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MHT CET 2026 (19 May, Shift 1) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2026 (19 May, Shift 1)

Q1MHT CET 2026 · 19 MAY · SHIFT 1Medium

Biprism, Thin Films & Other Devices

In a biprism experiment, using a light of wavelength $4800$ Å, a steady interference pattern was obtained on a screen $1$ m away from the source. Now the source is replaced by another of wavelength $6400$ Å and the screen is moved away from it by $20$ cm. The fringewidth will

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Answer: (c)

$\beta = \frac{\lambda D}{d} \Rightarrow \frac{\beta'}{\beta} = \frac{\lambda'}{\lambda}\cdot\frac{D'}{D}$ .
$\frac{\beta'}{\beta} = \frac{6400}{4800}\times\frac{1.2}{1} = 1.6$ .
So the answer is (c): fringe width increases by $60\%$ .

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