Q.A ray of light incident on one face of an equilateral glass prism (refracting angle $A=60^{\circ}$) suffers minimum deviation. If the refractive index of the glass is $1.5$, calculate the angle of minimum deviation.

At minimum deviation, the prism formula $n=\dfrac{\sin\left(\frac{A+D_m}2\right)}{\sin\left(\frac A2\right)}$ applies. With $A=60^{\circ}$, $\sin(A/2)=\sin30^{\circ}=0.5$; and $n=1.5$. So $\sin\left(\dfrac{A+D_m}2\right)=n\sin(A/2)=1.5\times0.5=0.75$, giving $\dfrac{A+D_m}2=\sin^{-1}(0.75)\approx48.59^{\circ}$, so $A+D_m …

Substitute $A$ and $n$ into the minimum-deviation prism formula, solve for $\sin\left(\frac{A+D_m}2\r …

• Forgetting to subtract $A$ after finding $(A+D_m)/2$ and its inverse-sine — a very common last-step slip. …