Critical Angle and Total Internal Reflection

PhysicsRay Optics And Optical InstrumentsFor NEET aspirants

The critical angle and total internal reflection decide whether light can leave a denser medium. When light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle $C$ , where $sin C = \frac{n _{rarer}}{n _{denser}}$ , no light is refracted: all of it is reflected back. Critical angle and total internal reflection explain optical fibres, sparkling diamonds, mirages and periscope prisms, and appear every year in JEE Main, JEE Advanced and NEET.

On this page1Critical angle2Conditions for TIR3Deviation graph4Circle of illuminance5Applications6Flowchart and map7Solved examples

Key Formulas - Quick Reference

★ Must learnCritical angle: $sin C = \frac{n _{rarer}}{n _{denser}}$ ; against air $sin C = \frac{1}{n}$
★ Must learnTIR needs both: light going denser $\to$ rarer, and $i > C$
Deviation: refraction $δ = r - i$ ( $i < C$ ); TIR $δ = 18 0^{\circ} - 2 i$ ( $i > C$ )
★ Must learnMaximum deviation $= 18 0^{\circ} - 2 C$ (just beyond the critical angle)
★ Must learnCircle of illuminance: $r = h tan C = \frac{h}{n ^{2} - 1}$
★ Must learnFibre in air traps every entering ray if $n \geq 2$
Numerical aperture (fibre in air): $sin θ_{a} = n_{1}^{2} - n_{2}^{2}$
Right-angled isosceles prism reflects totally only if $n > 2$

1. The Critical Angle

Send light from a denser medium (index $n_{d}$ ) into a rarer one ( $n_{r}$ ). By Snell's law $n_{d} sin i = n_{r} sin r$ , and since $n_{d} > n_{r}$ the refracted ray bends away from the normal: $r > i$ . As $i$ grows, $r$ reaches $9 0^{\circ}$ first. The angle of incidence at which this happens is the critical angle $C$ .

Critical angle: the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is

9 0^{\circ}

. Putting

r = 9 0^{\circ}

in Snell's law:

sin C = \frac{n _{r}}{n _{d}}

For a medium against air or vacuum,

sin C = \frac{1}{n}

Figure 1: Rays from

O

in water. Below

C

the ray refracts (with weak reflection); at

C = 48. 6^{\circ}

it grazes the surface (

r = 9 0^{\circ}

); beyond

C

all the light is reflected.

Medium (against air)	$n$	Critical angle $C$
Water	$1.33$	$48. 6^{\circ}$
Crown glass	$1.50$	$41. 8^{\circ}$
Dense flint glass	$1.65$	$37. 3^{\circ}$
Diamond	$2.42$	$24. 4^{\circ}$

Figure 2:

C = sin^{- 1} \frac{1}{n}

(denser medium against air). The denser the medium, the smaller the critical angle, so the easier total internal reflection becomes: diamond (

24. 4^{\circ}

) traps light far better than water (

48. 6^{\circ}

The critical angle belongs to a pair of media: glass against water has $sin C = \frac{4/3}{3/2} = \frac{8}{9}$ , $C = 62. 7^{\circ}$ , much larger than glass against air.
Since $n$ is larger for violet than for red, violet has the smaller critical angle: violet light is totally reflected first.
In terms of speeds, $sin C = \frac{v _{d}}{v _{r}}$ (speed in the denser medium over speed in the rarer one).

Key idea

The critical angle is the largest angle of incidence at which light can still escape from a denser medium:

sin C = \frac{n _{rarer}}{n _{denser}}

2. Total Internal Reflection

For $i > C$ , Snell's law would need $sin r > 1$ , which is impossible. No refracted ray exists; the whole beam is reflected back into the denser medium, obeying the ordinary laws of reflection. This is total internal reflection (TIR).

Two conditions for TIR, both necessary:
1. Light must travel from the optically denser medium towards the rarer medium.
2. The angle of incidence must exceed the critical angle,

i > C

Total internal reflection

At the boundary of a transparent medium. 100% of the light is reflected; no metal coating, no absorption, no ghost images. Needs denser to rarer and $i > C$ .

