Question

A young's double slit apparatus has slits separated by $0.25\text{mm}$ and a screen $48\text{cm}$ away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light $(\lambda =700\text{nm in vaccuum})$. Find the fringe width of the pattern formed on the screen $\left(\text{in mm}\right)$
$({\mu }_w=\frac{4}{3})$(Integer only)

Answer 1

Wavelength of light in water,
${\lambda }^{{}^{\prime }}=\frac{\lambda }{\mu }$

Fringe width, $\beta =\frac{{\lambda }^{{}^{\prime }}D}{d}=\frac{\lambda D}{\mu d}$

$\beta =\frac{(700\times {10}^{-9})\left(0.48\right)}{\left(\frac{4}{3}\right)(0.25\times {10}^{-3})}={10}^{-3}\text{m}=1\text{mm}$

Hence, $1$ is the correct answer.