91Ó°ÊÓ

We -45°-45°-90° as shown in the figure below. The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by the speed of light in the material. The prism is dropped into the water and as the ray is totally reflected, therefore, the incident angle is 90°.

Now, we can use Snell’s law in the next form

$n_{glass}\sin {\mathrm{θ}}_{glass}=n_{water}\sin {\mathrm{θ}}_{water}$

Where ${\mathrm{θ}}_{glass}$represents the critical angle which is 45° and . $n_{water}=1.33$. Now, solve Snell’s law for $n_{glass}$and plug the values for $n_{water},{\mathrm{θ}}_{glass},{\mathrm{θ}}_{water}$.

$$\begin{gathered} n_{glass}=\frac{n_{water}\sin {\mathrm{θ}}_{water}}{{\mathrm{θ}}_{glass}} \\ =\frac{\left(1.33\right)\left(\sin 90°\right)}{\sin 45°} \\ =1.88 \end{gathered}$$

Hence the refractive index of glass is 1.88.