In Young's double slit experiment, the aperture screen distance is `2m`. The fringe width is `1mm`. Light of `600nm` is used. If a thin plate of glass `(mu=1.5)` of thickness `0.06mm` is placed over one of the slits, then there will be a lateral displacement of the fringes by

Options

A. zero
B. 6 cm
C. 10 cm
D. 15 cm

Answer

B

Explanation

When a thin glass of thickness t is placed over one of the slits, then lateral displacement is given by
`x=((mu-1)tD)/(d)`
Given, `mu=1.5, t=0.06mm=6xx10^(-5)m` and
`D=2m, d=1mm=1xx10^(-3)m`

Putting the values in the above relation, we get
`x=((1.5-1)xx6xx10^(-5)xx2)/(1xx10^(-3))=0.5xx12xx10^(-2)=6cm`