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Q.

Light of wavelength 5000A° is diffracted by a slit. In diffraction pattern fifth minimum is at a distance of 5mm from central maximum. If the distance between the screen and the slit is 1m. The slit width is (in mm)

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answer is 0.5.

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Detailed Solution

$\sin θ^{'} \approx \tan θ^{'} \Rightarrow \frac{5 λ}{a} = \frac{y}{D}$

$a = \frac{5 λ D}{y}$

$= \frac{5 \times 5 \times 10^{- 7} \times 1}{5 \times 10^{- 3}}$

$= 5 \times 10^{- 4}$

$= 0.5 m m$

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