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MHT CET 2026 (19 May, Shift 1) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2026 (19 May, Shift 1)

Q1MHT CET 2026 · 19 MAY · SHIFT 1Medium

Biprism, Thin Films & Other Devices

In a biprism experiment, using a light of wavelength 48004800 Å, a steady interference pattern was obtained on a screen 11 m away from the source. Now the source is replaced by another of wavelength 64006400 Å and the screen is moved away from it by 2020 cm. The fringewidth will

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Answer: (c)

  1. β=λDd⇒β′β=λ′λ⋅D′D\beta = \frac{\lambda D}{d} \Rightarrow \frac{\beta'}{\beta} = \frac{\lambda'}{\lambda}\cdot\frac{D'}{D}.
  2. β′β=64004800×1.21=1.6\frac{\beta'}{\beta} = \frac{6400}{4800}\times\frac{1.2}{1} = 1.6.
  3. So the answer is (c): fringe width increases by 60%60\%.
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