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MHT CET 2019 (2 May, Shift 1) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2019 (2 May, Shift 1)

Q25MHT CET 2019 · 2 MAY · SHIFT 1Medium

Biprism, Thin Films & Other Devices

In biprism experiment, the distance between source and eyepiece is 1.2 m, the distance between two virtual sources is 0.84 mm. Then the wavelength of light used if eyepiece is to be moved transversely through a distance of 2.799 cm to shift 30 fringes is

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Answer: (a)

  1. 30β=2.799 cm⇒β=9.33×10−4 m30\beta = 2.799\ \text{cm} \Rightarrow \beta = 9.33\times10^{-4}\ \text{m}.
  2. λ=βdD=9.33×10−4×0.84×10−31.2\lambda = \frac{\beta d}{D} = \frac{9.33\times10^{-4}\times 0.84\times10^{-3}}{1.2}.
  3. λ≈6.53×10−7 m≈6531\lambda \approx 6.53\times10^{-7}\ \text{m} \approx 6531 Å, closest to option (a).
  4. So the answer is (a) 6533 Å.
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