Alright fiyin — let's do this. The question asks to calculate the angle of minimum deviation ($\delta_m$) for a prism. Given: • Refractive index of the prism, $n = 1.4$ • The "Reason" section specifies that it is an equilateral prism, which means the prism angle, $A = 60^\circ$. The formula for the refractive index of a prism in terms of the prism angle ($A$) and the angle of minimum deviation ($\delta_m$) is: $$n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ Step 1: Substitute the given values into the formula. $$1.4 = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}$$ Step 2: Simplify the denominator. $$\sin\left(\frac{60^\circ}{2}\right) = \sin(30^\circ) = 0.5$$ Step 3: Substitute the simplified denominator back into the equation. $$1.4 = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{0.5}$$ Step 4: Isolate the sine term. Multiply both sides by $0.5$: $$1.4 \times 0.5 = \sin\left(\frac{60^\circ + \delta_m}{2}\right)$$ $$0.7 = \sin\left(\frac{60^\circ + \delta_m}{2}\right)$$ Step 5: Find the inverse sine (arcsin) of $0.7$. $$\frac{60^\circ + \delta_m}{2} = \arcsin(0.7)$$ Using a calculator, $\arcsin(0.7) \approx 44.427^\circ$. Step 6: Solve for $\delta_m$. $$\frac{60^\circ + \delta_m}{2} = 44.427^\circ$$ Multiply both sides by $2$: $$60^\circ + \delta_m = 2 \times 44.427^\circ$$ $$60^\circ + \delta_m = 88.854^\circ$$ Subtract $60^\circ$ from both sides: $$\delta_m = 88.854^\circ - 60^\circ$$ $$\delta_m = 28.854^\circ$$ Step 7: Round the result to the nearest whole number. $$\delta_m \approx 29^\circ$$ The calculated angle of minimum deviation is $29^\circ$. This matches option A. The final answer is $\boxed{\text{A}}$. Send me the next one 📸