Total Internal Reflection PYQs
(SpeEdLabs)
1. Consider light travelling from a medium A to medium B separated by a plane
interface. If the light undergoes total internal reflection during its travel from
medium A to B and the speed of light in media A and B are 2.4 × 10^8 m/s and 2.7
× 10^8 m/s, respectively, then the value of the critical angle is:
A. cos⁻¹(8/9)
B. sin⁻¹(9/8)
C. cot⁻¹(3/√13)
D. tan⁻¹(8/√17)
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Solution:
For total internal reflection to occur, light must travel from a denser medium (lower speed) to a rarer medium
(higher speed).
The speed of light in medium A (denser) is
vₐ = 2.4 × 10⁸ m/s
The speed of light in medium B (rarer) is
vᵦ = 2.7 × 10⁸ m/s
Since vₐ < vᵦ, medium A is optically denser than medium B, satisfying the condition for total internal reflection.
The critical angle is determined using Snell’s law at the boundary where the angle of refraction is 90°:
nₐ sin(θc) = nᵦ sin(90°)
⇒ sin(θc) = nᵦ / nₐ
Since refractive index n = c / v,
nᵦ / nₐ = (c / vᵦ) / (c / vₐ) = vₐ / vᵦ
So,
sin(θc) = vₐ / vᵦ
= (2.4 × 10⁸) / (2.7 × 10⁸)
= 8/9
Now, construct a right triangle:
sin(θc) = opposite / hypotenuse = 8/9
⇒ adjacent side = √(9² − 8²) = √(81 − 64) = √17
Thus,
tan(θc) = opposite / adjacent = 8 / √17
⇒ θc = tan⁻¹(8 / √17)
Therefore, the critical angle is:
tan⁻¹(8 / √17)
Hence, the correct option is (D).
2. A ray of light entering from air into a denser medium of refractive index , as
shown in figure. The light ray suffers total internal reflection at the adjacent
surface as shown. The maximum value of angle should be equal to :
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A. sin⁻¹(√7 / 3)
B. sin⁻¹(√5 / 4)
C. sin⁻¹(√7 / 4)
D. sin⁻¹(√5 / 3)
Solution:
At maximum angle θ, the ray at point B goes into grazing emergence. For all smaller values of θ, total internal
reflection (TIR) occurs.
At point B:
(4/3) × sin(θ″) = 1 × sin(90°)
⇒ θ″ = sin⁻¹(3/4)
θ′ = (π/2 − θ″)
At point A:
1 × sin(θ) = (4/3) × sin(θ′)
sin(θ) = (4/3) × sin(π/2 − θ″)
sin(θ) = (4/3) × cos(θ″)
sin(θ) = (4/3) × cos[sin⁻¹(3/4)]
Now,
cos[sin⁻¹(3/4)] = √(1 − (3/4)²) = √(1 − 9/16) = √(7/16) = √7 / 4
So,
sin(θ) = (4/3) × (√7 / 4) = √7 / 3
⇒ θ = sin⁻¹(√7 / 3)
Final Answer: sin⁻¹(√7 / 3) (Option A)
3. The speed of light in media A and B are 2.0 × 10¹⁰ cm/s and 1.5 × 10¹⁰ cm/s,
respec vely. A ray of light enters from medium B to A at an incident angle θ. If the
ray suffers total internal reflec on, then:
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A. θ = sin⁻¹(3/4)
B. θ > sin⁻¹(2/3)
C. θ < sin⁻¹(3/4)
D. θ > sin⁻¹(3/4)
Solution:
μₐ = (3 × 10⁸) / (2 × 10⁸) = 1.5
μᵦ = (3 × 10⁸) / (1.5 × 10⁸) = 2
For total internal reflection (TIR):
θ > i_c
θ > sin⁻¹(μₐ / μᵦ)
θ > sin⁻¹(1.5 / 2)
θ > sin⁻¹(3 / 4)
Hence, option(D) is Correct
Sample Questions
1. A ray of light travels from a medium with refractive index μ₁ to a medium with
refractive index μ₂ (μ₁ > μ₂). If the critical angle for this interface is C, then the
value of sin C is:
A) μ₁ / μ₂
B) μ₂ / μ₁
C) √(μ₁μ₂)
D) 1 / (μ₁μ₂)
Solution:
By Snell’s Law, at the critical angle, the angle of refraction is 90°.
μ₁ sin C = μ₂ sin 90°
Since sin 90° = 1, we get sin C = μ₂ / μ₁.
Hence option B is correct.
2. An underwater swimmer at a depth h looks up at the surface of the water
(refractive index μ). The radius of the circular horizon through which the
outside world is visible is:
A) h / √(μ² - 1)
B) h√(μ² - 1)
C) μh / √(μ² - 1)
D) h / μ
Solution:
Light enters from the outside and reaches the eye within a cone of angle C.
The radius r is given by r = h tan C.
Using sin C = 1 / μ,
we find tan C = 1 / √(μ² - 1).
Thus, r = h / √(μ² - 1).
Hence Option A is correct.
3. A light ray is incident at an angle of 45° on a glass-air interface. If the
refractive index of glass is √2, what happens to the ray?
A) It undergoes Total Internal Reflection
B) It grazes the surface (angle of refraction = 90°)
C) It refracts into the air at 30°
D) It reflects back at 30°
Solution:
Critical angle C = sin⁻¹(1/μ)
C = sin⁻¹(1/√2) = 45°.
Since the incident angle (45°) is exactly equal to the critical angle,
the ray grazes the surface.
So, Option B is correct.
4. The velocity of light in a liquid is 1.5 x 10⁸ m/s and in air is 3 x 10⁸ m/s.
Calculate the critical angle for a ray passing from the liquid to air.
A) 30°
B) 45°
C) 60°
D) 90°
Solution:
μ_liquid = c / v_liquid = (3 x 10⁸) / (1.5 x 10⁸) = 2.
Sin C = 1 / μ
Sin C = 1/2.
Thus, C = sin⁻¹(1/2)
C = 30°.
Hence Option A is Correct.
5. A light ray enters a rectangular glass block of refractive index √3 at an angle of
60°. After traveling some distance, it hits the side face. Will it emerge from the
side face into the air?
A) Yes, at 30°
B) Yes, at 90°
C) No, it undergoes TIR
D) It travels straight through
Solution:
Using Snell's Law at the first surface: 1 * sin 60° = √3 * sin r (√3/2) = √3 * sin
r → sin r = 1/2 → r = 30°.
The angle of incidence at the side face i' = 90° - 30° = 60°.
Critical angle C = sin⁻¹(1/√3) ≈ 35.3°.
Since i' (60°) > C (35.3°),
the ray undergoes TIR and does not emerge.
Hence Option C is Correct.