Question

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

Answer 1

Firstly we need to draw a diagram to understand the situation clearly. We will use the equation to find the refractive index of a prism and substitute the given situation of angle of minimum deviation in the expression and hence obtain an equation for the refractive index. Then we will apply the limit from $0{}^\circ$ to $90{}^\circ$ and obtain the range of refractive index which we could have using the given angle of minimum deviation.

Formula used:
$\mu =\dfrac{\sin \dfrac{\left( A+{{\delta }_{m}} \right)}{2}}{\sin \dfrac{A}{2}}$

Complete step by step answer:
We will draw the diagram using the parameters, angle of prism as ($A$), angle of incidence ($i$), angle of refraction ($r$) and angle of minimum deviation (${{\delta }_{m}}$).

It is given that angle of minimum deviation to be equal to the refracting angle.
$\Rightarrow {{\delta }_{m}}=A$
Now, we will substitute this in the equation to find the refractive index of prism and it will become,
$\mu =\dfrac{\sin \dfrac{\left( A+A \right)}{2}}{\sin \dfrac{A}{2}}=\dfrac{\sin \left( A \right)}{\sin \dfrac{A}{2}}$
Now, we will use the trigonometric relation $\sin \left( x \right)=2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)$.
$\Rightarrow \mu =\dfrac{\sin \left( A \right)}{\sin \dfrac{A}{2}}=\dfrac{2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}{\sin \left( \dfrac{A}{2} \right)}$
$\mu =2\cos \left( \dfrac{A}{2} \right)$

So, we have the expression for a refractive index.
Now we will apply the limit for angle of prism from $0{}^\circ$ to $90{}^\circ$. Then, if $A$ is $0{}^\circ$,
$\mu =2\cos \left( \dfrac{0}{2} \right)=2$
And if $A$ is $90{}^\circ$, $\mu =2\cos \left( \dfrac{90}{2} \right)= 2\cos \left( 45 \right)=\sqrt{2}$
So, we can conclude that the prism could have a refractive index which lies between $2$ and $\sqrt{2}$.

So, the correct answer is “Option B”.

Note:
The soul of this question is situated at the part where we are applying the limit for $A$.
There we are taking the angles $0{}^\circ$ to $90{}^\circ$ because they are the maximum and minimum angles which a prism could have. We are taking ${{\delta }_{m}}=A$ because if the minimum deviation is equal to refracting angle, by geometry, the angle of prism and the refracting angles will be equal.