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The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length 1 m. The width of the slit is 0.3 mm. If the third minimum is formed at a distance of 5 mm from the central maximum then the wavelength of light will be ---

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8500 A°

5000 A°

2500 A°

7500 A°

answer is C.

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Detailed Solution

$\begin{array}{l} asin θ = nλ; a \frac{y}{f} = nλ; λ = \frac{0 .3 \times 10^{- 3} \times 5 \times 10^{- 3}}{(3) (1)} \\ = 5 \times 10^{- 7} m = 5000 A^{0} \end{array}$

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