Question

In a Young's double slit experiment, the fringes are displaced by a distance x when a glass plate of refractive index 1.5 is introduced in the path of one of the beams.Then this plate is replaced by another plate of the same thickness, the shift of fringes is 3/2 x. The refractive index of the second plate is

• A. 2.25
• B. 2
• C. 1.75
• D. 1.25

Answer 1

Answer: (C)

1.75

Given,

$x$ is displacement of first plate

$3/2x$ is displacement of Second plate

$\mu =1.5$

If thickness $t$ and refractive index is $\mu$ then path difference of plate is,

Separation between slits if path difference is $\lambda$ and phase shift is $\beta$ is

$\beta =\frac{\lambda D}{2d}\times 2d$

Here, path difference is $(\mu -1)t$ and $x$ is shift

$x=\frac{(\mu -1)t\beta }{\lambda }$

$\frac{({\mu }_2-1)}{({\mu }_1-1)}=\frac{\frac{3}{2}x}{x}$

Substitute all value in above equation

$\frac{({\mu }_2-1)}{(1.5-1)}=1.5({\mu }_2-1)=0.75{\mu }_2=1.75$

correct option is C.