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.
$n=\dfrac{\sin \dfrac{(A+{{S}_{m}})}{2}}{\sin \dfrac{A}{2}}$
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
$n=\dfrac{\sin \dfrac{(A+{{S}_{m}})}{2}}{\sin \dfrac{A}{2}}$
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 = $n=\dfrac{\sin \dfrac{(A+A)}{2}}{\sin \dfrac{A}{2}}=\dfrac{\sin \text{ }A}{\sin \dfrac{A}{2}}$
Use$\left[ \text{sin A=2 sin}\dfrac{A}{2}\text{cos }\dfrac{A}{2} \right]$
The refractive index is given by,
$n=\dfrac{2\operatorname{Sin}\dfrac{A}{2}\operatorname{Cos}\dfrac{A}{2}}{\operatorname{Sin}\dfrac{A}{2}}$
$n=2\operatorname{Cos}\dfrac{A}{2}$
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 \[\text{A}=\text{9}0{}^\circ \]
       $n=2\operatorname{Cos}\dfrac{90{}^\circ }{2}$ Use$\left[ \text{Cos 45}{}^\circ =\dfrac{1}{\sqrt{2}} \right]$
          = $2\operatorname{Cos}45{}^\circ $
${{n}_{\min }}=2\times \dfrac{1}{\sqrt{2}}\text{ = }\sqrt{2}$
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 =\[\left( n-1 \right)\]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.