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MHT CET 2024 (3 May, Shift 1) · Diffraction at a Single Slit

Diffraction at a Single Slit question from MHT CET 2024 (3 May, Shift 1)

Q38MHT CET 2024 · 3 MAY · SHIFT 1Medium

Diffraction at a Single Slit

A single slit diffraction pattern is formed with light of wavelength $6195\text{ Å}$ . The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' $\lambda_0$ '. The value of ' $\lambda_0$ ' is

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Answer: (b)

$n$ th secondary maximum: $a\sin\theta = \dfrac{(2n+1)\lambda}{2}$
$5 \times 6195 = 7\lambda_0$ (2nd for 6195 Å, 3rd for $\lambda_0$ )
$\lambda_0 = \dfrac{30975}{7} = 4425$ Å. So the answer is (b).

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