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. $0.400\text{mm}$
• C. $0.125\text{mm}$
• D. $0.600\text{mm}$

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 $(\mu =\frac{4}{3})$

Since, $\frac{{\mu }_1}{{\mu }_2}=\frac{{\lambda }_2}{{\lambda }_1}$

${\lambda }_{\text{new}}=\frac{\lambda }{\mu }=\frac{6\times {10}^{-7}}{\left(\frac{4}{3}\right)}\text{m}=4.5\times {10}^{-7}\text{m}$

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

Hence, option (b) is the correct answer.