Question

Young's double-slit experiment is first performed in air and then in a medium other than air. It is found that the $8^{\mathrm{th}}$ bright fringe in the medium lies where the $5^{\mathrm{th}}$ dark fringe lies in the air. The refractive index of the medium is near:

Answer 1

Step 1: Given and assumption

Let $\mathrm{\lambda }$ be the wavelength, $\mathrm{\mu }$ be the refractive index, d is fringe width and D be the distance of slits from the screen. The eighth bright fringe in the medium lies in the same area where the fifth dark fringe lies in the air.

Step 1: Finding the refractive index

In a different medium, we know that,

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{medium}}=\frac{{\mathrm{\lambda }}_{\mathrm{air}}}{\mathrm{\mu }} \\ \mathrm{\lambda }\text{'}=\frac{\mathrm{\lambda }}{\mathrm{\mu }} \end{gathered}$$

For air, $5^{\mathrm{th}}$ dark fringe,

$$\begin{gathered} \mathrm{y}=(2\mathrm{m}-1)\frac{\mathrm{\lambda D}}{2\mathrm{d}} \\ \mathrm{m}=5 \\ \mathrm{y}=\left(2\right(5)-1)\frac{\mathrm{\lambda D}}{2\mathrm{d}} \\ \mathrm{y}=\frac{9\mathrm{\lambda D}}{2\mathrm{d}} \end{gathered}$$

Similarly for bright fringe in medium,

$$\begin{gathered} \mathrm{y}=\mathrm{n}\frac{\mathrm{\lambda }\text{'}\mathrm{D}}{\mathrm{d}} \\ \mathrm{\lambda }\text{'}=\frac{\mathrm{\lambda }}{\mathrm{\mu }} \\ \mathrm{y}=\frac{8\mathrm{\lambda D}}{\mathrm{\mu d}} \end{gathered}$$

Now, equating the fringe widths,

$$\begin{gathered} \frac{9\mathrm{\lambda D}}{2\mathrm{d}}=\frac{8\mathrm{\lambda D}}{\mathrm{\mu d}} \\ \mathrm{\mu }=\frac{16}{9} \\ \mathrm{\mu }=1.78 \end{gathered}$$

Hence, the refractive index is equal to $1.78$.