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Interference and Young’s double-slit experiment: NEET MCQs with answers

90 questions on this topic of Optics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Two coherent sources of light have an intensity ratio of 9 : 1. In the interference pattern obtained, the ratio of maximum to minimum intensity (I_max / I_min) is:

  1. 9 : 1
  2. 4 : 1
  3. 16 : 1
  4. 25 : 1
Show answer and explanation

Answer: (B) 4 : 1

The amplitudes are in the ratio a₁/a₂ = √(I₁/I₂) = √(9/1) = 3/1. Thus, I_max / I_min = (a₁ + a₂)² / (a₁ - a₂)² = (3 + 1)² / (3 - 1)² = 16 / 4 = 4 : 1.

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2. In a Young's double-slit setup, the path difference between two interfering beams at a point P on the screen is λ/6. The ratio of the intensity at point P to the maximum intensity on the screen is:

  1. 3 / 4
  2. 1 / 2
  3. 1 / 4
  4. √3 / 2
Show answer and explanation

Answer: (A) 3 / 4

Phase difference ϕ = (2π / λ) × Δx = (2π / λ) × (λ/6) = π/3 rad. The intensity is I = I_max cos²(ϕ/2) = I_max cos²(π/6). Since cos(π/6) = √3/2, we have I / I_max = (√3/2)² = 3/4.

hard

3. In a standard Young's double-slit experiment, let the slit separation be d and the screen distance be D (where D >> d >> λ). The minimum distance from the central maximum where the intensity drops to half of its maximum value is:

  1. λD / (2d)
  2. λD / (4d)
  3. λD / (3d)
  4. λD / (8d)
Show answer and explanation

Answer: (B) λD / (4d)

Intensity is I = I₀ cos²(Δϕ/2). Setting I = I₀ / 2 gives cos²(Δϕ/2) = 1/2, so cos(Δϕ/2) = 1/√2, which implies Δϕ/2 = π/4, hence Δϕ = π/2 rad. The path difference is Δx = (λ/2π) · Δϕ = λ/4. Since Δx = y d / D, we obtain y = λD / (4d).

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4. In Young’s double-slit experiment, λ = 600 nm, screen distance D = 1.5 m and slit separation d = 1.5 mm. The fringe width is:

  1. 0.6 mm
  2. 1.2 mm
  3. 0.3 mm
  4. 6e-7 mm
Show answer and explanation

Answer: (A) 0.6 mm

β = λD/d = (600×10⁻⁹×1.5)/(1.5×10⁻³) = 0.6 mm.

Quick tip: β = λD/d. Convert nm and mm before substituting.

easy

5. In Young's double-slit experiment, the intensity at the central maximum is I₀. What is the intensity at a point on the screen where the path difference between the interfering waves is λ/4?

  1. I₀ / 4
  2. I₀ / √2
  3. 3 I₀ / 4
  4. I₀ / 2
Show answer and explanation

Answer: (D) I₀ / 2

Path difference Δx = λ/4 corresponds to a phase difference Δϕ = (2π/λ) · (λ/4) = π/2 rad. The resultant intensity is given by I = I₀ cos²(Δϕ/2) = I₀ cos²(π/4) = I₀ (1/√2)² = I₀ / 2.

hard

6. In a Young's double-slit experiment, when a thin transparent mica sheet of thickness t and refractive index μ is placed in front of one of the slits, the central fringe shifts by a distance equal to 5 fringe widths. The thickness t of the sheet is (where λ is the wavelength of light):

  1. 5λ / (μ - 1)
  2. λ / [5(μ - 1)]
  3. 5(μ - 1) / λ
  4. 2.5λ / (μ - 1)
Show answer and explanation

Answer: (A) 5λ / (μ - 1)

The shift in fringe pattern due to a thin sheet of thickness t is given by Δy = (D/d)(μ - 1)t. The problem states Δy = 5β = 5(λD/d). Equating both: (D/d)(μ - 1)t = 5(λD/d), which simplifies to (μ - 1)t = 5λ, hence t = 5λ / (μ - 1).

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7. In Young's double-slit experiment, light of wavelength 400 nm is used, the slits are 1 mm apart and the screen is 2 m away. The fringe width is:

  1. 8 mm
  2. 1.6 mm
  3. 0.8 mm
  4. 0.4 mm
Show answer and explanation

Answer: (C) 0.8 mm

β = λD/d = 400 × 10⁻⁹ × 2/(1 × 10⁻³) m = 0.8 mm.

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8. A Young’s double-slit setup uses light of wavelength 450 nm, with the slits 1 mm apart and the screen 1.5 m away. The fringe width is:

  1. 0.338 mm
  2. 0.675 mm
  3. 1.35 mm
  4. 3e-7 mm
Show answer and explanation

Answer: (B) 0.675 mm

β = λD/d = (450×10⁻⁹×1.5)/(1×10⁻³) = 0.675 mm.

Quick tip: β = λD/d. Convert nm and mm before substituting.

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9. In Young's double-slit experiment, light of wavelength 500 nm is used, the slits are 0.25 mm apart and the screen is 1 m away. The fringe width is:

  1. 2 mm
  2. 4 mm
  3. 1 mm
  4. 20 mm
Show answer and explanation

Answer: (A) 2 mm

β = λD/d = 500 × 10⁻⁹ × 1/(0.25 × 10⁻³) m = 2 mm.

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10. In Young's double-slit experiment, the distance between the two slits is d and distance to screen is D. The minimum distance from the central bright fringe to a point where the intensity is half of the maximum intensity is:

  1. λD / (4d)
  2. λD / (3d)
  3. λD / (2d)
  4. λD / (8d)
Show answer and explanation

Answer: (A) λD / (4d)

Maximum intensity is I_max = 4I₀. For I = I_max / 2, we have cos²(ϕ/2) = 1/2, giving ϕ/2 = π/4, so ϕ = π/2. Path difference Δx = (λ / 2π) × ϕ = λ/4. Since Δx = yd / D, the position y = λD / (4d).

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11. In Young's double-slit experiment, the ratio of maximum to minimum intensity in the fringe system is 25 : 9. The ratio of the amplitudes of the two interfering waves is:

  1. 4 : 1
  2. 5 : 3
  3. 16 : 1
  4. 25 : 1
Show answer and explanation

Answer: (A) 4 : 1

Given I_max / I_min = 25/9. Since √(I_max / I_min) = (A₁ + A₂) / (A₁ - A₂), we have (A₁ + A₂) / (A₁ - A₂) = 5/3. Cross-multiplying gives 3A₁ + 3A₂ = 5A₁ - 5A₂, which simplifies to 2A₁ = 8A₂, so A₁ / A₂ = 4 / 1.

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12. A Young’s double-slit setup uses light of wavelength 700 nm, with the slits 1 mm apart and the screen 2.5 m away. The fringe width is:

  1. 2.8e-7 mm
  2. 1.75 mm
  3. 0.875 mm
  4. 3.5 mm
Show answer and explanation

Answer: (B) 1.75 mm

β = λD/d = (700×10⁻⁹×2.5)/(1×10⁻³) = 1.75 mm.

Quick tip: β = λD/d. Convert nm and mm before substituting.

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