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 $\mu$ :

• A. $\mu <1$
  
• B. $1<\mu <\sqrt{2}$
• C. $\sqrt{2}<\mu <2$
• D. $\mu >\sqrt{2}$

Answer 1

The correct option is C $\sqrt{2}<\mu <2$

In minimum deviation case,

$\hat{i}=e$

and,

$A+ 5 \ min= \hat {i}+e$given,$A= \hat {i} \ (as\ 5\ min =A)$

Now, angle of incidence can vary from $0^o$ to ${90}^o$.

thus, A varies $0^o\to {90}^o\to \left(1\right)$

From minimum deviation formulas,

$\mu =\frac{\sin \left(\frac{A+5min}{2}\right)}{s\in \left(A/2\right)}$

$A=5min$ given :

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

So, $\mu =\cos A/2$

When $A=0^o$ (minimum), $\mu =2$ (maximum )

Also $A={90}^o$ (max) given $\mu =2cos\left(\frac{{90}^o}{2}\right)=\sqrt{2}$ (min)

$\therefore$ Range for $\mu :\sqrt{2}<\mu <2$.