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 $ $1.732=\frac{sin A}{sin ⁡ \frac{A}{2}}$
$\therefore $ $A=60^\circ $