Question

For a prism of refractive index, $1.732$ , the angle of minimum deviation is equal to the angle of prism. Then the angle of the prism is

• A. $60^{\circ }$
• B. $70^{\circ }$
• C. $50^{\circ }$
• D. none of these

Answer 1

Answer: (A)

We know $\mu$= $\frac{\sin \left(\frac{{\delta }_{\min }+A}{2}\right)}{\sin \frac{A}{2}}$
$A={\delta }_{min}$
$\therefore$ $1.732=\frac{sinA}{sin\frac{A}{2}}$
$\therefore$ $A=60^{\circ }$