Question

For any prism, prove that $'n'$ or $\mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$ where the terms have their usual meaning.

Answer 1

Hint: In this solution we will use Snell's law which states that when a ray incident on the substance and refracted from the substance then the sines of the incidence angle and refracted angle will be constant in the given medium.


Complete step by step solution
Let us assume that the refractive index of the prism is $\mu $.

Express the formula of Snell’s law.

${n_{air}}\sin i = \mu \;\sin r$ …….(1)

Here,

${n_{air}}$ is the incident index of air which is equal to $1$.
$\mu $ is the refracted index.
$i$ is the incident angle.
$r$ is the refracted angle.

Express the formula of deviation by prism.

$\delta = i + e - A$

Where, $A = {r_1} + {r_2}$

Here,

$A$ is the angle of prism.
$\delta $ is the deviation of prism.
${r_1}$ and ${r_2}$ are the refracted angle for medium $1$ and $2$.

The conditions for minimum deviation can be given as,

 $i = e$ and ${r_1} = {r_2}$

Therefore,

  $\begin{array}{c}
A = 2{r_1}\\
{r_1} = \dfrac{A}{2}
\end{array}$ ……(2)

The angle of minimum deviation is given by,

$\begin{array}{c}
{\delta _m} = 2i - A\\
i = \dfrac{{A + {\delta _m}}}{2}
\end{array}$ …….(3)

Substitute the values of equation (2) and (3) in equation (1) to obtain the refractive index.
$\mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$


Note: While solving this type of formula we will use the minimum deviation condition otherwise we will not be able to meet with the desired solution.