A double-slit apparatus is immersed in a liquid of refractive index \( \mu = \displaystyle\frac{4}{3} \). It has slit separation of \( 10^{-3}\,\mathrm{m} \) and distance between the plane of slits and screen is 133 cm. The slits are illuminated by a parallel beam of light of wavelength 630 nm in air. What is the fringe width?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.63 mm
The fringe width in Young's double-slit experiment is \( \beta = \frac{\lambda D}{d} \), where \( \lambda \) is the wavelength in the medium in which the slits and screen are placed. Light of wavelength \( \lambda_0 \) in air has the shorter wavelength \( \lambda = \frac{\lambda_0}{\mu} \) in the liquid.
\( \displaystyle \lambda = \frac{630\,\mathrm{nm}}{4/3} = 472.5\,\mathrm{nm} = 4.725\times10^{-7}\,\mathrm{m} \)
Distance to the screen: \( D = 133\,\mathrm{cm} = 1.33\,\mathrm{m} \).
\( \displaystyle \beta = \frac{\lambda D}{d} = \frac{4.725\times10^{-7}\,\mathrm{m}\times1.33\,\mathrm{m}}{10^{-3}\,\mathrm{m}} \)
\( \beta = 6.28\times10^{-4}\,\mathrm{m} \approx 0.63\,\mathrm{mm} \)