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A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Width of the slit is

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$2 \times 10^{- 6} m$

$20 \times 10^{- 6} m$

$200 \times 10^{- 6} m$

$2000 \times 10^{- 6} m$

answer is C.

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Detailed Solution

$λ = 500 \times 10^{- 9} m; D = 1 m$

Distance of the first minimum from the centre of the screen x = 2.5 mm for first minimum n = 1 from $n λ = \frac{x d}{D}$

$d = \frac{n λ D}{x} = \frac{500 \times 10^{- 9}}{2.5 \times 10^{- 3}} = 200 \times 10^{- 6} m$

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