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Class 12 Physics Chapter 10: Wave Optics case study question with answers

One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
In a physics laboratory, Meera sets up a double-slit experiment, as shown. A narrow slit S, lit by monochromatic light of wavelength 600 nm, illuminates two parallel slits S₁ and S₂ separated by d = 0.25 mm. Fringes are seen on a screen at D = 1.0 m from the slits; O is the point on the screen opposite the midpoint of the slits. She moves a fine probe across the screen and marks a point P at which the light from S₂ has travelled 1.5 μm farther than the light from S₁.
screenSS₁S₂POdDx
Part 1 · 1 mark
The fringe width on the screen is
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Answer: (c) 2.4 mm

β = λD/d = 600 × 10⁻⁹ × 1.0/(0.25 × 10⁻³) = 2.4 × 10⁻³ m = 2.4 mm.

Part 2 · 1 mark
The point P lies on
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Answer: (b) the third dark fringe from O

Path difference/λ = 1.5 μm/0.60 μm = 2.5 = (n + 1/2) with n = 2: destructive interference, the third dark fringe (n = 0, 1, 2 giving the first, second and third).

Part 3 · 2 marks
Find the distance OP.
Answer and marking scheme

Path difference = xd/D, so x = (1.5 × 10⁻⁶ × 1.0)/(0.25 × 10⁻³) = 6.0 × 10⁻³ m = 6.0 mm. As a check, 2.5β = 2.5 × 2.4 = 6.0 mm.

  • path difference = xd/D (1)
  • OP = 6.0 mm (1)
Or
Meera now fills the space between the slits and the screen with water of refractive index 4/3. Find the new fringe width, and the new distance of the third dark fringe from O.
Answer and marking scheme

In water λ becomes λ/n, so β = 2.4/(4/3) = 1.8 mm. The third dark fringe is at 2.5β = 4.5 mm.

  • β/n = 1.8 mm (1)
  • 2.5 × 1.8 = 4.5 mm (1)
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