Question

Angle of minimum deviation is equal to the angle of prism A of an equilateral glass prism. The angle of incidence at which minimum deviation will be obtained is

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. $si{n}^{-1}\left(\frac{2}{3}\right)$
• D. ${60}^{\circ }$

Answer 1

The correct option is D

${60}^{\circ }$

$A={\delta }_m={60}^{\circ }$
At minimum deviation $i=\left(\frac{A+{\delta }_m}{2}\right)={60}^{\circ }$

We know that,
${\delta }_{min}=2i-A\Rightarrow i=\left(\frac{A+{\delta }_{min}}{2}\right)$
Now, ${\delta }_{min}=A={60}^{\circ }$ (ABC is an equilateral $\Delta$)
$i=\frac{A+{\delta }_{min}}{2}=\frac{{60}^{\circ }+{60}^{\circ }}{2}={60}^{\circ }$
Here, $i={60}^{\circ }$