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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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For the first dark band, `sin theta_(1) = (lambda)/(d)`
But `sin theta_(1) approx theta_(1) approx tan theta_(1) = (2.5 xx 10^(-3))/(1) = 2.5 xx 10^(-3)`
`therefore "d" = (lambda)/(sin theta_(1)) = (500xx10^(-9))/(2.5xx10^(-3)) = 0.2 mm`
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