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A parallel beam of light of 500 nm falls...

A parallel beam of light of 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. Calculate the width of the slit.

Text Solution

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`theta=(y)/(D), theta= (2.5xx10^(-3))/(1)` radian. Now, `d sin theta = n lambda`
Since `theta` is very small, therefore `sin theta = theta`.
or `d= (n lambda)/(theta)=(1xx500xx10^(-9))/(2.5xx10^(-3))m = 2xx 10^(-4)m= 0.2 mm`.
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