Prism — derive minimum deviation formula and find refractive index

Question

Derive the relation $\mu = \dfrac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ for a prism, where $A$ is the angle of the prism and $D_m$ is the angle of minimum deviation. Find $\mu$ if $A = 60°$ and $D_m = 30°$ .

(JEE Main 2022 / CBSE 12th standard)

Solution — Step by Step

Establish the basic prism relations

For a prism with angle $A$ , when a ray passes through:

$A = r_1 + r_2$ (angles of refraction at the two surfaces)
$D = (i_1 - r_1) + (i_2 - r_2) = (i_1 + i_2) - A$ (total deviation)

where $i_1, i_2$ are the angles of incidence at the two faces.

Condition for minimum deviation

At minimum deviation, the ray passes symmetrically through the prism. This means:

$i_1 = i_2 = i$ and $r_1 = r_2 = r$

From $A = r_1 + r_2$ : $A = 2r$ , so $r = A/2$ .

From $D_m = 2i - A$ : $i = (A + D_m)/2$ .

Apply Snell's law

At the first surface: $\mu = \frac{\sin i}{\sin r}$

Substituting $i = (A + D_m)/2$ and $r = A/2$ :

$\mathbf{\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}}$

Numerical calculation

Given $A = 60°$ and $D_m = 30°$ :

$\mu = \frac{\sin\left(\frac{60° + 30°}{2}\right)}{\sin\left(\frac{60°}{2}\right)} = \frac{\sin 45°}{\sin 30°} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \mathbf{\sqrt{2} \approx 1.414}$

Why This Works

At minimum deviation, the light ray inside the prism is parallel to the base. The symmetry condition ( $i_1 = i_2$ ) minimises deviation because any asymmetry adds extra bending.

This can be verified by calculus: treating $D$ as a function of $i_1$ and using $dD/di_1 = 0$ at the minimum. The condition yields $i_1 = i_2$ .

The minimum deviation formula is practically important because it gives the most accurate method to measure the refractive index of a glass prism — spectrometers in physics labs use exactly this principle.

Alternative Method

For a thin prism (small $A$ ), use the approximation $\sin\theta \approx \theta$ (in radians):

$\mu \approx \frac{(A + D_m)/2}{A/2} = \frac{A + D_m}{A}$

This gives $D_m = (\mu - 1)A$ , which is the thin prism formula. For $A = 60°$ , the thin prism approximation is not very accurate, but it is useful for quick estimates in JEE.

💡 Expert Tip

The combination $A = 60°, D_m = 30°$ giving $\mu = \sqrt{2}$ is the most commonly tested numerical in both CBSE boards and JEE. The second most common: $A = 60°, D_m = 60°$ , giving $\mu = \sin 60°/\sin 30° = \sqrt{3}$ .

Common Mistake

⚠️ Common Mistake

Students mix up the angle of the prism $A$ with the angle of deviation $D$ . The prism angle $A$ is the angle between the two refracting surfaces — it is a fixed property of the prism. The deviation $D$ depends on the angle of incidence. At minimum deviation ( $D_m$ ), the path is symmetric. Keep these two angles clearly labelled in your diagram.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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