Reflection at a mirror

At a silvered surface. Any angle of incidence works, but only about 90 to 95% is reflected; the rest is absorbed, and a back-silvered glass mirror gives faint double images.

Exam Trick

Check the direction before the angle. Light going from air into glass, or from water into glass, can never be totally reflected, however large $i$ is. If the question sends light into the denser medium, answer with Snell's law and stop.

2.1 Deviation against angle of incidence

For light going from a denser to a rarer medium the deviation behaves in two different ways:

$i < C$ (refraction): $δ = r - i = sin^{- 1} (\frac{n _{1}}{n _{2}} sin i) - i$ , rising from $0$ to $9 0^{\circ} - C$ .
$i > C$ (TIR): $δ = 18 0^{\circ} - 2 i$ , falling from $18 0^{\circ} - 2 C$ to $0$ at grazing incidence.

Figure 3: Glass to air (

n = 1.5

). Refraction gives

δ = r - i

, rising to

9 0^{\circ} - C = 48. 2^{\circ}

; just beyond

C

total internal reflection gives the maximum

δ = 18 0^{\circ} - 2 C = 96. 4^{\circ}

. Any

δ

below

48. 2^{\circ}

occurs at two angles of incidence (for

3 0^{\circ}

38. 3^{\circ}

and

7 5^{\circ}

Exam Trick

Biggest deviation is just past the critical angle: $18 0^{\circ} - 2 C$ . The graph jumps at $i = C$ . Any deviation between $0$ and $9 0^{\circ} - C$ is produced at two angles of incidence (one refracted, one totally reflected); deviations between $9 0^{\circ} - C$ and $18 0^{\circ} - 2 C$ come only from TIR.

Quick Recall: tap to check

Can light going from water into glass be totally reflected?

No. TIR needs light going from the denser medium to the rarer one; water to glass is rarer to denser.

Critical angle of glass (

n = 1.5

) against air?

C = sin^{- 1} \frac{2}{3} \approx 41. 8^{\circ}

Maximum deviation of light at a glass-air surface, from inside the glass?

18 0^{\circ} - 2 C = 96. 4^{\circ}

, at

i

just above

41. 8^{\circ}

Key idea

TIR happens only when light goes from denser to rarer with

i > C

; then 100% of it is reflected.

3. Circle of Illuminance and Snell's Window

A point source at depth $h$ in a liquid sends rays in all directions. Only the rays striking the surface at less than $C$ escape. They lie inside a cone of half-angle $C$ , which cuts the surface in a bright circle, the circle of illuminance.

Figure 4: Light escapes only through a circle of radius

r = h tan C = \frac{h}{n ^{2} - 1}

. For water

r = \frac{3 h}{7} \approx 1.13 h

; an opaque disc of this radius floating above the lamp hides it completely.

The boundary ray strikes the surface at exactly $C$ , a horizontal distance $r$ from the point above the source: $tan C = \frac{r}{h}$ .
With $sin C = \frac{1}{n}$ , $tan C = \frac{1}{n ^{2} - 1}$ .
Hence
$r = h tan C = \frac{h}{n ^{2} - 1}$
For water, $r = \frac{3 h}{7} \approx 1.13 h$ .

An opaque disc of radius $r$ floating directly above the source hides it from every observer in air. Running the rays backwards gives Snell's window: an eye under water sees the entire sky squeezed into a cone of half-angle $C$ ; beyond it the surface is a perfect mirror showing the pool floor.

Figure 5: Snell's window. Sky rays from every direction, even grazing ones, reach an underwater eye within a cone of half-angle

C

(

9 7^{\circ}

across for water). Outside the cone the surface acts as a perfect mirror.

Exam Trick

Water: $r \approx 1.13 h$ , window $\approx 9 7^{\circ}$ . Remember $tan C = \frac{1}{n ^{2} - 1}$ : for $n = \frac{4}{3}$ it is $\frac{3}{7}$ , for $n = 1.5$ it is $\frac{2}{5}$ , for $n = 2$ exactly $1$ ( $C = 4 5^{\circ}$ ).

Key idea

A source at depth

h

lights only a circle of radius

\frac{h}{n ^{2} - 1}

; the same cone is the underwater eye's window to the sky.

