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A parallel beam of light of wavelength 5...

A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is obtained on a screen 1 m away. If the first minimum is formed at a distance of 2.5 mm from the centre of the screen, find the
(i) width of the slit, and
(ii) distance of first secondary maximum from the centre of the screen.

Text Solution

Verified by Experts

Here wavelength of light waves `lambda=500 nm=500xx10^(-9)m`, distance of screen from the slit D = 1m, distance of first minima from the centre of screen `x_(1)=2.5m m=2.5xx10^(-3)m`.
(i) `because x_(1)=D theta_(1)= (D lambda)/(a)`, where a = slit width
`rArr a=(Dlambda)/(x_(1))=(1xx500xx10^(-9))/(2.5xx10^(-3))=2xx10^(-4)m or 0.2`mm
(ii) The distance of first secondary maxima from the centre of the screen
`x._(1) =(3)/(2)(Dlambda)/(a)=(3)/(2) x_(1)=(3)/(2) xx 2.5xx10^(-3)=3.75xx10^(-3)m` or 3.75 mm
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