Question

In Young's double slit experiment, the fringe width is $\beta$, If the entire arrangement is placed in a liquid of refractive index $n$, the fringe width becomes:

• A. $n\beta$
• B. $\frac{\beta }{n+1}$
• C. $\frac{\beta }{n-1}$
• D. $\frac{\beta }{n}$

Answer 1

The correct option is D $\frac{\beta }{n}$

Fringe width $\beta =\frac{\lambda D}{d}$ where all the symbols have their usual meanings.

Now the entire apparatus is placed in liquid of refractive index $n$.

Thus the wavelength of the incident wave changes.

$⟹$ ${\lambda }^{\prime }=\frac{\lambda }{n}$

$\therefore$ New fringe width ${\beta }^{\prime }=\frac{{\lambda }^{\prime }D}{d}$

$⟹$ ${\beta }^{\prime }=\frac{\lambda }{n}\frac{D}{d}=\frac{\beta }{n}$