JEE Main 2025 Jan 24 Shift 2, Physics Q1: Young’s Double Slit Experiment

When a YDSE apparatus is immersed in a medium of refractive index $\mu$, the wavelength inside the medium becomes $\lambda =\frac{{\lambda }_0}{\mu }$. The fringe-width is therefore:

$\beta =\frac{\lambda D}{d}=\frac{{\lambda }_{0}D}{\mu d}$

Substituting ${\lambda }_0=690\text{nm}=690\times 10^{-9}\text{m}$, $D=0.72\text{m}$, $\mu =1.44$, $d=1.5\text{mm}=1.5\times 10^{-3}\text{m}$:

$\beta =\frac{(690\times 10^{-9})(0.72)}{(1.44)(1.5\times 10^{-3})}=\frac{4.968\times 10^{-7}}{2.16\times 10^{-3}}\approx 2.3\times 10^{-4}\text{m}=0.23\text{mm}$

Correct option: (1)