Question

For the angle of minimum deviation of a prism is equal to its refracting angle, the prism must be made of a material whose refractive index:
(A) Between 2 and $\sqrt{2}$
(B) Is less than 1
(C) Is greater than 2
(D) Lies between $\sqrt{2}$ and 1

Answer 1

Hint : We can find the refractive index corresponding to minimum deviation by using a prism formula.
$n={\displaystyle \frac{\sin {\displaystyle \frac{(A+S_m)}{2}}}{\sin {\displaystyle \frac{A}{2}}}}$
Here, n is refractive index of material,
A is the angle of the prism.
$S_m$ is the angle of minimum deviation
Use the condition, refractive index is minimum when angle of prism is $$\text{9}0{}^{\circ }$$
Refractive index is maximum when the angle of the prism is $$0{}^{\circ }$$.

Complete step by step answer
The prism formula is given by
$n={\displaystyle \frac{\sin {\displaystyle \frac{(A+S_m)}{2}}}{\sin {\displaystyle \frac{A}{2}}}}$
Here n is a refractive index.
We know deviation will be minimum, angle of incidence is equal to angle of prism.
$S_m=2i-A$
$S_m=\text{2}AA=A$
From prism formula, put the value of minimum deviation,
n = $n={\displaystyle \frac{\sin {\displaystyle \frac{(A+A)}{2}}}{\sin {\displaystyle \frac{A}{2}}}}={\displaystyle \frac{\sin A}{\sin {\displaystyle \frac{A}{2}}}}$
Use$\left[\text{sin A=2 sin}{\displaystyle \frac{A}{2}}\text{cos }{\displaystyle \frac{A}{2}}\right]$
The refractive index is given by,
$n={\displaystyle \frac{2\mathrm{Sin}{\displaystyle \frac{A}{2}}\mathrm{Cos}{\displaystyle \frac{A}{2}}}{\mathrm{Sin}{\displaystyle \frac{A}{2}}}}$
$n=2\mathrm{Cos}{\displaystyle \frac{A}{2}}$
This is the formula for prism,
Now, we have to find a minimum refractive index.
Angle of prism A varies from $$0{}^{\circ }$$to $$\text{9}0{}^{\circ }$$
For $n_{min}$, minimum value of refractive index is given by
Put $$\text{A}=\text{9}0{}^{\circ }$$
       $n=2\mathrm{Cos}{\displaystyle \frac{90{}^{\circ }}{2}}$ Use$[\text{Cos 45}{}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}]$
          = $2\mathrm{Cos}45{}^{\circ }$
$n_{min}=2\times {\displaystyle \frac{1}{\sqrt{2}}}\text{ = }\sqrt{2}$
For$n_{max}$, maximum value of refractive index
        A= 0°
$$\text{n}=\text{2 Cos }0{}^{\circ }=\text{2}$$ (1) $$[\text{Cos }0{}^{\circ }=\text{ 1}]$$
$n_{max}=2$
Hence, the value of refractive index varies between 2 and $\sqrt{2}$

Note
When refracting angle is small, the deviation is calculated from,
S =$$(n-1)$$A
When refraction angle is bigger for the prism the deviation is calculated from, S = $(i_1+i_2)-A$where $i_1$and $i_2$are angle of incidence on different faces of prism.