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MHT CET 2026 (11 Apr, Shift 2) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from MHT CET 2026 (11 Apr, Shift 2)

Q7MHT CET 2026 · 11 APR · SHIFT 2Medium

Biprism, Thin Films & Other Devices

In biprism experiment, the $4^{th}$ dark band is formed opposite to one of the slits. The wavelength of light used is (D=distance between source and screen, d= the distance between the slits)

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Answer: (d)

Point opposite a slit: $y = \frac{d}{2}$ , path difference $= \frac{yd}{D} = \frac{d^2}{2D}$ .
4th dark band: $\frac{d^2}{2D} = \frac{7\lambda}{2}$ .
$\lambda = \frac{d^2}{7D}$ . So the answer is (d).

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