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

Options

A. `1000 mum`
B. `500 mum`
C. `250 mu` mu
D. `125 mu`m

Answer

D

Explanation

`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`