Question

In a Young’s double slit experiment, the fringe width is found to be $0.4\text{mm}$. If the whole apparatus is immersed in water of refractive index $\left(4/3\right)$, without disturbing the geometrical arrangement, the new fringe width will be -

• A. $0.30\text{mm}$
• B. $0.40\text{mm}$
• C. $0.53\text{mm}$
• D. $450$ microns

Answer 1

The correct option is A $0.30\text{mm}$

Fringe width $\beta =0.4\text{mm}$
${\mu }_w=4/3$
Fringe width in water $=?$

We know that $\beta =\frac{D\lambda }{d}$ i.e. $\beta \propto \lambda$

${\lambda }_w=\frac{{\lambda }_{air}}{{\mu }_w}=\frac{\lambda }{4/3}=\frac{3}{4}\lambda .$

$\frac{{\beta }_1}{{\beta }_w}=\frac{{\lambda }_1}{{\lambda }_w}$

$\Rightarrow$ $\frac{0.4}{{\beta }_w}=\frac{\lambda }{\frac{3}{4}\lambda }$
${\beta }_w=0.3\text{mm}$