In Young's experiment the slits, separated by `d = 0.8 mm`, are illuminated with light of wavelength 7200 Å, Interference pattern is obtained on a screen `D = 2 m` from the slits. Find minimum distance from central maximum at which the a average intensity is 50% of maximum?

Answer

`0.45`

Explanation

Resultant intensity at a point due to superposition of coherent waves of intensities `I_(1) and I_(2)` phase difference `phi` can be expressed as
`I_(R) = I_(1) + I_(2) + sqrt( I_(1) I_(2)) cos phi`
Here,` I_(1) = I_(2) = I`
`I_(R) = 2I + 2I cos phi = 2I (1 + cos phi)`
`I_(R) = 4I cos ^(2) (phi//2)`
At the position of interference maxima
`phi = 2 n pi`
`cos^(2) (phi//2) = cos^(2) n pi = 1 ` [ n is integer]
`I_("max") = 4I rArr I_(R) = I_("max") cos ^(2) (phi //2)`
Here, `I_(R) = (I_("max"))/(2) ` [50% of maxima]
`(I_("max"))/(2) = I_("max") cos^(2) "" (phi)/(2) rArr cos ^(2) "" (phi)/(2) = (1)/(2)`
`cos "" (phi)/(2) = (1)/( sqrt(2)) rArr (phi)/( 2) = (pi)/(4) or phi = pi // 2 `
Therefore, phase difference at a point at which resultant intensity is `50%` of maximum is ` pi//2`
Now, phase difference `phi = (2 pi // lambda)` (path difference )
`:.` Path difference ` = (lambda)/( 2 pi) xx (pi)/( 2) = (lambda) /(4)` . . . (i)
`= d (y)/(D) `
So, to obtain the desired distance
`(y)/( D) = (lambda)/(4) rArr y = (lambdaD)/( 4 d)`
Given `lambda = 7200 Å = 7200 ~~ 10^(-10) m and D = 2 m , d = 0. 8 m m `
`:. y = (7200 xx 10^(-10) xx 2)/( 4 xx 0.8 xx 10 ^(-3)) = 4 . 5 rArr y = 10^(-5) ` m
`= 0.45 m m `