n = sin((A+δₘ)/2) / sin(A/2). Worked example with 60° prism.

Prism — Angle of Minimum Deviation

Question

A glass prism has an apex angle of $A = 60°$ . When a monochromatic ray passes through the prism at minimum deviation, the angle of minimum deviation is $\delta_m = 60°$ . Find the refractive index of the glass.

This is a direct JEE Main 2023 style problem — the formula is standard, but students lose marks on the substitution step.

Solution — Step by Step

Recall the minimum deviation condition

At minimum deviation, the refracted ray inside the prism travels parallel to the base. This means the ray is symmetric — the angle of incidence equals the angle of emergence. The prism formula at this condition is:

$n = \frac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$

Plug in the values

We have $A = 60°$ and $\delta_m = 60°$ . Substitute directly:

$n = \frac{\sin\left(\dfrac{60° + 60°}{2}\right)}{\sin\left(\dfrac{60°}{2}\right)} = \frac{\sin(60°)}{\sin(30°)}$

Evaluate the trig values

Both angles are standard ones from the table every JEE student should know cold:

$n = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

State the final answer

The refractive index of the glass is $n = \sqrt{3} \approx 1.732$ .

Answer: $n = \sqrt{3}$

Why This Works

The minimum deviation condition is special because the ray path becomes symmetric about the prism's bisector. At this exact angle, the ray inside travels parallel to the base, which means both refracting surfaces contribute equally to the total bending.

This symmetry is why the formula simplifies so cleanly. The angle of refraction inside the prism becomes exactly $r = A/2$ , and the angle of incidence becomes $i = (A + \delta_m)/2$ . Snell's law applied at the first surface then gives us the relation directly.

For a $60°$ equilateral prism, if $\delta_m = A = 60°$ , the math always collapses to $n = \sqrt{3}$ . This is worth memorising as a result — it appears in MCQs where the answer needs to be identified quickly.

Prism Formula at Minimum Deviation

$n = \frac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$

Valid only at minimum deviation. For any other angle of incidence, use $n_1 \sin i = n_2 \sin r$ at each surface separately.

Alternative Method

We can verify using the critical angle approach as a cross-check. If $n = \sqrt{3}$ , the critical angle $C$ is:

$\sin C = \frac{1}{n} = \frac{1}{\sqrt{3}} \implies C \approx 35.26°$

The angle of refraction inside the prism at minimum deviation is $r = A/2 = 30°$ . Since $30° < 35.26°$ , total internal reflection does NOT occur — the ray exits cleanly. This consistency check confirms $n = \sqrt{3}$ is physically valid.

In NEET, they sometimes ask whether TIR is possible for a given prism-and-refractive-index combination. Running this quick check takes 20 seconds and can earn you a 4-mark question.

💡 Expert Tip

For a 60° prism, the three "magic" combinations to remember:

$\delta_m = 30°$ → $n = 1$ (trivial, no refraction)
$\delta_m = 60°$ → $n = \sqrt{3} \approx 1.73$ (crown glass)
$\delta_m = 45°$ → $n \approx 1.55$ (appears in NEET numericals)

Common Mistake

⚠️ Common Mistake

The most common error is writing the formula as $n = \sin(A + \delta_m) / \sin(A)$ — forgetting the factor of $\frac{1}{2}$ in both arguments. This gives $n = \sin(120°)/\sin(60°) = 1$ , which makes no physical sense. The denominator and numerator both need the half-angle. Write the formula out fully before substituting — never shortcut the structure under exam pressure.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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