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MHT CET 2026 (19 Apr, Shift 2) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2026 (19 Apr, Shift 2)

Q4MHT CET 2026 · 19 APR · SHIFT 2Medium

Biprism, Thin Films & Other Devices

In a biprism experiment, a steady interference pattern is observed on the screen kept at a distance of 100 cm using a light of wavelength 50005000 Å. Without changing the distance between the virtual images of the slit, the source of light is replaced by a source of wavelength 64006400 Å. Now, to reduce the fringe width by 20%20\% of its initial value, the screen should be moved

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Answer: (a)

  1. β=λDd\beta = \frac{\lambda D}{d}, dd fixed. Required: β′=0.8β\beta' = 0.8\beta.
  2. 6400 D′=0.8×5000×100⇒D′=62.5 cm6400\,D' = 0.8 \times 5000 \times 100 \Rightarrow D' = 62.5\ \text{cm}.
  3. Screen must move 100−62.5=37.5100 - 62.5 = 37.5 cm closer.
  4. So the answer is (a) towards the source by 37.5 cm.
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