In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is 6600A⁰, then wavelength of first maximum will be :

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5500A⁰

4400A⁰

3300A⁰

6600A⁰

answer is B.

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Detailed Solution

$\begin{array}{l} (1) λ_{1} = (2 (1) + 1) \frac{λ_{2}}{2}; Δx = nλ (min) \\ Δx = (2 n + 1) \frac{λ}{2} (max) ∴ λ_{2} = \frac{2}{3} λ_{1} = \frac{2}{3} \times 6600 \\ 4400 A^{0} \end{array}$

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