Question

In a Young's double slit experiment for interference of light, the slits are $0.2\text{cm}$ apart and are illuminated by yellow light$(\lambda =600\text{nm})$. What would be the fringe width on a screen placed $1\text{m}$ from the plane of slits if the whole system is immersed in water of refractive index $\frac{4}{3}.$

• A. $0.225\text{mm}$
• B. $2.25\text{mm}$
• C. $0.125\text{mm}$
• D. $1.25\text{mm}$

Answer 1

The correct option is A $0.225\text{mm}$

Given,
$\lambda =600\text{nm}=6\times {10}^{-7}\text{m}$
$D=1\text{m}$
$d=0.2\text{cm}=2\times {10}^{-3}\text{m}$

As the apparatus is dipped in water$(\mu =\frac{4}{3})$ :
${\lambda }_{new}=\frac{\lambda }{\mu }=\frac{6\times {10}^{-7}}{\frac{4}{3}}\text{m}=4.5\times {10}^{-7}\text{m}$

$\Rightarrow {\beta }_{new}=\frac{{\lambda }_{new}D}{d}=\frac{4.5\times {10}^{-7}\times 1}{2\times {10}^{-3}}=2.25\times {10}^{-4}\text{m}=0.225\text{mm}$

Hence, $\left(C\right)$ is the correct answer.