Question

For the angle of minimum deviation of a prism to be equal to prism 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 $1$.
• D. is greater than $2$.

Answer 1

The correct option is B lies between $2$ and $\sqrt{2}$.

Let the angle of incidence, angle of emergent and angle of prism are $i,e$ and $A$ respectively.

Deviation produced by prism is,
$\delta =i+e-A$ ...(i)

For minimum deviation,
$\delta ={\delta }_{min}$ when $i=e$
$\Rightarrow {\delta }_{min}=2i-A$

Since, ${\delta }_{min}=A$ (given)
$\Rightarrow A=2i-A$
or, $2A=2i$
$\therefore i=A$

At minimum deviation,

$\mu =\frac{\sin \left(\frac{A+{\delta }_{min}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow \frac{\sin \left(\frac{2A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}$

or, $\mu =2\cos \left(\frac{A}{2}\right)$

$\Rightarrow \mu =2\cos \left(\frac{i}{2}\right)$

Maximum possible angle of incidence is $i={90}^{\circ }$
Thus,
${\mu }_{min}=2\cos \left(\frac{i_{max}}{2}\right)$

or, ${\mu }_{min}=2\cos {45}^{\circ }=\sqrt{2}$

Similarly, ${\mu }_{max}=2\cos \left(\frac{i_{min}}{2}\right)$

$\Rightarrow {\mu }_{max}=2\cos 0^{\circ }=2$

$\therefore$ Refractive index lies between $\sqrt{2}$ and $2$.

Why this question? Tip: While dealing with the relation $\mu =2\cos \frac{i}{2},$ it should be noted that ${\mu }_{max}$ will occur when angle of incidence is minimum. It holds because $cos\theta$ is a decreasing function in the range $[0,\frac{\pi }{2}]$.