Question

Minimum deviation for an equilateral prism is 30°, refractive index is:

• A. \( \sqrt{2} \)
• B. \( \sqrt{\dfrac{3}{2}} \)
• C. 2
• D. 4

Hint:

For an equilateral prism the angle of the prism is always $A = 60^\circ$, so start from that number before doing anything else. At minimum deviation the ray path inside the prism is symmetric, which means the angle of refraction at each face is exactly half the prism angle, not half the angle of deviation. Fix the refraction angle first using this fact, then bring in the given minimum deviation value to complete the refractive index formula.

Answer 1

The Correct Option is A

Answer 2

Concept:

• Trace the ray step by step through the prism using the refraction angle at each surface, instead of jumping straight to the formula. This makes it clear why $r=A/2$ and $i=(A+\delta_m)/2$ hold at minimum deviation.

Step 1: Set up the two refracting surfaces.
A prism has two surfaces meeting at angle $A = 60^\circ$. The ray refracts once entering (angle $r_1$) and once leaving (angle $r_2$), with $r_1 + r_2 = A$.

Step 2: Use the symmetry of minimum deviation.
At minimum deviation, the ray inside the prism runs parallel to the base, which forces $r_1 = r_2$. Combined with $r_1+r_2=60^\circ$, this gives $r_1 = r_2 = 30^\circ$.

Step 3: Find the incidence angle from the total deviation.
Total deviation $\delta_m = (i_1 - r_1) + (i_2 - r_2)$. Since $i_1 = i_2 = i$ and $r_1=r_2=r$ at minimum deviation, this becomes $\delta_m = 2i - 2r + (r_1+r_2) - (r_1+r_2)$, which simplifies to $\delta_m = 2i - A$.
So $i = \dfrac{\delta_m+A}{2} = \dfrac{30^\circ+60^\circ}{2} = 45^\circ$.

Step 4: Apply Snell's law at the point of entry.
$\mu = \dfrac{\sin i}{\sin r_1} = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{\sqrt{2}/2}{1/2} = \sqrt{2}$

Final Answer: $\mu = \sqrt{2}$