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MHT CET 2026 (20 Apr, Shift 1) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2026 (20 Apr, Shift 1)

Q3MHT CET 2026 · 20 APR · SHIFT 1Medium

Biprism, Thin Films & Other Devices

In a biprism experiment, the distance between 4th4^{th} and 13th13^{th} bright band on the same side is 'yy' when light of wavelength 60006000 Å is used. For a light of wavelength 'λ\lambda' the distance between 6th6^{th} and 16th16^{th} bright band on the same side is again 'yy'. The value of 'λ\lambda' in Å units is

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Answer: (c)

  1. Distance between mmth and nnth bright bands =(n−m)β= (n-m)\beta, with β∝λ\beta \propto \lambda.
  2. 9β1=y9\beta_1 = y and 10β2=y⇒9×6000=10λ10\beta_2 = y \Rightarrow 9 \times 6000 = 10\lambda.
  3. λ=5400\lambda = 5400 Å. So the answer is (c) 5400.
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