When the Young's double slit apparatus is immersed in a medium of refractive index $\mu$, the wavelength of light decreases to:
$$\lambda' = \frac{\lambda}{\mu}$$
Since fringe width $\beta = \frac{\lambda D}{d}$, the new fringe width in the medium becomes:
$$\beta' = \frac{\beta}{\mu}$$
Given $\beta = 0.4\,\text{mm}$ and $\mu = \frac{4}{3}$:
$$\beta' = \frac{0.4}{4/3} = 0.4 \times \frac{3}{4} = 0.3\,\text{mm}$$
Hence, option 3 is the correct answer.
जब उपकरण को $\mu$ अपवर्तनांक वाले माध्यम में डुबोया जाता है, तो प्रकाश की तरंगदैर्घ्य घट जाती है:
$$\lambda' = \frac{\lambda}{\mu}$$
अतः माध्यम में नई फ्रिंज चौड़ाई:
$$\beta' = \frac{\beta}{\mu}$$
मान रखने पर:
$$\beta' = \frac{0.4}{4/3} = 0.4 \times \frac{3}{4} = 0.3\,\text{mm}$$
अतः विकल्प 3 सही उत्तर है।
