Definition
Total internal reflection
Occurs when light goes from denser to rarer medium and angle of incidence exceeds critical angle θ_c = sin⁻¹(1/n). Applications: optical fibres, prisms in periscopes.
-- NCERT Class 12 Physics, Ch. 9, p. 230Total Internal Reflection
8 MCQs1 revision card9-step worked example
Source: NCERT OpticsPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 26 Sep 2026
Total Internal Reflection, explained for NEET
The trap in total internal reflection is the direction of the ratio. The critical angle satisfies sin θ_c = n₂/n₁, with n₁ the denser medium the light is already travelling in and n₂ the rarer one it is trying to enter. For glass in air that is sin θ_c = 1/1.5, giving about 42°. Write sin θ_c = n₁/n₂ instead and you get sin θ_c = 1.5 — impossible, but under exam pressure the arcsine key still produces something and the nearest option absorbs it.
Two conditions must both hold, and NEET stems routinely give you only one. Light must travel from denser to rarer, and the angle of incidence must exceed θ_c. Denser-to-rarer alone is not enough; a large angle of incidence in the wrong direction is not enough either. Light going air-to-glass never undergoes total internal reflection at any angle, because the refracted ray always exists.
NCERT Class 12 Physics Part 2, Chapter 9, page 230 defines the effect: at incidence beyond the critical angle no refracted ray is produced at all and the entire incident energy returns into the denser medium. That word total is literal — unlike ordinary partial reflection at a boundary, nothing is transmitted. This is why an optical fibre carries light around bends with almost no loss, and why a diamond (n ≈ 2.42, θ_c ≈ 24°) traps light through many internal bounces before releasing it.
The relation is inverse: a larger refractive index gives a smaller critical angle. Denser material, easier to trap light. Aspirants who reason "bigger n, bigger angle" invert every comparison question in the set.
Watch-out for the exam hall: when a stem names two media by refractive index without saying which the light starts in, the answer depends on that reading. Identify the denser medium first, confirm the ray begins there, and only then compute. If the light begins in the rarer medium, total internal reflection is off the table regardless of the angle quoted.
Can you answer these Total Internal Reflection MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Total internal reflection can occur only when light travels
Show answer and why every option is right or wrong
Answer: B. Both conditions are required together — denser to rarer, and incidence beyond the critical angle, as defined in NCERT Class 12 Physics Part 2, Chapter 9, page 230.
Why A is wrong: A is wrong because light entering a denser medium is always refracted; a refracted ray exists at every angle of incidence, so no total reflection is possible.
Why C is wrong: C is wrong because it states the angle condition correctly but the direction backwards; rarer-to-denser never produces total internal reflection.
Why D is wrong: D is wrong because it reverses the angle condition; below the critical angle the ray is partly refracted out and partly reflected, which is not total.
MCQ 2Easy RecallPractice
For light passing from a medium of refractive index n₁ into a medium of refractive index n₂, with n₁ > n₂, the critical angle θ_c is given by
Show answer and why every option is right or wrong
Answer: B. The critical angle is the incidence angle for which the refraction angle reaches 90°, giving sin θ_c = n₂/n₁ with the denser index in the denominator (NCERT Class 12 Physics Part 2, Chapter 9, page 230).
Why A is wrong: A is wrong because it inverts the ratio; with n₁ > n₂ this gives sin θ_c > 1, which has no solution.
Why C is wrong: C is wrong because the tangent form belongs to the polarizing-angle relation, not to the critical angle.
Why D is wrong: D is wrong because the boundary condition is the refraction angle reaching 90°, and the sine of the incidence angle — not its cosine — carries that ratio.
MCQ 3Easy RecallPractice
At the critical angle of incidence, the angle of refraction in the rarer medium is
Show answer and why every option is right or wrong
Answer: C. The critical angle is defined as the incidence angle at which refraction reaches exactly 90°, so the refracted ray just grazes the boundary (NCERT Class 12 Physics Part 2, Chapter 9, page 230).
