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

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`.