If a transparent medium of refractive index `mu=1.5` and thickness `t=2.5xx10^-5m` is inserted in front of one of the slits of Young's Double Slit experiment, how much will be the shift in the interference patten? The distance between the slits is `0.5mm` and that between slits and screen is `100cm`

Options

A. 5 cm
B. 2.5 cm
C. 0.25 cm
D. 0.1 cm

Answer

B

Explanation

Shift= `((mu-1)tD)/(d)= ((1.5-1) xx 2.5 xx 10^(-5) xx 100 xx 10^(-2))/(0.5 xx 10^(-3))= 2.5 cm`