In Young's double slit experiment `(d)/(D)=10^(-4)` (d=distance between slit , D=distance of screen from the slits ). At a point P on the screen resulting intesity is equal to the intensity due to individual slit `I_(0)`. The the distance of point P from the central maximum is `(lambda=6000Å)`

Options

A. 2 mm
B. 1 mm
C. 0.5 mm
D. 4 mm

Answer

a

Explanation

`I = 4 I_(0) cos^(2) ((phi)/(2))`
`implies I_(0) = 4 I_(0) cos^(2) ((phi)/(2))`
`implies cos ((phi)/(2)) = (1)/(2)`
or `(phi)/(2) = (pi)/(3)`
or `phi = (2 pi)/(3) = ((2 pi)/(lambda)) Delta x`
or `(1)/(3) = ((1)/(lambda)) y (d)/(D) (Delta x = (yd)/(D))` ltbr. `:. y = (lambda)/(3 xx (d)/(D)) = (6 xx 10^(-7))/(3 xx 10^(-4))`
`:. y=(lamda)/(3 xx (d)/(D))=(6 xx10^(-7))/(3 xx 10^(-4))`
`= 2 xx 10^(-3) m = 2 mm`.