Question

What is speed of light in a denser medium of Polarising angle ${30}^{\circ }$?

Answer 1

Step 1: Given data

Polarising angle, $\mathrm{\theta }={30}^{\circ }$

Let the speed of light be $\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

Step 2: Formula used

$$\begin{gathered} \mathrm{refractive}\mathrm{index}=\tan \left(\mathrm{polarising}\mathrm{angle}\right) \\ \mathrm{\mu }=\mathrm{tan\theta } \end{gathered}$$

$$\begin{gathered} \mathrm{refractive}\mathrm{index}=\frac{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{vacuum}}{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{denser}\mathrm{medium}} \\ \mathrm{\mu }=\frac{\mathrm{c}}{\mathrm{v}} \end{gathered}$$

Step 3: Calculation for the refractive index

$$\begin{gathered} \mathrm{refractive}\mathrm{index}=\tan \left(\mathrm{polarising}\mathrm{angle}\right) \\ \mathrm{\mu }=\mathrm{tan\theta } \\ \mathrm{\mu }=\tan {30}^{\circ } \\ \mathrm{\mu }=\frac{1}{\sqrt{3}} \end{gathered}$$

Step 4: Calculation for the speed of light in a denser medium

$$\begin{gathered} \mathrm{refractive}\mathrm{index}=\frac{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{vacuum}}{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{denser}\mathrm{medium}} \\ \mathrm{\mu }=\frac{\mathrm{c}}{\mathrm{v}} \\ \mathrm{v}=\frac{\mathrm{c}}{\mathrm{\mu }} \\ \mathrm{v}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{{\displaystyle \frac{1}{\sqrt{3}}}} \\ \mathrm{v}=\sqrt{3}\times 3\times {10}^8 \\ \mathrm{v}=5.2\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of light in a denser medium is $5.2\times {10}^8\mathrm{m}/\mathrm{s}$