A double-slit experiment is immersed in a liquid of refractive index 1.33. The separation between the slits is 1.0 mm and the distance between the slits and the screen is 1.33 m. If the slits are illuminated by a parallel beam of light of wavelength 6300 Å, the fringe width will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.63 mm
In a medium of refractive index \( n \), the wavelength of light becomes \( \lambda/n \). The slit separation and the screen distance do not change. So the fringe width \( \beta=\lambda D/d \) becomes \( \lambda D/(nd) \) in the liquid.
In SI units the data are:
\( \lambda=6300\,\mathrm{\mathring{A}}=6.3\times10^{-7}\,\mathrm{m} \)
\( D=1.33\,\mathrm{m} \)
\( d=1.0\,\mathrm{mm}=1.0\times10^{-3}\,\mathrm{m} \)
\( n=1.33 \)
The fringe width in the liquid is:
\( \begin{aligned} \beta' &= \frac{\lambda D}{nd} \\ &= \frac{6.3\times10^{-7}\,\mathrm{m}\times1.33\,\mathrm{m}}{1.33\times1.0\times10^{-3}\,\mathrm{m}} \\ &= 6.3\times10^{-4}\,\mathrm{m} \\ &= 0.63\,\mathrm{mm} \end{aligned} \)