In a YDSE experiment, d = 1mm, `lambda`= 6000Å and D= 1m. The minimum distance between two points on screen having 75% intensity of the maximum intensity will be

Options

A. 0.50mm
B. 0.40mm
C. 0.30 mm
D. 0.20 mm

Answer

D

Explanation

`d = 1 mm ,lambda = 6000 Å, D = 1 m`
`I_(max) = 4I_(0)`
`rArr 75%` of Imax = `3 I_(0)` ltbr ` 3I_(0) = I_(0) + I+ 2 sqrt(I_(0)I_(0)) cos (Delta theta)`
`I_(0) = 2 I_(0) cosDelta theta`
`cos Delta theta = (1)/(2)`
`Delta theta = (1)/(2)`
` Delta theta = pm pi//3 , pm 2pi//3`
`(2pi)/(lambda) Deltax = pi //3 rArr Deltax = lambda//6`
`(2pi)/(lambda) Delta x = 2pi //3 rArr Deltax = lambda //3`
`(y_(1)d)/(D) Deltax = 2pi//3 rArr Delta x = lambda//3`
`(y_(2)d)/(D) = lambda rArr y_(2) = (lambdaD)/(3D)`
`y_(2) - y_(1) = (lambdaD)/(3d) - (lambdaD)/(6d) = (lambdaD)/(6d)`
`y_(2) - y_(1) = (6 xx 10^(-7) xx1)/(9xx 10^(-3)) = 10^(-4) m`
0.1 mm