JEE Mains Physics · Ray Optics

Prisms: Minimum Deviation, Grazing Emergence and Dispersion

Inside a prism r₁ + r₂ = A and the deviation is i + e − A; at minimum deviation the ray passes symmetrically and μ = sin((A + δm)/2)/sin(A/2), while a thin prism deviates by (μ − 1)A.

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Why this matters

Twenty-nine PYQs, twenty-three of them multiple choice, and nine from 2026. Fifteen are about minimum deviation: the formula, what is true at the minimum, the shape of the deviation graph, and prisms whose index is given as a speed or as a trig function of the prism angle. Five have the ray graze out of the second face, sometimes a coated one. Nine are thin prisms and colour: deviation without dispersion, dispersion without deviation, and the colours of a prism, a rainbow and a lens.

Concept 1 of 3: Prism and minimum deviation

A ray through a prism bends at both faces. The two refraction angles inside always add up to the prism angle A, and the total deviation is i + e − A. As i grows, the deviation first falls and then rises. At the bottom of that curve the ray passes symmetrically: i = e, and inside an isosceles prism it runs parallel to the base. That symmetric ray gives the formula that measures μ.

Definition

$r_1 + r_2 = A$ , $\delta = i + e - A$ .
At minimum deviation: $i = e$ , $r_1 = r_2 = A/2$ , $\delta_m = 2i - A$ , so $i = \dfrac{A + \delta_m}{2}$ .
$\mu = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$ . A liquid in a hollow prism is the prism: use the liquid's μ.
Deviation against i: one minimum. Every larger deviation is reached at two values of i, which swap roles as i and e.
μ given as a speed: $\mu = c/v$ . μ given as a trig function of A: write $\sin\frac{A + \delta_m}{2} = \mu\sin\frac{A}{2}$ and compare the two sides.
For the same A, a larger μ gives a larger $\delta_m$ .

Prism and minimum deviation

r_1 + r_2 = A, \qquad \delta = i + e - A, \qquad \mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}

Worked example

A prism of angle

60^{\circ}

gives a minimum deviation of

40^{\circ}

. Find the angle of incidence at minimum deviation, the angle of refraction inside, and the refractive index. (

\sin 50^{\circ} = 0.766

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q15Moderate

Example 1 · Ray Optics · Prisms: Minimum Deviation, Grazing Emergence and Dispersion

A ray of light passing through an equilateral prism is having velocity

2.12 \times10^{8}\text{ }m/s

in the prism material, then the minimum angle of deviation is

\_\_\_\_

degrees.

Use A/2 inside, not A

At minimum deviation each internal angle is A/2. Putting A into Snell's law at the first face doubles the refraction angle.

Through a prism the deviation is i + e − A

i − r is the bending at one face only. A prism bends the ray at both faces, and i + e − A adds the two.

Two angles of incidence give the same deviation

Except at the minimum, each deviation occurs for a pair of values of i, the ray and its reverse. The graph of deviation against i is a curve with one minimum, not a straight line.

Drill 14 more on prism and minimum deviation

Concept 2 of 3: Grazing emergence from a prism

Make a ray meet the second face more and more steeply, and at some point it can no longer get out: it would need to emerge at 90°. That grazing ray marks the limit. At the second face the angle inside is then the critical angle, and r₁ = A − r₂ fixes the first face. Coat the second face and the critical angle there changes.

Definition

Grazing emergence ( $e = 90^{\circ}$ ): $\sin r_2 = 1/\mu$ , so $r_2 = C$ .
Then $r_1 = A - C$ and $\sin i = \mu\sin(A - C)$ . This is the smallest i for which light gets out of the second face.
Grazing incidence ( $i = 90^{\circ}$ ) at the first face gives $r_1 = C$ . If $A > 2C$ , no ray gets out of the second face at all.
Exit face coated with a film of index $n_2$ : the critical angle there is $\sin C' = n_2/\mu$ . Total reflection at that face needs $r_2 > C'$ , that is $r_1 < A - C'$ .
A ray parallel to the base meets the first face at an angle set by the prism's shape. Find i from the geometry first.

