Question

A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light $\left(\lambda =700\mathrm{nm}\mathrm{in}\mathrm{vacuum}\right)$. Find the fringe-width of the pattern formed on the screen.

Answer 1

Given:
Separation between two slits, $d=0.28\mathrm{mm}=0.28\times {10}^{-3}\mathrm{m}$
Distance between screen and slit (D) = 48 cm = 0.48 m
Wavelength of the red light, ${\lambda }_a=700\mathrm{nm}\mathrm{in}\mathrm{vaccum}=700\times {10}^{-9}\mathrm{m}$
Let the wavelength of red light in water = ${\lambda }_{\omega }$
We known that refractive index of water (μ_w =4/3),
μ_w = $\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{vacuum}}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{water}}$
$$\begin{gathered} \mathrm{So},{\mathrm{\mu }}_{\mathrm{w}}=\frac{v_a}{v_{\omega }}=\frac{{\lambda }_a}{{\lambda }_{\omega }} \\ \Rightarrow \frac{4}{3}=\frac{{\lambda }_a}{{\lambda }_{\omega }} \\ \Rightarrow {\lambda }_{\omega }=\frac{3{\lambda }_a}{4}=\frac{3\times 700}{4}=525\mathrm{nm} \end{gathered}$$
So, the fringe width of the pattern is given by
$$\begin{gathered} \mathrm{\beta }=\frac{{\lambda }_{\omega }D}{d} \\ =\frac{525\times {10}^{-9}\times \left(0.48\right)}{\left(0.28\right)\times {10}^{-3}} \\ =9\times {10}^{-4}=0.90\mathrm{mm} \end{gathered}$$

Hence, fringe-width of the pattern formed on the screen is 0.90 mm.