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MHT CET 2026 (18 Apr, Shift 2) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2026 (18 Apr, Shift 2)

Q6MHT CET 2026 · 18 APR · SHIFT 2Medium

Biprism, Thin Films & Other Devices

In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index 'μ\mu' is placed in one of the interfering paths. Now 7th7^{th} bright fringe is obtained at the same point. If 'λ\lambda' is the wavelength of light used, the thickness of film is equal to

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Answer: (d)

  1. 5th dark fringe: Δ=(2×5−1)λ2=4.5λ\Delta = (2\times5-1)\frac{\lambda}{2} = 4.5\lambda. 7th bright: Δ=7λ\Delta = 7\lambda.
  2. Extra path added by the film: (μ−1)t=7λ−4.5λ=2.5λ(\mu-1)t = 7\lambda - 4.5\lambda = 2.5\lambda.
  3. t=2.5λμ−1t = \frac{2.5\lambda}{\mu - 1}. So the answer is (d).
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