Why A is wrong: A is wrong because equal incidence and refraction angles occur only when the two refractive indices are equal, so no critical angle exists.
Why B is wrong: B is wrong because refraction along the normal happens at normal incidence, which is the opposite extreme of the critical angle.
Why D is wrong: D is wrong because the refracted ray vanishes only beyond the critical angle; at the critical angle itself it still exists, grazing the surface.
MCQ 4Direct ApplicationPractice
A transparent medium has a refractive index of 2.00 and is surrounded by air of refractive index 1.00. The critical angle for the medium–air boundary is
Show answer and why every option is right or wrong
Answer: C. sin θ_c = n₂/n₁ = 1.00/2.00 = 0.500, so θ_c = 30.0° (critical-angle relation, NCERT Class 12 Physics Part 2, Chapter 9, page 230).
Why A is wrong: A is wrong because 60.0° is the angle whose cosine is 0.500; the relation uses the sine of the critical angle.
Why B is wrong: B is wrong because 45.0° would require sin θ_c = 0.707, corresponding to a refractive index of about 1.41, not 2.00.
Why D is wrong: D is wrong because it halves the correct 30.0° result, treating the index ratio as if it scaled the angle directly rather than its sine.
MCQ 5Direct ApplicationPractice
Light inside a glass block of refractive index 1.50 strikes the glass–air boundary. Taking the critical angle as 41.8°, total internal reflection occurs for an angle of incidence of
Show answer and why every option is right or wrong
Answer: A. Total internal reflection requires incidence strictly greater than the critical angle, and only 50.0° exceeds 41.8° (NCERT Class 12 Physics Part 2, Chapter 9, page 230).
Why B is wrong: B is wrong because 41.0° falls just short of 41.8°; a refracted ray still emerges, close to grazing but present.
Why C is wrong: C is wrong because 20.0° is well below the critical angle and most of the light is transmitted into the air.
Why D is wrong: D is wrong because 30.0° is below the critical angle, so the ray is partly refracted into the air and partly reflected.
MCQ 6Direct ApplicationPractice
Light travelling in water of refractive index 1.33 meets a boundary with a denser glass of refractive index 1.50. For this water-to-glass passage, the critical angle is
Show answer and why every option is right or wrong
Answer: D. A critical angle exists only for denser-to-rarer passage; here light moves from water into the denser glass, so a refracted ray always exists and total internal reflection is impossible (NCERT Class 12 Physics Part 2, Chapter 9, page 229).
Why A is wrong: A is wrong because 90.0° is the angle of refraction at the critical angle in the reverse passage, not the critical angle itself.
Why B is wrong: B is wrong because it applies the critical-angle relation to a rarer-to-denser passage where the concept does not apply; the numerical value has no physical meaning here.
Why C is wrong: C is wrong on two counts: the direction is rarer-to-denser, and the ratio is inverted — sin θ_c = 1.50/1.33 exceeds 1 and has no solution.
MCQ 7Concept TrapPractice
Three transparent materials in air have refractive indices 1.33, 1.50 and 2.42. Arranged in order of increasing critical angle at their boundary with air, the sequence is
Show answer and why every option is right or wrong
Answer: D. Since sin θ_c = 1/n, the critical angle decreases as n increases, so the largest index gives the smallest angle (NCERT Class 12 Physics Part 2, Chapter 9, page 230).
Why A is wrong: A is wrong because it assumes the critical angle grows with refractive index; the relation is inverse, so this ordering is exactly reversed.
Why B is wrong: B is wrong because it misorders the first two indices and still treats the relationship as increasing rather than inverse.
Why C is wrong: C is wrong because although it starts with the largest index, it then places 1.33 before 1.50, whereas 1.33 gives the largest critical angle of the three and must come last.
MCQ 8CalculationPractice
A ray inside a medium of refractive index 1.60 strikes a boundary with a second medium of refractive index 1.20. Taking sin 48.6° = 0.750, the ray will undergo total internal reflection at an angle of incidence of
Show answer and why every option is right or wrong
Answer: A. sin θ_c = 1.20/1.60 = 0.750, so θ_c = 48.6°; total internal reflection then requires incidence above 48.6°, and only 55.0° qualifies (NCERT Class 12 Physics Part 2, Chapter 9, page 230).
