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Visible light of wavelength 500 nm falls...

Visible light of wavelength 500 nm falls normally on a single slit and produces a diffraction pattern. It is found that the diffraction pattern is on a screen 1 m away from slit. If the first minimum is produced at a distance of 2.5 mm from the centre of screen, then the width of the slit is

A

`0.1 mm`

B

`0.4 mm`

C

`0.3 mm`

D

`0.2 mm`

Text Solution

Verified by Experts

The correct Answer is:
D
`lamda = 500xx10^(-9) m`
So, for first minimum, `s sin theta _(1)= lamda`
`implies sin theta _(1) ~~ tan theta _(1) = (lamda)/(d) implies y/D(lamda)/(d)`
So, `d = (lamda D)/(y ) = (5000xx10^(-9)xx1)/(2.5 xx10^(-3))m`
`d = 2 xx 10^(-4) m`
` d =0.2 mm`
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