In Young's double slit experiment separation between slits is 1mm, distance of screen from slits is 2m. If wavelength of incident light is 500 nm. Determine
(i) fringe width
(ii) angular fringe width
(iii) distance between 4 th bright fringe and 3rd dark fringe
(iii) If whole arrangement is immersed in water `(mu_(w)=4//3)`, new angular fringe width.

Explanation

Given, `d=1mm, D=2m, lambda=500 nm = 500xx10^(-9)m`
(i) Fringe width,
`beta=(Dlambda)/d=(2xx500xx10^(-9))/(1xx10^(-3))=10^(-3) m=1mm`
(ii) Angular fringe width,
`theta_(0)=lambda/d=(500xx10^(-9))/(1xx10^(-3))=5xx10^(-4)"rad"`
(iii) Distance of 4 th bright fringe from centre,
`y_(4)=(nDlambda)/d=(4Dlambda)/d`
Distance of 3rd dark fringe from centre,
`y'_(3)=((2n+1)Dlambda)/d=((2xx2+1)Dlambda)/(2d)=(5Dlambda)/(2d)`
Distance between 4th bright fringe and 3rd dark fringe,
`y_(4)-y_(3)'=(4Dlambda)/d-(5Dlambda)/(2d)=(3Dlambda)/(2d)`
`=3/2beta=3/2xx1=3/2mm`
(iv) Wavelength in water, `lambda_(w)=lambda/mu_(w)=lambda/(4//3)=3/4lambda`
`(theta_(0))_(w)=(theta_(0))/(mu_(w))=(5xx10^(-4))/(4//3)=15/4xx10^(-4)"rad"`