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Refraction at plane/spherical surfaces and total internal reflection: NEET MCQs with answers

43 questions on this topic of Optics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A ray of light traveling inside a denser medium of refractive index √2 strikes a plane interface with a rarer medium at an angle of incidence of 45°. If the angle of refraction is 90°, what is the speed of light in the rarer medium? (Speed of light in vacuum c = 3 × 10⁸ m/s)

  1. 1.50 × 10⁸ m/s
  2. 2.12 × 10⁸ m/s
  3. 2.00 × 10⁸ m/s
  4. 3.00 × 10⁸ m/s
Show answer and explanation

Answer: (D) 3.00 × 10⁸ m/s

By Snell's law, μ₁ sin 45° = μ₂ sin 90°. Thus, √2 × (1/√2) = μ₂ × 1, which gives μ₂ = 1. A refractive index of 1 corresponds to vacuum or air, where the speed of light is 3.00 × 10⁸ m/s.

medium

2. A glass hemisphere of radius R and refractive index μ = 1.5 is placed on a table with its flat face downwards. A small mark on the tabletop at the center of the flat face is viewed from directly above through the curved surface. The mark appears to be at a distance:

  1. 4R/3 below the top vertex
  2. 2R/3 below the top vertex
  3. R/3 below the top vertex
  4. R below the top vertex
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Answer: (D) R below the top vertex

Light rays starting from the mark at the center of the flat face travel radially outwards towards the curved spherical surface. Because each ray strikes the curved surface along its normal (the center of curvature of the curved surface is the mark itself), no refraction or bending occurs at the surface (angle of incidence = 0°). Hence, the rays emerge undeviated, and the mark appears at its actual position, which is at the center, a distance R below the top vertex.

hard

3. A parallel beam of light is incident from air on a solid glass sphere of radius R and refractive index μ = 1.5. At what distance from the back surface (second pole) of the sphere does the beam come to a focus?

  1. R / 2
  2. R / 4
  3. R
  4. 2 R
Show answer and explanation

Answer: (A) R / 2

For the first refraction at pole P₁: n₁ = 1, n₂ = 1.5, u₁ = -∞, R₁ = +R. Using n₂/v₁ - n₁/u₁ = (n₂ - n₁)/R₁ gives 1.5/v₁ - 0 = (1.5 - 1)/R = 1/(2R), which yields v₁ = 3R from P₁. Since the diameter of the sphere is 2R, this virtual object for the second surface is at u₂ = +(3R - 2R) = +R from the second pole P₂. For the second surface: n₁ = 1.5, n₂ = 1, R₂ = -R. Refraction formula: 1/v₂ - 1.5/(+R) = (1 - 1.5)/(-R) = +1/(2R). Therefore, 1/v₂ = 1/(2R) + 3/(2R) = 4/(2R) = 2/R, giving v₂ = R/2 from the second surface.

medium

4. Consider the following two statements: Statement I: The twinkling of stars is primarily caused by total internal reflection occurring across turbulent layers of the atmosphere. Statement II: Mirages observed in deserts on hot summer days are due to total internal reflection occurring in progressively warmer, less dense air layers near the ground. Choose the correct option:

  1. Statement I is false but Statement II is true
  2. Both Statement I and Statement II are true
  3. Statement I is true but Statement II is false
  4. Both Statement I and Statement II are false
Show answer and explanation

Answer: (A) Statement I is false but Statement II is true

Twinkling of stars is caused by atmospheric refraction through layers of fluctuating optical density, not total internal reflection. Mirages are indeed caused by total internal reflection when light bends away from warmer, rarer air layers near the ground until the angle of incidence exceeds the critical angle.

easy

5. A microscope is focused on an ink dot at the bottom of a beaker. When a liquid of refractive index 1.6 is poured into the beaker to a depth of 8.0 cm, by what distance must the microscope be raised to bring the ink dot back into sharp focus?

  1. 2.0 cm
  2. 5.0 cm
  3. 3.0 cm
  4. 4.8 cm
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Answer: (C) 3.0 cm

The apparent shift Δt = t(1 - 1/μ) = 8.0 × (1 - 1/1.6) = 8.0 × (1 - 5/8) = 8.0 × 3/8 = 3.0 cm. Thus, the microscope must be raised by 3.0 cm.

hard

6. A light ray enters a rectangular glass block (μ = √3) from air at an angle of incidence θ. What is the maximum value of θ such that the ray undergoes total internal reflection at the adjacent perpendicular vertical face of the block?