4. Applications of Total Internal Reflection

4.1 Optical fibres

An optical fibre is a thin, long strand of glass or plastic. Light entering one end is totally reflected at the walls again and again and comes out at the far end, even when the fibre is bent, with very little loss. Fibres carry telephone and internet signals and are used in endoscopes.

Figure 6: Fibre in air,

n = 1.5

i = 6 0^{\circ}

r = 35. 3^{\circ}

, wall angle

9 0^{\circ} - r = 54. 7^{\circ} > C

, so the ray is totally reflected 5 times here. Every entering ray is trapped if

n \geq 2

A ray enters the flat end at incidence $i$ and refracts at $r$ : $sin i = n sin r$ .
It meets the side wall at incidence $9 0^{\circ} - r$ and is trapped if $9 0^{\circ} - r > C$ .
The worst case is grazing entry, $i = 9 0^{\circ}$ , which gives the largest $r = C$ . The condition becomes $9 0^{\circ} - C > C$ , so $C < 4 5^{\circ}$ .
So $sin C = \frac{1}{n} < \frac{1}{2}$ : every ray that enters is trapped if $n > 2$ (in practice $n \geq 2$ ).

JEE Advanced

Clad fibre and numerical aperture. Real fibres have a core ( $n_{1}$ ) inside a cladding of slightly lower index ( $n_{2}$ ), so TIR happens at the core-cladding boundary, protected from dirt and scratches. A ray entering from air at $θ$ is guided only if its wall angle exceeds $sin^{- 1} \frac{n _{2}}{n _{1}}$ , which gives

sin θ_{m a x} = n_{1}^{2} - n_{2}^{2}

This is the numerical aperture; rays outside the acceptance cone $θ_{m a x}$ leak into the cladding (Figure 7).

Figure 7: Clad fibre. Rays within the acceptance cone,

sin θ_{a} = n_{1}^{2} - n_{2}^{2} = 0.539

(

θ_{a} = 32. 6^{\circ}

), are guided; steeper rays leak out.

n_{1}^{2} - n_{2}^{2}

is the numerical aperture.

4.2 Totally reflecting prisms

A right-angled isosceles glass prism has $4 5^{\circ}$ angles. Light entering normally through one face meets the next face at $4 5^{\circ}$ , which exceeds $C = 41. 8^{\circ}$ for glass, so it is totally reflected. Such prisms turn a beam through $9 0^{\circ}$ (periscopes) or $18 0^{\circ}$ (binoculars, reflectors) and invert images, better than mirrors because nothing is lost.

Figure 8: Right-angled isosceles prisms (

4 5^{\circ}

incidence

> C = 41. 8^{\circ}

). Left: one total reflection turns the beam

9 0^{\circ}

(periscopes). Right: two total reflections send it back,

18 0^{\circ}

(binoculars, reflectors). Works only if

n > 2

How large can the angle $θ$ be before light leaks out of the long face? For a right-angled prism entered normally, the condition is worked out in Solved Example 7:

Figure 9: Solved Example 7. Entering normally at

A C

, the ray meets

A B

i = 9 0^{\circ} - θ

. Total internal reflection needs

9 0^{\circ} - θ > C

, i.e.

θ < cos^{- 1} \frac{2}{3} = 48. 2^{\circ}

; drawn for

θ = 4 0^{\circ}

4.3 Mirage

On a hot day the air near a road is hotter and optically rarer than the air above. A ray from a distant object travelling downwards meets layers of ever smaller $n$ , bends further and further from the normal and finally turns upward when it grazes a layer. The observer sees an inverted image as if reflected from a pool of water: a mirage. The same bending over cold ground or sea (colder, denser air below) lifts images upward instead, an effect called looming.

Figure 10: Mirage (index gradient exaggerated for clarity). The ray bends continuously, turns just above the road (a gradual total internal reflection) and reaches the eye from below the horizontal. Tracing it straight back, the eye sees an inverted tree under the road, which looks like a reflection in water.

