In Young's double slit experiment when light of wavelength 600 nm is used the fringe width is 0.1 mm. Calculate the fringe width when the entire apparatus is immersed in a medium of refractive index \( \displaystyle\frac{4}{3} \) by keeping the source of light in air
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.075 mm
The fringe width is \( \beta = \lambda D/d \). In a medium of refractive index \( \mu \) the frequency of light stays the same but its wavelength becomes \( \lambda/\mu \), while \( D \) and \( d \) do not change.
\( \displaystyle \beta' = \frac{(\lambda/\mu)\,D}{d} = \frac{\beta}{\mu} \)
\( \displaystyle \beta' = \frac{0.1\,\mathrm{mm}}{4/3} = 0.1 \times \frac{3}{4}\,\mathrm{mm} = 0.075\,\mathrm{mm} \)