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 $\mu$ is :

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

Answer 1

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

R.E.F image

i = angle of incidence

e = angle of emergence

In minimum deviation case,

i = e

and ,

A+Smin = i+e given, A = i (as Smin = A)

Now,angle of incidence can vary from $0^{\circ }$ to ${90}^{\circ }$

thus, A varies $0^{\circ }\to {90}^{\circ }$ $\to$ (1)

From minimum deviation formula,

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

A = Smini given

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

so, $\mu =2cosA/2$

when $A=0^{\circ }$ (minimum),$\mu =2$ (maximum)

also $A={90}^{\circ }$ (max) gives $\mu =2cos\left(\frac{{90}^{\circ }}{2}\right)=\sqrt{2}\left(min\right)$

$\therefore$ Range for $\mu :\sqrt{2}<\mu <2$ (Ans)