JEE Mains Physics · Ray Optics

Critical Angle and Total Internal Reflection

Light going from a denser to a rarer medium is totally reflected when its angle of incidence exceeds the critical angle C, where sin C = n(rarer)/n(denser).

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Why this matters

Thirteen PYQs, eleven of them multiple choice, and two from 2026. Seven find the critical angle from refractive indices, from the speeds of light in two media or from a dielectric constant, and the options often write it as a tan or a cos instead of a sin. Six use it in a shape: the bright circle above a lamp under water, a coin seen over the rim of a bowl, a ray trapped in a block, or a prism face that is painted, dipped in a liquid or reached by three colours at once.

Concept 1 of 2: Critical angle

Going from a denser to a rarer medium, a ray bends away from the normal. As the angle of incidence grows, the refracted ray swings towards the surface. At the critical angle it grazes along the surface; beyond it no light gets out and all of it is reflected. Light going from rarer to denser always gets through, so total internal reflection never happens that way.

Definition

Two conditions: the light is in the denser medium, and $i > C$ .
$\sin C = \dfrac{n_{\text{rarer}}}{n_{\text{denser}}} = \dfrac{v_{\text{denser}}}{v_{\text{rarer}}}$ . Into air: $\sin C = 1/\mu$ .
The medium in which light is slower is the denser one, since $n = c/v$ .
From $\sin C = p/q$ , draw a right triangle: $\cos C = \dfrac{\sqrt{q^{2} - p^{2}}}{q}$ , $\tan C = \dfrac{p}{\sqrt{q^{2} - p^{2}}}$ . Options often give C as a tan or a cos.
For an electromagnetic wave in a medium, $n = \sqrt{\mu_r\varepsilon_r}$ .
At $i = C$ the ray grazes the surface; total reflection needs $i$ strictly larger.

Critical angle

\sin C = \frac{n_{\text{rarer}}}{n_{\text{denser}}} = \frac{v_{\text{denser}}}{v_{\text{rarer}}}, \qquad n = \sqrt{\mu_r\varepsilon_r}

Worked example

Light travels at

2.25 \times 10^{8}

m/s in water and at

1.8 \times 10^{8}

m/s in a plastic. In which medium must the light start for total internal reflection at their boundary? Write the critical angle as a sin, a cos and a tan.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q97Moderate

Example 1 · Ray Optics · Critical Angle and Total Internal Reflection

Light travels in two media

M_{1}

and

M_{2}

with speeds

1.5 \times10^{8}{\text{ }ms}^{- 1}

and

2.0 \times10^{8}{\text{ }ms}^{- 1}

respectively. The critical angle between them is:

No total reflection from rarer to denser

Light entering a denser medium bends towards the normal and always gets through. Total internal reflection needs the light to start in the denser medium.

The slower medium is the denser one

Given speeds, the medium where light is slower has the larger refractive index. The critical angle has the slower speed on top: sin C = v(slow)/v(fast).

At i = C the light is not yet trapped

At exactly the critical angle the ray grazes along the surface. Total internal reflection needs an angle of incidence greater than C, so the condition is a strict inequality.

Drill 6 more on critical angle

Concept 2 of 2: Total internal reflection in tanks, blocks and prisms

A critical-angle question with a shape is solved in two steps. First find, from the geometry, the angle at which the ray meets the surface in question. Then compare it with the critical angle for the two media on either side of that surface. A coating or a liquid outside a face changes the critical angle at that face.

Definition

A lamp at depth h under a liquid of index μ: light escapes only through a circle of radius $r = h\tan C = \dfrac{h}{\sqrt{\mu^{2} - 1}}$ . Rays meeting the surface farther out are reflected back.
A ray entering one face of a rectangular block at angle θ meets the adjacent face at $90^{\circ} - r$ , where $\sin\theta = \mu\sin r$ . Total reflection there needs $\sin(90^{\circ} - r) = \cos r > \sin C$ .
A face coated with, or dipped in, a medium of index $n_2 < n$ : $\sin C = n_2/n$ . A larger $n_2$ makes total reflection harder, so the boundary value of $n_2$ is the largest that still allows it.
A ray entering a prism face normally goes straight on and meets the next face at an angle set by the prism's shape: $60^{\circ}$ in an equilateral prism, $45^{\circ}$ on the long face of a right-angled isosceles prism.
Different colours have different μ and so different critical angles: violet, with the largest μ, is the first to be trapped; red is the last.

Circle of light and a coated face

r = h\tan C = \frac{h}{\sqrt{\mu^{2} - 1}}, \qquad \sin C = \frac{n_2}{n}\ (\text{coated face})

Worked example

A small lamp lies 3 m deep in a liquid of refractive index 1.25. What is the area of the liquid surface through which its light escapes?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q118Moderate

Example 2 · Ray Optics · Critical Angle and Total Internal Reflection

A small bulb is placed at the bottom of a tank containing water to a depth of

\sqrt{7}\text{ }m

. The refractive index of water is

\frac{4}{3}

. The area of the surface of water through which light from the bulb can emerge out is

x\pi m^{2}

. The value of

x

\_\_\_\_

The circle's radius uses tan C

The edge ray leaves the lamp at C to the vertical, so it travels h tan C sideways before reaching the surface. Using sin C gives a circle that is too small.

A coating raises the critical angle

With a film of index n₂ on a face, sin C = n₂/n, which is larger than 1/n. A ray that was totally reflected in air may escape through the coated face.

A 'minimum index for total reflection' may be a maximum

Total reflection at a coated face gets harder as n₂ rises. The boundary value is the largest n₂ that still works; check the direction of the inequality before choosing.

Drill 5 more on total internal reflection in tanks, blocks and prisms

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 (2)

Critical angle
Critical angle
$\sin C = \frac{n_{\text{rarer}}}{n_{\text{denser}}} = \frac{v_{\text{denser}}}{v_{\text{rarer}}}, \qquad n = \sqrt{\mu_r\varepsilon_r}$
Total internal reflection in tanks, blocks and prisms
Circle of light and a coated face
$r = h\tan C = \frac{h}{\sqrt{\mu^{2} - 1}}, \qquad \sin C = \frac{n_2}{n}\ (\text{coated face})$

Watch out for (6)

No total reflection from rarer to denser
→ Critical angle
The slower medium is the denser one
→ Critical angle
At i = C the light is not yet trapped
→ Critical angle
The circle's radius uses tan C
→ Total internal reflection in tanks, blocks and prisms
A coating raises the critical angle
→ Total internal reflection in tanks, blocks and prisms
A 'minimum index for total reflection' may be a maximum
→ Total internal reflection in tanks, blocks and prisms

Test yourself on Ray Optics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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