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In a single slit diffraction experiment first minima for $λ_{1} = 660 nm$ coincide with first maxima for wavelength $λ_{2}$ . Calculate $λ_{2}$

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440 nm

445 nm

435 nm

442 nm

answer is A.

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Detailed Solution

First minima $sinθ = \frac{λ_{1}}{d}$
First maxima ${sinθ}^{1} = \frac{3 λ_{2}}{2 d}$
$\frac{λ_{1}}{d} = \frac{3 λ_{2}}{2 d}$
$λ_{2} = \frac{2 λ_{1}}{3} = \frac{2 \times 660}{3} = 440 nm$

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