In Young's double slit expriment, separation between slits is `1 mm`, distance of screen from slits is `2.5 m`. If wavelength of incident light is `400 nm`. Determine
(a) Fringe width.
(b) Angular frige width.
(c ) Distance between `4^(th)` bright fringe and `3^(rd)` dark fringe.
(d) If whole arrangement is immersed in water `(mu_(omega)=4//3)`, new angular fringe width.

Explanation

Given: `d=1 mm, D=2.5 m, lambda=400 nm=400xx10^(-9) m`
(a) Fringe width:
`beta=(Dlambda)/d=(2.5xx400xx10^(-3))/(1xx10^(-3))=10^(-3) m=1 mm`
(b) Angular fringe width:
`theta_(0)=lambda/d=(400xx10^(-9))/(1xx10^(-3))=4xx10^(-4) rad`
(c ) Distance of `4^(th)` bright fringe from centre
`y_(3)^(')=((2n+1)Dlambda)/d=((2xx2+1)Dlambda)/(2d)=(5Dlambda)/(2d)`
`y_(4)-y_(3)^(')=(4Dlambda)/d-(5Dlambda)/(2d)=(3Dlambda)/(2d)`
`=3/2 beta=3/2xx1=3/2 mm`
(d) Wavelength in water, `lambda_(omega)=lambda/mu_(omega)=lambda/(4//3)=3/4 lambda`
`(theta_(0))_(omega)=theta_(0)/mu_(omega)=(4xx10^(-4))/(4//3)=3xx10^(-4) rad`