Angle of minimum deviation for a prism of refractive index 1.5, is equal to the angle of the prism. Then the angle of the prism is : (Given $cos41^{\circ }=0.75$)

Let ${\delta }_m$ be the angle of minimum deviation $A={\delta }_m$

$A={\delta }_m$

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

$1.5=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$1.5=\frac{\sin 2\left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \left(\frac{A}{2}\right)}$

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

$0.75=\cos \frac{A}{2}$

$\frac{A}{2}={\cos }^{-1}(0.75)$

$\frac{A}{2}=41^o$

$\boxed{A=82^o}$