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Light of wavelength $λ = 5000 Å$ falls normally on a narrow slit. A screen placed at a distance of 1 m from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is

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0.1 mm

1.0 mm

0.5 mm

0.2 mm

answer is A.

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Detailed Solution

$Position of n^{th} minima, x_{n} = \frac{nλD}{d}$

$\begin{array}{l} \Rightarrow 5 \times 10^{- 3} = \frac{1 \times 5000 \times 10^{- 10} \times 1}{d} \\ \Rightarrow d = 10^{- 4} m = 0 .1 mm \end{array}$

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