4.4 Brilliance of diamonds

Diamond has $C = 24. 4^{\circ}$ . Its facets are cut so that light entering from the top meets the lower faces at more than $24. 4^{\circ}$ and is totally reflected several times before leaving through the top, which makes a cut diamond sparkle. A glass imitation ( $C \approx 41. 8^{\circ}$ ) lets most light escape through its back.

Figure 11: Exact trace through the same cut. (a) Diamond: the ray meets the pavilion facets at

40. 7^{\circ}

and

57. 9^{\circ}

, both above

C = 24. 4^{\circ}

, so it is totally reflected twice and comes back out of the top. (b) Glass:

40. 7^{\circ} < C = 41. 8^{\circ}

, so the light escapes through the back and the stone looks dull.

Quick Recall: tap to check

Why is the core of an optical fibre given a cladding of lower refractive index?

TIR then happens at the core-cladding boundary, which stays clean; the cladding also fixes the acceptance cone through

n_{1}^{2} - n_{2}^{2}

Minimum refractive index for a

4 5^{\circ}

4 5^{\circ}

9 0^{\circ}

prism to turn light through

9 0^{\circ}

by TIR?

2 \approx 1.414

, so that

C < 4 5^{\circ}

Key idea

Fibres, reflecting prisms, mirages and diamonds all use the same rule: keep the angle inside the denser medium above

C

5. Flowchart and Mind Map

Every question on this topic starts with the direction of the light, then compares the angle of incidence with the critical angle.

Figure 12: Problem-solving flowchart. Check the direction first (only denser to rarer can give total internal reflection), then compare

i

with

C

Figure 13: Critical angle and total internal reflection on one page. Revise from the map, then test yourself on the solved examples.

6. Solved Examples

Solved Example 1

Find the largest angle of incidence inside glass (

n = 1.5

) at which light can still pass into vacuum.

Solution:

At the limit $r = 9 0^{\circ}$ : $1.5 sin C = 1$ , so $sin C = \frac{2}{3}$ .

Answer: $C = sin^{- 1} \frac{2}{3} \approx 41. 8^{\circ}$ .

Solved Example 2

Light travelling in vacuum strikes a medium of refractive index

2

at an angle of incidence equal to twice the critical angle of that medium. Find the angle of refraction.

Solution:

$C = sin^{- 1} \frac{1}{2} = 3 0^{\circ}$ , so $i = 6 0^{\circ}$ . The light goes from rarer to denser, so TIR cannot occur. Snell's law: $sin 6 0^{\circ} = 2 sin r$ , $sin r = \frac{3}{4}$ .

Answer: $r = sin^{- 1} \frac{3}{4} \approx 25. 7^{\circ}$ .

Solved Example 3

Total internal reflection is possible for light travelling from
(A) air to water
(B) water to glass
(C) glass to water
(D) air to glass

Solution:

Only denser to rarer qualifies: glass ( $1.5$ ) to water ( $1.33$ ). Its critical angle is $sin^{- 1} \frac{4/3}{3/2} = sin^{- 1} \frac{8}{9} \approx 62. 7^{\circ}$ .

Answer: (C).

Solved Example 4

Light from a denser medium strikes a rarer medium at incidence

i

, and the reflected and refracted rays are perpendicular to each other. Find the critical angle.

Solution:

Reflected at $i$ , refracted at $r$ , with $i + 9 0^{\circ} + r = 18 0^{\circ}$ , so $r = 9 0^{\circ} - i$ . Snell's law, with $n = \frac{n _{D}}{n _{R}}$ : $n sin i = sin (9 0^{\circ} - i) = cos i$ , so $n = cot i$ .

Answer: $sin C = \frac{1}{n} = tan i$ , $C = sin^{- 1} (tan i)$ .

Solved Example 5

A lamp is

1 m

below the surface of a pond (

n = \frac{4}{3}

). The radius of the circle on the surface through which light emerges is
(A)

0.75 m

(B)

1.13 m

(C)

1.33 m

(D)

1.51 m

Solution:

$r = \frac{h}{n ^{2} - 1} = \frac{1}{16/9 - 1} = \frac{3}{7} \approx 1.13 m$ . An opaque disc of this radius (diameter $2.27 m$ ) centred above the lamp hides it completely.

