Answer:
[tex]\large\text{$48.59^\circ$}[/tex]
Explanation:
[tex]\large\text{Solution: $}[/tex]
[tex]\large\text{Minimum angle of deviation $(\delta_m)= ?$}[/tex]
[tex]\large\text{Angle of incidence$(i)$ = ?}[/tex]
[tex]\large\text{Refractive index of glass$(\mu)$ = 1.5}[/tex]
[tex]\large\text{Angle of deviation$(A)\ = \ 60^\circ$ because the prism is equilateral.}[/tex]
[tex]\large\text{Using formula,$}[/tex]
[tex]\large\text{$\dfrac{\sin i}{\sin r}=\mu$}[/tex]
[tex]\large\text{In the case of minimum deviation,$}[/tex]
[tex]\large\text{$\dfrac{\sin \bigg(\dfrac{A+\delta_m}{2}\bigg)}{\sin \bigg(\dfrac{A}{2}\bigg)}=\mu$}[/tex]
[tex]\large\text{Substitute the given values:$}[/tex]
[tex]\large\text{$\dfrac{\sin \bigg(\dfrac{60^\circ+\delta_m}{2}\bigg)}{\sin\bigg(\dfrac{60}{2}\bigg)^\circ}=1.5$}[/tex]
[tex]\large\text{$\dfrac{\sin\bigg(30^\circ+\dfrac{\delta_m}{2}\bigg)}{\sin 30^\circ}=1.5$}[/tex]
[tex]\large\text{$\sin\bigg(30^\circ+\dfrac{\delta_m}{2}\bigg)=0.75$}[/tex]
[tex]\large\text{$30^\circ+\dfrac{\delta_m}{2}=\sin^{-1}(0.75)$}[/tex]
[tex]\large\text{$\delta_m=2(\sin^{-1}(0.75)-30^\circ)$}[/tex]
[tex]\boxed{\large\text{$\delta_m=37.18^\circ$}}[/tex]
[tex]\large\text{In case of minimum deviation, we have, $}[/tex]
[tex]\large\text{Angle of incidence$(i)=\dfrac{A+\delta_m}{2}=\dfrac{60+37.18}{2}=48.59^\circ$}[/tex]
