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

Hint: 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:
$$\mu ={\displaystyle \frac{\sin {\displaystyle \frac{(A+{\delta }_m)}{2}}}{\sin {\displaystyle \frac{A}{2}}}}$$

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,
$$\mu ={\displaystyle \frac{\sin {\displaystyle \frac{(A+A)}{2}}}{\sin {\displaystyle \frac{A}{2}}}}={\displaystyle \frac{\sin \left(A\right)}{\sin {\displaystyle \frac{A}{2}}}}$$
Now, we will use the trigonometric relation $$\sin \left(x\right)=2\sin \left({\displaystyle \frac{x}{2}}\right)\cos \left({\displaystyle \frac{x}{2}}\right)$$.
$$\Rightarrow \mu ={\displaystyle \frac{\sin \left(A\right)}{\sin {\displaystyle \frac{A}{2}}}}={\displaystyle \frac{2\sin \left({\displaystyle \frac{A}{2}}\right)\cos \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}}$$
$$\mu =2\cos \left({\displaystyle \frac{A}{2}}\right)$$

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 $$0{}^{\circ }$$,
$$\mu =2\cos \left({\displaystyle \frac{0}{2}}\right)=2$$
And if $$A$$ is $$90{}^{\circ }$$, $$\mu =2\cos \left({\displaystyle \frac{90}{2}}\right)=2\cos \left(45\right)=\sqrt{2}$$
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.