  1. 60°
  2. 45°
  3. 90°
  4. 30°
Show answer and explanation

Answer: (C) 90°

For total internal reflection at the vertical face, the angle of incidence inside must be r' ≥ θc, where sin θc = 1/μ = 1/√3. Since r + r' = 90°, we need r = 90° - r' ≤ 90° - θc, so sin r ≤ cos θc = √(1 - 1/3) = √(2/3). By Snell's law at the front face, sin θ = μ sin r ≤ √3 × √(2/3) = √2 > 1. Since sin θ cannot exceed 1, the condition is satisfied for all possible angles of incidence from 0° to 90°. Thus, the maximum value of θ is 90°.

easy

7. A ray of light is incident from a denser medium 1 into a rarer medium 2. If the speed of light in medium 1 is v₁ and in medium 2 is v₂, what is the critical angle θc for this pair of media?

  1. sin⁻¹(v₁ / v₂)
  2. sin⁻¹(v₂ / v₁)
  3. cos⁻¹(v₁ / v₂)
  4. tan⁻¹(v₂ / v₁)
Show answer and explanation

Answer: (A) sin⁻¹(v₁ / v₂)

By definition, sin θc = n₂ / n₁. Since refractive index n is inversely proportional to wave speed v (n = c / v), we get n₂ / n₁ = v₁ / v₂. Thus, θc = sin⁻¹(v₁ / v₂).

easy

8. A light ray travels from medium A into medium B. The velocity of light in medium A is 2.0 × 10⁸ m/s and in medium B is 2.5 × 10⁸ m/s. What is the critical angle for a ray attempting to cross from medium A into medium B?

  1. sin⁻¹(0.80)
  2. sin⁻¹(0.75)
  3. sin⁻¹(0.60)
  4. sin⁻¹(0.50)
Show answer and explanation

Answer: (A) sin⁻¹(0.80)

Medium A is denser because v_A < v_B. The refractive index of medium A relative to B is μ = v_B / v_A = (2.5 × 10⁸)/(2.0 × 10⁸) = 1.25. The critical angle C satisfies sin C = 1/μ = v_A / v_B = 2.0/2.5 = 0.80. Thus, C = sin⁻¹(0.80).

easy

9. Consider the following two statements regarding optical fibres: Statement I: Light is guided along an optical fibre by undergoing successive total internal reflections at the core-cladding boundary. Statement II: The refractive index of the cladding material must be strictly greater than that of the core material. In the light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Light propagates through an optical fibre via multiple total internal reflections at the interface between the core and the cladding. For total internal reflection to occur, light must travel from a denser to a rarer medium, meaning the core must have a higher refractive index than the cladding (n_core > n_cladding). Therefore, Statement I is correct and Statement II is incorrect.

medium

10. A vessel of depth 21 cm is half filled with an oil of refractive index 1.4 and the remaining upper half is filled with water of refractive index 1.33. When viewed normally from above, what is the apparent depth of the bottom of the vessel?

  1. 15.4 cm
  2. 16.2 cm
  3. 14.5 cm
  4. 13.8 cm
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Answer: (A) 15.4 cm

Each liquid layer has an actual depth d = 21/2 = 10.5 cm. Apparent depth d' = (d₁/μ₁) + (d₂/μ₂) = (10.5 / 1.333) + (10.5 / 1.4) = (10.5 × 3/4) + (10.5 / 1.4) = 7.875 + 7.5 = 15.375 cm ≈ 15.4 cm.

medium

11. An optical fiber consists of a core of refractive index n₁ = 1.50 and a cladding of refractive index n₂ = 1.40. The maximum angle of incidence θ in air at the core end face for which the guided rays undergo total internal reflection inside the fiber is:

  1. sin⁻¹(0.40)
  2. sin⁻¹(0.70)
  3. sin⁻¹(0.29)
  4. sin⁻¹(0.54)
Show answer and explanation

Answer: (D) sin⁻¹(0.54)

The acceptance angle is given by sin θ = √(n₁² - n₂²). Here √(1.50² - 1.40²) = √(2.25 - 1.96) = √0.29 ≈ 0.5385 ≈ 0.54. Hence, θ = sin⁻¹(0.54).

easy

12. A glass slab of refractive index 1.5 and thickness 9 cm has a tiny air bubble trapped inside it. When viewed normally from one face, it appears at a depth of 4 cm, and when viewed normally from the opposite face, it appears at a depth of 2 cm. What is the actual distance of the bubble from the first face?

  1. 3 cm
  2. 6 cm
  3. 5 cm
  4. 4.5 cm
Show answer and explanation

Answer: (B) 6 cm

Let the actual distance from the first face be x. Then the distance from the second face is 9 - x. The apparent depth from the first face is x/μ = 4 cm, which gives x = 4 × 1.5 = 6 cm.

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