Fringe width in Young’s double slit experiment

The fringe width is given by $\beta = \frac{\lambda D}{d}$ where:

• $\lambda$ = wavelength in the medium
• $D = 0.72\,\text{m}$
• $d = 1.5\,\text{mm} = 1.5 \times 10^{-3}\,\text{m}$

Wavelength in the liquid

Since the apparatus is immersed in a liquid of refractive index $\mu = 1.44$, $\lambda' = \frac{\lambda_{air}}{\mu}$

Given: $\lambda_{air} = 690\,\text{nm} = 690 \times 10^{-9}\,\text{m}$

Therefore, $\lambda' = \frac{690 \times 10^{-9}}{1.44}$ $\lambda' \approx 479.17 \times 10^{-9}\,\text{m}$

Calculate fringe width

$\beta = \frac{\lambda' D}{d}$ $\beta = \frac{(479.17 \times 10^{-9})(0.72)}{1.5 \times 10^{-3}}$

First, $(479.17 \times 10^{-9})(0.72) = 345.0 \times 10^{-9}$

Now divide by $1.5 \times 10^{-3}$: $\beta = \frac{345.0 \times 10^{-9}}{1.5 \times 10^{-3}}$ $\beta = 230 \times 10^{-6}\,\text{m}$ $\beta = 2.30 \times 10^{-4}\,\text{m}$

Convert to mm

$\beta = 2.30 \times 10^{-4}\,\text{m} = 0.23\,\text{mm}$

Match with options

The correct option is: $\boxed{\text{D: }0.23\,\text{mm}}$