Answer: (B).

Solved Example 6

Light inside glass (

n = 1.5

) meets the glass-air surface. At which angles of incidence is it deviated by exactly

3 0^{\circ}

Solution:

TIR branch ( $i > 41. 8^{\circ}$ ): $18 0^{\circ} - 2 i = 3 0^{\circ}$ gives $i = 7 5^{\circ}$ .

Refraction branch ( $i < 41. 8^{\circ}$ ): need $r = i + 3 0^{\circ}$ with $sin r = 1.5 sin i$ . Expanding, $sin i cos 3 0^{\circ} + cos i sin 3 0^{\circ} = 1.5 sin i$ , so $tan i = \frac{0.5}{1.5 - 0.866} = 0.789$ and $i = 38. 3^{\circ}$ (below $C$ , so valid).

Answer: $i = 38. 3^{\circ}$ (refracted) or $i = 7 5^{\circ}$ (totally reflected), the two points on Figure 3.

Solved Example 7

Light enters face

A C

of a right-angled prism (

n = \frac{3}{2}

, right angle at

C

, angle

θ

B

) normally. For what

θ

does no light cross the face

A B

Solution:

Normal entry means no bending at $A C$ . The ray meets $A B$ at $i = 9 0^{\circ} - θ$ (Figure 9). No light crosses if $i > C$ : $sin (9 0^{\circ} - θ) > sin C$ , i.e. $cos θ > \frac{2}{3}$ .

Answer: $θ < cos^{- 1} \frac{2}{3} \approx 48. 2^{\circ}$ .

Solved Example 8

4 5^{\circ}

4 5^{\circ}

9 0^{\circ}

glass prism (

n = 1.5

) turns a beam through

9 0^{\circ}

in air. If the prism is placed in water (

n = \frac{4}{3}

) it
(A) still turns the beam through

9 0^{\circ}

(B) turns it through

18 0^{\circ}

Solution:

In water the critical angle becomes $sin^{- 1} \frac{4/3}{3/2} = 62. 7^{\circ}$ . The beam still meets the hypotenuse at $4 5^{\circ}$ , now below $C$ , so it refracts out (with only a weak reflection).

Answer: (C). TIR depends on the surrounding medium as well as on the glass.

Solved Example 9

Light enters the flat end of a long transparent cylinder in air. Find the least refractive index for which every ray that enters is totally reflected at the curved wall.

Solution:

Wall angle $= 9 0^{\circ} - r$ , smallest when $r$ is largest. Grazing entry ( $i = 9 0^{\circ}$ ) gives $r_{m a x} = C$ . Condition $9 0^{\circ} - C > C$ , so $C < 4 5^{\circ}$ and $\frac{1}{n} < \frac{1}{2}$ .

Answer: $n > 2 \approx 1.414$ .

Solved Example 10

A fibre has a core of index

1.50

and a cladding of index

1.40

. Find its numerical aperture and the acceptance angle in air.

Solution:

$NA = n_{1}^{2} - n_{2}^{2} = 2.25 - 1.96 = 0.29 = 0.539$ . Acceptance angle $θ_{a} = sin^{- 1} 0.539 = 32. 6^{\circ}$ : a cone of full angle $6 5^{\circ}$ (Figure 7).

Answer: NA $= 0.539$ , $θ_{a} \approx 32. 6^{\circ}$ .

