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 lies between:

• A. $\sqrt{2}and2$
• B. $1and\sqrt{2}$
• C. $\sqrt{3}and2$
• D. $\sqrt{5}and2$

Answer 1

The correct option is A $\sqrt{2}and2$

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

In the given situation,

${\mu }_{min}=2\cos \left(\frac{90°}{2}\right)=\sqrt{2}$

${\mu }_{max}=2\cos \left(0°\right)=2$