has a shorter wavelength than yellow, so the fringe width decreases.

Question 12

Previous-year question
In Young's double slit experiment carried out with light of wavelength λ = 5000 Å, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as the zeroth maximum) will be at x equal to [AIPMT 1992]
  1. 1.67 cm
  2. 1.5 cm
  3. 0.5 cm
  4. 5.0 cm
Show answer
Answer: (B) 1.5 cm
Explanation: For the nth maximum, x = nλD/d = (3 × 5000 × 10⁻¹⁰ × 2)/(0.2 × 10⁻³) = 3 × 10⁻⁶/(2 × 10⁻⁴) = 1.5 × 10⁻² m = 1.5 cm.

Question 13

Previous-year question
Interference was observed in an interference chamber where air was present. Now the chamber is evacuated, and if the same light is used, a careful observer will see [AIPMT 1993]
  1. no interference
  2. interference with brighter bands
  3. interference with dark bands
  4. interference with larger bands
Show answer
Answer: (D) interference with larger bands
Explanation: Since β ∝ λ and the wavelength in vacuum is slightly greater than in air (μ_air ≈ 1.0003), evacuating the chamber makes the wavelength — and hence the fringe width — increase slightly. The observer therefore sees marginally larger bands.

Question 14

Previous-year question
Colours of thin soap bubbles are due to [AIPMT 1999]
  1. refraction
  2. dispersion
  3. interference
  4. diffraction
Show answer
Answer: (C) interference
Explanation: White light reflected from the upper and lower surfaces of the thin soap film superposes. Depending on the film thickness and the wavelength, certain colours satisfy the condition for constructive interference and are reinforced while others are cancelled, producing the shifting iridescent colours.

Question 15

Previous-year question
In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits is reduced to half. As a result the fringe width [NEET 2013]
  1. is halved
  2. becomes four times
  3. remains unchanged
  4. is doubled
Show answer
Answer: (B) becomes four times
Explanation: Since β = λD/d, doubling D and halving d gives β′ = λ(2D)/(d/2) = 4λD/d = 4β — the fringe width becomes four times its original value.

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