Grazing emergence

\sin r_2 = \frac{1}{\mu}\ (e = 90^{\circ}), \qquad r_1 = A - r_2, \qquad \sin i = \mu\sin r_1

Worked example

A prism of angle

60^{\circ}

is made of glass of refractive index 1.5. Find the smallest angle of incidence for which light can come out of the second face. (

\sin^{-1}(2/3) = 41.8^{\circ}

\sin 18.2^{\circ} = 0.312

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q15Moderate

Example 2 · Ray Optics · Prisms: Minimum Deviation, Grazing Emergence and Dispersion

Consider an equilateral prism (refractive index

\sqrt{2}

). A ray of light is incident on its one surface at a certain angle

i

. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to

\_\_\_\_

Grazing emergence fixes r₂, not i

e = 90° sets the angle inside the second face to the critical angle. The first-face angles then follow from r₁ = A − r₂ and Snell's law.

A coating changes the critical angle at that face only

A film of index n₂ on the exit face makes sin C' = n₂/μ there. The entry face still has the air critical angle.

Drill 4 more on grazing emergence from a prism

Concept 3 of 3: Thin prisms and dispersion

For a prism of small angle the deviation hardly depends on the angle of incidence: δ = (μ − 1)A. Because μ depends on colour, so does δ: violet is bent most, red least. Two thin prisms placed opposite ways can cancel the mean deviation and keep the colours spread, or cancel the spread and keep a deviation.

Definition

Thin prism, small angle of incidence: $\delta = (\mu - 1)A$ .
Angular dispersion $\delta_v - \delta_r = (\mu_v - \mu_r)A$ ; dispersive power $\omega = \dfrac{\mu_v - \mu_r}{\mu_y - 1}$ , with $\mu_y$ the mean (yellow) index.
Dispersion without deviation, prisms placed opposite ways: $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$ , with the mean indices.
Deviation without dispersion: $(\mu_{v1} - \mu_{r1})A_1 = (\mu_{v2} - \mu_{r2})A_2$ ; the net deviation is then $(\mu_1 - 1)A_1 - (\mu_2 - 1)A_2$ .
Two glasses in contact do not bend a colour at their common face when their indices for that colour are equal.
Red has the longest wavelength and the smallest μ, so it is deviated least. A lens has a slightly different focus for each colour: chromatic aberration.
Primary rainbow: one internal reflection in each drop, red on the outside (top), violet inside. The secondary bow has two reflections, reversed colours, and is fainter.

Thin prism and dispersion

\delta = (\mu - 1)A, \qquad (\mu_1 - 1)A_1 = (\mu_2 - 1)A_2, \qquad \omega = \frac{\mu_v - \mu_r}{\mu_y - 1}

Worked example

A thin prism of angle

8^{\circ}

and mean refractive index 1.5 is combined with a thin prism of mean refractive index 1.8, placed the opposite way, to give dispersion without deviation. Find the deviation produced by the first prism alone and the angle of the second.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q20Moderate

Example 3 · Ray Optics · Prisms: Minimum Deviation, Grazing Emergence and Dispersion

A thin prism

P_{1}

with angle

4^{\circ}

made of glass having refractive index 1.54 , is combined with another thin prism

P_{2}

made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism

P_{2}

in degrees is

The prisms face opposite ways

Dispersion without deviation needs the two prisms placed opposite ways. The condition equates the sizes of their mean deviations, (μ₁ − 1)A₁ = (μ₂ − 1)A₂.

δ = (μ − 1)A is for thin prisms only

It holds for small prism angles and small angles of incidence. For a 60° prism use the minimum-deviation formula instead.

Red bends least

Red has the longest wavelength and the smallest refractive index in glass. It is deviated least, focuses farthest from a lens, and sits on top of the primary rainbow.

Drill 8 more on thin prisms and dispersion

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Prism and minimum deviation
Prism and minimum deviation
$r_1 + r_2 = A, \qquad \delta = i + e - A, \qquad \mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$
Grazing emergence from a prism
Grazing emergence
$\sin r_2 = \frac{1}{\mu}\ (e = 90^{\circ}), \qquad r_1 = A - r_2, \qquad \sin i = \mu\sin r_1$
Thin prisms and dispersion
Thin prism and dispersion
$\delta = (\mu - 1)A, \qquad (\mu_1 - 1)A_1 = (\mu_2 - 1)A_2, \qquad \omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$

Watch out for (8)

Use A/2 inside, not A
→ Prism and minimum deviation
Through a prism the deviation is i + e − A
→ Prism and minimum deviation
Two angles of incidence give the same deviation
→ Prism and minimum deviation
Grazing emergence fixes r₂, not i
→ Grazing emergence from a prism
A coating changes the critical angle at that face only
→ Grazing emergence from a prism
The prisms face opposite ways
→ Thin prisms and dispersion
δ = (μ − 1)A is for thin prisms only
→ Thin prisms and dispersion
Red bends least
→ Thin prisms and dispersion

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