a) Define Snell's Law. Using a neat, labelled diagram derive an expression for the refractive index of the material of an equilateral prism.
b) A ray of light, after passing through a medium, meets the surface separating the medium from air at an angle of 45° and is not just refracted. Find the refractive index of medium ?

a) Snell's Law: The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. The angles of incidence (i) and refraction (r) are the angles that the incident and its refracted ray make with the normal.

$\frac{\sin \mathrm{i}}{\sin \mathrm{r}}{=}_1{\mathrm{\mu }}_2=\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1}\Rightarrow \sin \mathrm{i}={\mathrm{\mu }}_2\sin \mathrm{r}$
${\mathrm{\mu }}_1^2$ is called is a constant, called the refractive index of the second medium with respect to the first medium is known as Snell's law of refraction.
i) Let ABC be the principal section of a prism. Its refracting angle is A. The refractive index of the material of the prism is μ.
The prism is kept in air which is a rarer medium with respect to the material of the prism.
ii) Let PQ be the incident ray on the first refracting surface AB of the prism. 
When it enters into the prism it bends towards the normal TN₁ as ray QR, as the light ray travels from rarer to denser medium. Let the angle of incidence be i₁ and the angle of refraction be r₁.
iii) After traveling through the prism, it falls on the second refracting surface AC. The light ray bends away from the normal TN₂ and travels along RS which is called emergent ray.
iv) The angle between the normal and the emergent ray is called angle of emergence denoted by i₂. The angle between the normal TN₂ and the refracted ray is r₂.
v) Angle of deviation is also defined as the angle between the incident ray produced forwards and the emergent ray produced backwards. This is shown in Fig.

$\text{ vi) In the }\mathrm{\Delta }\text{ QRT }{\mathrm{r}}_1+{\mathrm{r}}_2+\widehat{\mathrm{T}}={180}^{\circ }\text{. ----(1)}$
In the quadrilateral AQTR
$\begin{array}{l}\mathrm{A}+{90}^{\circ }+\mathrm{T}+{90}^{\circ }={360}^{\circ }\\ \Rightarrow \mathrm{T}={180}^{\circ }-\mathrm{A}\to \left(2\right)\end{array}$
Substituting in (2) in (1)
${\mathrm{r}}_1+{\mathrm{r}}_2+{180}^{\circ }-\mathrm{A}={180}^{\circ }\Rightarrow {\mathrm{r}}_1+{\mathrm{r}}_2=\mathrm{A}$
In the $∆$ QUR
$\begin{array}{l}{\mathrm{i}}_1-{\mathrm{r}}_1+{\mathrm{i}}_2-{\mathrm{r}}_2+{180}^{\circ }-\mathrm{D}={180}^{\circ }\\ \left({\mathrm{i}}_1+{\mathrm{i}}_2\right)-\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)=\mathrm{D}\\ {\mathrm{i}}_1+{\mathrm{i}}_2-\mathrm{A}=\mathrm{D}\Rightarrow {\mathrm{i}}_1+{\mathrm{i}}_2=\mathrm{A}+\mathrm{D}\end{array}$
vii) Minimum Deviation : As the angle of incidence is increased gradually, it is found experimentally that the angle of deviation decreases till it reaches a minimum value and then it increases. This least value of deviation is called angle of minimum deviation denoted by $\mathrm{\delta }$.
viii)The graph between the angle of incidence (i) and the angle of deviation (D) is as shown in Fig. When D decreases the two angles i₁ and i₂ become closer to each other and at the angle of minimum deviation, the two angles of incidence are same i.e., i₁= i₂ then r₁= r₂, taking i₁= i₂= i, r₁= r₂= r, then
$\begin{array}{l}2\mathrm{r}=\mathrm{A}\Rightarrow \mathrm{r}=\frac{\mathrm{A}}{2}\\ \mathrm{i}+\mathrm{i}=\mathrm{A}+\mathrm{\delta }\\ \Rightarrow \mathrm{i}=\frac{\mathrm{A}+\mathrm{\delta }}{2}\end{array}$

ix) By Snell’s law of refraction, the refractive index of the material of the prism is
$\mathrm{\mu }=\frac{\sin {\mathrm{i}}_1}{\sin {\mathrm{r}}_1}=\frac{\sin {\mathrm{i}}_2}{\sin {\mathrm{r}}_2}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\left(\because {\mathrm{i}}_1={\mathrm{i}}_2=\mathrm{i}\text{ and }{\mathrm{r}}_1={\mathrm{r}}_2=\mathrm{r}\right)$
$\mathrm{\mu }=\frac{\sin \left(\frac{\mathrm{A}+\mathrm{\delta }}{2}\right)}{\sin \frac{\mathrm{A}}{2}}$
It is called prism formula.
b) As the light ray is travelling from denser to rarer medium
$\begin{array}{l}\mathrm{\mu }=\frac{\sin \mathrm{r}}{\sin \mathrm{i}}=\frac{\sin {90}^{\circ }}{\sin {45}^{\circ }}\\ \mathrm{\mu }=\frac{1}{1/\sqrt{2}}=\sqrt{2}=1.414\end{array}$