Practice Questions

Find the critical angle of diamond ( $n = 2.42$ ) against air.Answer: $sin^{- 1} \frac{1}{2.42} \approx 24. 4^{\circ}$
Light travels in a medium at $1.8 \times 1 0^{8} m s^{- 1}$ . Find its critical angle against air.Answer: $n = \frac{5}{3}$ , $C = sin^{- 1} 0.6 \approx 36. 9^{\circ}$
Find the critical angle for light going from glass ( $n = 1.6$ ) into water ( $n = \frac{4}{3}$ ).Answer: $sin^{- 1} \frac{4/3}{1.6} = sin^{- 1} 0.833 \approx 56. 4^{\circ}$
A lamp is $4 m$ deep in water. Find the smallest opaque disc on the surface that hides it.Answer: Radius $\frac{3 \times 4}{7} \approx 4.54 m$ , centred above the lamp
A diver $1 m$ under water looks up. What is the radius of the circle through which she sees the whole sky, and the angle of that cone?Answer: $1.13 m$ ; full cone angle $2 C \approx 9 7^{\circ}$
A transparent rod is used as a light pipe under water. Find the least index for which all entering rays are trapped.Answer: $n > \frac{4}{3} 2 \approx 1.89$
A $4 5^{\circ}$ - $4 5^{\circ}$ - $9 0^{\circ}$ prism of index $1.4$ receives light normally on a short face. Does the light leave through the hypotenuse? If so, at what angle?Answer: Yes, $C = 45. 6^{\circ} > 4 5^{\circ}$ ; it emerges at $sin^{- 1} (1.4 sin 4 5^{\circ}) \approx 81. 9^{\circ}$

Common Mistakes to Avoid

Watch out

Applying TIR to light going from a rarer to a denser medium. It can happen only from denser to rarer.
Using $sin C = \frac{1}{n}$ when the outer medium is not air. In general $sin C = \frac{n _{rarer}}{n _{denser}}$ .
Treating $i = C$ as total internal reflection. At $C$ the ray grazes the surface; TIR needs $i > C$ .
Measuring the angle of incidence from the surface instead of from the normal.
Writing the circle of illuminance radius as $h sin C$ or $\frac{h}{n}$ . It is $h tan C = \frac{h}{n ^{2} - 1}$ .
In fibre problems using the entry angle at the wall. The wall angle is $9 0^{\circ} - r$ , where $r$ is the refraction angle at the end face.
Assuming a glass prism still reflects totally in water. The critical angle grows when the surroundings are denser.
Thinking the maximum deviation occurs at grazing incidence. It occurs just above $i = C$ and equals $18 0^{\circ} - 2 C$ .

Frequently Asked Questions

What is the critical angle?

The critical angle is the angle of incidence in the denser medium for which the refracted ray just grazes the boundary, making 90 degrees with the normal. It is given by sin C equals n rarer divided by n denser. For glass in air it is about 41.8 degrees and for water about 48.6 degrees.

What are the conditions for total internal reflection?

Two conditions must both hold. Light must travel from an optically denser medium towards a rarer one, and the angle of incidence must be greater than the critical angle for that pair of media. Then no light is refracted and all of it is reflected back into the denser medium.

Why is total internal reflection better than reflection from a mirror?

In total internal reflection all of the light is reflected, with no absorption by a metal coating and no faint double images from the glass front surface. Mirrors reflect only about 90 to 95 percent. This is why periscopes and binoculars use totally reflecting prisms.

How does an optical fibre work?

Light entering one end strikes the side walls at angles larger than the critical angle, so it is totally reflected again and again and travels along the fibre, even around bends, with very little loss. A cladding of lower refractive index around the core keeps the reflections clean.

Why does a diamond sparkle?

Diamond has a very high refractive index of 2.42, so its critical angle is only 24.4 degrees. Its facets are cut so that light entering from the top is totally reflected several times inside before leaving through the top, which concentrates the light and makes the stone sparkle.

What causes a mirage on a hot road?

Air near a hot road is hotter and optically rarer than the air above. Light from the sky or a distant object bends away from the normal as it travels down through these layers and finally turns upward. The eye traces it back and sees an inverted image that looks like water.

Is total internal reflection important for JEE Main and JEE Advanced?

Yes. JEE Main asks critical angle, circle of illuminance and prism conditions regularly. JEE Advanced adds optical fibre acceptance angle, the deviation against incidence graph, TIR in prisms with changing surroundings and combinations with refraction at curved surfaces.

Which questions on total internal reflection come in NEET?

NEET mostly asks the definition and formula of the critical angle, the conditions for total internal reflection, and applications such as optical fibres, mirages, diamonds and totally reflecting prisms. Remember sin C equals 1 over n and that the light must go from denser to rarer.

Previous year questions on Critical Angle and Total Internal Reflection

11 questions from past papers, each with a step-by-step solution.

Show all 11 questions

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