Question

In YDSE, if the thickness of a glass slab $(\mu =1.5)$ which should be placed before the upper slit $S1$ so that the central maximum now lies at a point where 5th bright fringe was lying earlier (before inserting the slab) is $Y\ast 10,000\mathring{A}$. (Wavelength of light used is $5000\mathring{A}$.) Find Y?

Answer 1

Given:

Refractive index $\mu =1.5$

Thickness of the slab, $t=Y\times 10,000\AA m$

Wavelength, $\lambda =5000\AA m$

To find:

Y

We know,

Path difference of glass slab, $\Delta x=(\mu -1)t=(1.5-1)t=0.5t$

Here it is given, there is a shift of 5 fringes.

$\therefore \Delta x=5\lambda$

$or,0.5t=5\times 5000$Å

Hence the thickness of the glasss slab, $t=50,000$Åm$=5\times 10,000$Åm

But it is given the thickness of glass slab is $Y\times 10,000$Åm.

Comparing both these, we get Y=5.