Question

In a Young's double-slit experiment, a thin sheet of glass is placed over one of the two slits. As a result, the center of the fringe pattern (on the screen) shifts so that the center is now occupied by what was the 30th dark band. The wavelength of the light in this experiment is 600 nm and the index of the glass is 1.50.

a) The glass thickness is____________.

Answer 1

0.036mm

Step-by-step explanation:

We are given that

Wavelength,
`\lambda=600nm=600* 10^(-9) m`

`1nm=10^(-9) m`

Index of glass=n=1.5

m=30

We have to find the thickness of glass.

Thickness of glass,t=
`(m\lambda)/(n-1)`

Substitute the values

`t=(30* 600* 10^(-9))/(1.5-1)`

`t=0.036* 10^(-3) m`

`1mm=10^(-3) m`

`t=0.036mm`