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In YDSE a = 2mm, D = 2m, lambda = 500 mm...

In YDSE `a = 2mm`, D = 2m, `lambda = 500` mm. Find distance of point on screen from central maxima where intensity becomes `50%` of central maxima

A

`1000 mum`

B

`500 mum`

C

`250 mu` mu

D

`125 mu`m

Text Solution

Verified by Experts

The correct Answer is:
D
`I = I_(0) cos^(2) '(phi)/(2) = (I_(0))/(2)`
`cos'(phi)/(2) = 1/(sqrt(2))`
`phi/2 = pi/4`
`phi = pi/2`
`Deltax = lambda/4`
`y = (beta)/(4) = (lambdaD)/(4d) = (500 xx 10^(-9) xx 2)/(4xx2xx10^(-3)) = 125 xx 10^(-6) m`
`y = 125 mum`
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