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MHT CET 2024 (15 May, Shift 1) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2024 (15 May, Shift 1)

Q13MHT CET 2024 · 15 MAY · SHIFT 1Easy

Biprism, Thin Films & Other Devices

In biprism experiment, the fringe width is 0.6 mm. The distance between 66th dark fringe and 88th bright fringe on the same side of central bright fringe is

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Answer: (c)

  1. nnth dark fringe: y=(2n−1)β2y = (2n-1)\frac{\beta}{2}, so 6th dark at 5.5β5.5\beta.
  2. 8th bright fringe at 8β8\beta.
  3. Distance =8β−5.5β=2.5β=2.5×0.6=1.5 mm= 8\beta - 5.5\beta = 2.5\beta = 2.5 \times 0.6 = 1.5\ \text{mm}.
  4. So the answer is (c) 1.5 mm.
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