Wave Optics ; Diffraction – Diffraction due to a single slit: NEET MCQs with Answers

15 free questions on Wave Optics ; Diffraction – Diffraction due to a single slit 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: Diffraction due to a single slit, width of central maximum.

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Question 1

Previous-year question
Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of a telescope whose objective has a diameter of 2 m is [NEET 2020]
  1. 1.83 × 10⁻⁷ rad
  2. 7.32 × 10⁻⁷ rad
  3. 6.00 × 10⁻⁷ rad
  4. 3.66 × 10⁻⁷ rad
Show answer
Answer: (D) 3.66 × 10⁻⁷ rad
Explanation: Δθ = 1.22λ/D = 1.22 × 600 × 10⁻⁹/2 = 3.66 × 10⁻⁷ rad.

Question 2

Previous-year question
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) [NEET 2019]
  1. 0.266°
  2. 0.15°
  3. 0.05°
  4. 0.1°
Show answer
Answer: (B) 0.15°
Explanation: Inside water the wavelength becomes λ/µ, so every diffraction angle is reduced by the factor µ. The new angular width is 0.2° × 3/4 = 0.15°.

Question 3

Previous-year question
The ratio of the resolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is [NEET 2017]
  1. 9 : 4
  2. 3 : 2
  3. 16 : 81
  4. 8 : 27
Show answer
Answer: (B) 3 : 2
Explanation: Resolving power of a microscope is 2µ sin θ/λ, so it varies as 1/λ. The ratio is therefore λ₂ : λ₁ = 6000 : 4000 = 3 : 2.

Question 4

Previous-year question
In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle of 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of [NEET 2016]
  1. sin⁻¹(1/4)
  2. sin⁻¹(2/3)
  3. sin⁻¹(1/2)
  4. sin⁻¹(3/4)
Show answer
Answer: (D) sin⁻¹(3/4)
Explanation: From a sin 30° = λ we get a = 2λ. The first secondary maximum satisfies a sin θ = 3λ/2, so sin θ = 3/4.

Question 5

Previous-year question
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the midpoint of the slit is [AIPMT 2015]
  1. π/2 rad
  2. π rad
  3. π/8 rad
  4. π/4 rad
Show answer
Answer: (B) π rad
Explanation: At the first minimum the edge-to-edge path difference is λ, so the edge-to-midpoint path difference is λ/2 and the corresponding phase difference is π radians.

Question 6

Previous-year question
A beam of light of λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is [AIPMT 2014]
  1. 1.2 cm
  2. 1.2 mm
  3. 2.4 cm
  4. 2.4 mm
Show answer
Answer: (D) 2.4 mm
Explanation: y = λD/a = 1.2 mm on each side, so the separation between the two first minima is 2.4 mm.

Question 7

Previous-year question
The diameter of the human eye lens is 2 mm. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 m from the eye? (The wavelength of light is 5000 Å.) [AIPMT 2002]
  1. 2.32 m
  2. 4.28 mm
  3. 1.25 cm
  4. 12.48 cm
Show answer
Answer: (C) 1.25 cm
Explanation: The angular resolution of the eye lens is dθ = λ/d = (5000 × 10⁻¹⁰)/(2 × 10⁻³) = 2.5 × 10⁻⁴ rad. At r = 50 m the minimum separation is s = r dθ = 50 × 2.5 × 10⁻⁴ = 1.25 × 10⁻² m = 1.25 cm.
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Question 8

Previous-year question
A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of [AIPMT 2004]
  1. 0.5 m
  2. 5 m
  3. 5 mm
  4. 5 cm
Show answer
Answer: (C) 5 mm
Explanation: The angular resolution is dθ = 1.22λ/D = (1.22 × 5000 × 10⁻¹⁰)/0.1 = 6.1 × 10⁻⁶ rad. The linear separation at 1 km is x = dθ × d = 6.1 × 10⁻⁶ × 10³ ≈ 6 × 10⁻³ m, i.e. of the order of 5 mm.

Question 9

Previous-year question
The angular resolution of a 10 cm diameter telescope at a wavelength of 5000 Å is of the order of [AIPMT 2005]
  1. 10⁶ rad
  2. 10⁻² rad
  3. 10⁻⁴ rad
  4. 10⁻⁶ rad
Show answer
Answer: (D) 10⁻⁶ rad
Explanation: Δθ = 1.22λ/D = (1.22 × 5000 × 10⁻¹⁰)/0.1 = 6.1 × 10⁻⁶ rad, which is of the order of 10⁻⁶ rad.

Question 10

Previous-year question
A beam of light of λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is [AIPMT 2014]
  1. 1.2 cm
  2. 1.2 mm
  3. 2.4 cm
  4. 2.4 mm
Show answer
Answer: (D) 2.4 mm
Explanation: The separation between the two first minima is the width of the central maximum, 2λD/a = (2 × 600 × 10⁻⁹ × 2)/(1 × 10⁻³) = 2.4 × 10⁻³ m = 2.4 mm.

Question 11

Previous-year question
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygens' wavelet from the edge of the slit and the wavelet from the midpoint of the slit is [AIPMT 2015]
  1. π/4 rad
  2. π/2 rad
  3. π rad
  4. π/8 rad
Show answer
Answer: (C) π rad
Explanation: The first minimum occurs when a sin θ = λ. The edge and the midpoint are separated by a/2, so their path difference is (a/2) sin θ = λ/2. The corresponding phase difference is (2π/λ)(λ/2) = π rad — exactly the condition for those two wavelets to cancel.

Question 12

Previous-year question
For a parallel beam of monochromatic light of wavelength λ, diffraction is produced by a single slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be [AIPMT 2015]
  1. 2Dλ/a
  2. Dλ/a
  3. Da/λ
  4. 2Da/λ
Show answer
Answer: (A) 2Dλ/a
Explanation: The first minima lie at sin θ = ±λ/a, so for small angles the distance of each from the centre is Y = λD/a. The central maximum extends between them, so its full width is 2Y = 2Dλ/a.

Question 13

Previous-year question
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10⁻⁵ cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is [NEET 2016]
  1. 0.10 cm
  2. 0.25 cm
  3. 0.20 cm
  4. 0.15 cm
Show answer
Answer: (D) 0.15 cm
Explanation: The position of the first minimum is Y = λD/a, where D is the focal length here. Hence Y = (5 × 10⁻⁵ × 60)/0.02 = 3 × 10⁻³/0.02 = 0.15 cm.

Question 14

Previous-year question
In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of [NEET 2016]
  1. sin⁻¹(2/3)
  2. sin⁻¹(1/2)
  3. sin⁻¹(3/4)
  4. sin⁻¹(1/4)
Show answer
Answer: (C) sin⁻¹(3/4)
Explanation: For the first minimum, a sin 30° = λ, so a = 2λ. The first secondary maximum satisfies a sin θ = 3λ/2, giving sin θ = 3λ/(2 × 2λ) = 3/4. Hence θ = sin⁻¹(3/4).

Question 15

Previous-year question
The ratio of resolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is [NEET 2017]
  1. 8 : 27
  2. 9 : 4
  3. 3 : 2
  4. 16 : 81
Show answer
Answer: (C) 3 : 2
Explanation: The resolving power of a microscope is inversely proportional to the wavelength, so RP₁/RP₂ = λ₂/λ₁ = 6000/4000 = 3/2, i.e. 3 : 2.

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