Why B is wrong: B is wrong in its arithmetic: 1.20/1.60 = 0.750, not 0.600, so the critical angle is 48.6° and not 36.9°; at 40.0° the ray is still refracted out.
Why C is wrong: C is wrong because the condition is strict — at 45.0°, below the critical angle, a refracted ray emerges; proximity to the critical angle does not satisfy it.
Why D is wrong: D is wrong because it checks only the direction condition and ignores the angle condition; 30.0° is far below the critical angle, so light is transmitted.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Total Internal Reflection: quick recall before you leave
How do you solve a Total Internal Reflection question? A worked example
- 1
Given
• Refractive index of slab, n₁ = 1.25• Refractive index of air, n₂ = 1.00 (exact)• Angle of incidence to test, i = 50.0°• Supplied trigonometric value: sin 53.1° = 0.800
- 2
Required
• The critical angle θ_c at the slab–air boundary.• Whether total internal reflection occurs at i = 50.0°.
- 3
Concept
Light is passing from the denser slab into rarer air, so a critical angle exists. Beyond it, no refracted ray is produced and all the incident energy returns into the slab (NCERT Class 12 Physics Part 2, Chapter 9, page 230). Below it, the ray is partly refracted out.
- 4
Formula
sin θ_c = n₂/n₁
- 5
Substitution
sin θ_c = 1.00/1.25
- 6
Calculation
sin θ_c = 0.800
Therefore θ_c = 53.1°, using the supplied value sin 53.1° = 0.800.
The refractive index of air, n₂ = 1.00, is an exact defined reference value and does not limit the significant-figure count; the precision of the answer is set by n₁ = 1.25, which carries three significant figures.
Condition check: the angle of incidence is 50.0°, and 50.0° < 53.1°. - 7
Final answer
The critical angle is 53.1°. A ray incident at 50.0° is not totally internally reflected — it falls below the critical angle, so part of the light refracts out into the air and part reflects back.
- 8
Common trap
The direction of the index ratio. Writing sin θ_c = n₁/n₂ = 1.25 produces a quantity greater than 1, which no angle satisfies — but a rushed candidate reads it as 1.25 radians or rounds toward the nearest option instead of recognising the impossibility. The denser index always sits in the denominator. The second trap here is answering "yes, TIR occurs" the moment the stem confirms denser-to-rarer travel, without comparing 50.0° against the computed 53.1°. Both conditions must be checked.
- 9
Similar NEET-style question
A ray inside a liquid of refractive index 1.40 meets the liquid–air boundary. Given sin 45.6° = 0.714, find the critical angle and state whether a ray incident at 60.0° emerges into the air. *(Answer: θ_c = 45.6°; 60.0° > 45.6°, so the ray is totally internally reflected and does not emerge.)*
What to remember before solving Total Internal Reflection questions
Which Total Internal Reflection formulas do you need for NEET?
1 formula — click to collapse
Critical angle (TIR)
Beyond this angle of incidence, total internal reflection occurs.
| Symbol | Quantity | SI Unit |
|---|---|---|
| theta_c | critical angle | rad |
| n1 | denser medium index | - |
| n2 | rarer medium index | - |
Valid when
- Light going from denser to rarer medium
- Plane interface
More in Optics: 9 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Total Internal Reflection questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys. — click to collapse
NEET 2023
1sin⁻¹(10t₂/t₁)
2sin⁻¹(t₁/(10 t₂))
3sin⁻¹(10 t₁/t₂)
4sin⁻¹(t₂/t₁)
NTA Answer: Option 3(final)
How does NEET ask about Total Internal Reflection?
1 recurring pattern from past papers — click to collapse
Critical angle: sin(θ_c) = 1/n. Apply to given material; check TIR condition.
Direct ApplicationEasy
Common distractors
inverts ratio
Uses n instead of 1/n
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →