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Q.

A screen is placed $50 c m$ from a single slit, which is illuminated with $6000 Å$ light. If distance between the first and third minima in the diffraction pattern is $3 m m,$ the width of the slit is (in mm)

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Detailed Solution

$P o s i t i o n o f n t h m i n i m a y_{n} = \frac{n λ D}{d} D i s \tan c e b e t w e e n t h e f i r s t a n d t h i r d m i n i m a = (y_{3} - y_{1}) = \frac{λ D}{d} (3 - 1) = \frac{2 λ D}{d} ∴ 3 \times 10^{- 3} = \frac{2 \times 6000 \times 10^{- 10} \times 0.5}{d} \Rightarrow d = 0.2 \times 10^{- 3} m = 0.2 m m$

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