Question

In a Young's double slit experiment, the fringe width is found to be $0.4mm.$ If the whole apparatus is immersed in water of refractive index $\frac{4}{3}$ without disturbing the geometrical arrangement, the new fringe width will be

• A. $0.30mm$
• B. $0.40mm$
• C. $0.53mm$
• D. $450\mu m$

Answer 1

Answer: (A)

$0.30mm$

The formula for the fringe width is :

$\beta =\frac{\lambda D}{d};$

$\therefore B\propto \lambda$

We also know that the wavelength is inversely proportional to the refractive index of the medium.

So, ${\lambda }^{{}^{\prime }}=\frac{\lambda }{\mu }$

So ,${\lambda }^{{}^{\prime }}=\frac{3\lambda }{4}$

So, the ratio can be obtained as:

$\frac{{\beta }^{{}^{\prime }}}{\beta }=\frac{{\lambda }^{{}^{\prime }}}{\lambda }$

${\beta }^{{}^{\prime }}=\frac{0.4}{4/3}$

$\Rightarrow {\beta }^{{}^{\prime }}=0.3mm$