Question

Angle of minimum deviation for a prism of refractive index $1.5$ is equal to the angle of prism of given prism. Then the angle of prism is __________ $(\sin {48}^{\circ }{36}^{\prime }=0.75)$.

• A. ${41}^{\circ }{24}^{\prime }$
• B. ${80}^{\circ }$
• C. ${60}^{\circ }$
• D. ${82}^{\circ }{48}^{\prime }$

Answer 1

The correct option is D ${82}^{\circ }{48}^{\prime }$

$\mu =\frac{sin\frac{A+{\delta }_m}{2}}{sin\frac{A}{2}}$

$\because A={\delta }_m,$ $\mu =\frac{sin\frac{2A}{2}}{sin\frac{A}{2}}$

$\therefore \mu =2cos\frac{A}{2}$ ( $sin2x=2sinxcosx$)

$\therefore \frac{1.5}{2}=cos\frac{A}{2}$

$\therefore \frac{A}{2}=co{s}^{-1}0.75={41}^o{24}^{\prime }$

$\therefore A={82}^o{48}^{\prime }$