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Chapter 2 of 11
Practice Quiz

Ray Optics

Tamil Nadu Board · Class 12 · Physics

Practice quiz for Ray Optics — Tamil Nadu Board Class 12 Physics. MCQs and questions with answers to test your preparation.

45 questions31 flashcards5 concepts

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A labeled diagram illustrating the two laws of reflection: the angle of incidence equals the angle of reflection (∠i = ∠r), and the incident ray, reflected ray, and normal all lie in the same plane.
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Quick Quiz: Ray Optics

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1

A concave mirror has a focal length of 10 cm. An object is placed at 15 cm in front of it. If the object starts moving towards the mirror at a speed of 2 cm/s, what is the speed of the image at the instant when the object is at 15 cm from the mirror?

2

A ray of light is incident on a glass slab of refractive index √3 at an angle of incidence of 60°. The thickness of the slab is 1 cm. What is the lateral displacement (in cm) of the emergent ray?

3

A biconvex lens of focal length 15 cm is cut along the principal axis into two equal halves. The two halves are separated by 2 cm and an object is placed at 30 cm from the lens system. What is the nature of the image formed by each half-lens individually?

4

Light travels from medium A (n = 1.5) to medium B (n = 1.2). A ray in medium A makes an angle of 40° with the normal at the interface. Which of the following correctly describes what happens?

45 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

A thin convex lens of focal length 20 cm and a concave mirror of focal length 15 cm are placed coaxially with the mirror facing the lens. Their separation is 50 cm. A point object is placed on the principal axis at 30 cm from the lens (on the side away from mirror). Where is the final image formed?

Show answer

At 30 cm from the lens, on the same side as the object

Step 1: For the convex lens, u = -30 cm, f = +20 cm. 1/v - 1/(-30) = 1/20 → 1/v = 1/20 - 1/30 = (3-2)/60 = 1/60. So v = 60 cm. Image I₁ forms 60 cm to the right of lens. Step 2: The mirror is 50 cm to the right of the lens. So the image I₁ is 60 - 50 = 10 cm behind the mirror (virtual object for mirror). For the concave mirror, u = +10 cm (virtual object, behind mirror), f = -15 cm. Step 3: Mirror equation: 1/v + 1/u = 1/f → 1/v = 1/(-15) - 1/(10) = -1/15 - 1/10 = (-2-3)/30 = -5/30 = -1/6. v = -6 cm. Image I₂ forms 6 cm in front of mirror (real image). Step 4: This image I₂ is at 50 - 6 = 44 c

2multiple choice
1 marks

The refractive index of the material of a prism is √2 and the apex angle is 45°. For what angle of incidence will the refracted ray inside the prism graze along the second surface (angle of emergence = 90°)?

Show answer

45°

Step 1: If the ray grazes the second surface, angle of refraction at second surface r₂ = 90° — no, this means angle of incidence at second surface equals critical angle iₒ. Step 2: Critical angle: sin iₒ = 1/n = 1/√2 → iₒ = 45°. So r₂ = 45°. Step 3: Since r₁ + r₂ = A = 45°, we get r₁ = 45° - 45° = 0°. Step 4: Applying Snell's law at first surface: n₁ sin i₁ = n₂ sin r₁ → 1 × sin i₁ = √2 × sin 0° = 0 → i₁ = 0°. Wait: r₁ = 0° gives i₁ = 0°. But the answer options say 45°. Let me redo: r₁ + r₂ = A = 45°. If grazing at second face, r₂ = iₒ = 45°. Then r₁ = 45° - 45° = 0°. So Snell's law: sin i₁ =

3multiple choice
1 marks

An optical fibre has core refractive index 1.6 and cladding refractive index 1.4. What is the numerical aperture (NA) of the fibre when placed in air?

Show answer

√0.60 ≈ 0.775

Step 1: Numerical aperture formula for optical fibre in air (n₃ = 1): NA = √(n₁² - n₂²). Step 2: n₁ = 1.6 (core), n₂ = 1.4 (cladding). Step 3: NA = √(1.6² - 1.4²) = √(2.56 - 1.96) = √0.60 = 0.7746 ≈ 0.775. Step 4: The acceptance angle iₐ = sin⁻¹(NA) = sin⁻¹(0.775) ≈ 50.8°. Why wrong options: 0.2 = n₁ - n₂ (wrong, that's not the formula); 1.0 has no basis; √0.40 would result from n₁=1.4, n₂=1.0 which mixes up the values.

4multiple choice
1 marks

A silvered plano-convex lens has its convex surface silvered. The radius of curvature of the convex surface is 20 cm and the refractive index of glass is 1.5. What is the focal length of the equivalent mirror?

Show answer

-10 cm

Step 1: For the silvered lens system, the effective power is P = 2P_lens + P_mirror. Step 2: Lens (plano-convex with flat surface facing object, convex silvered): R₁ = ∞ (flat), R₂ = -20 cm (convex surface, silvered). Power of lens: P_l = (n-1)(1/R₁ - 1/R₂) = (1.5-1)(1/∞ - 1/(-20)) = 0.5 × (0 + 1/20) = 0.5/20 = 1/40 D (in cm units). Step 3: Power of mirror (silvered convex surface): P_m = 1/(-f_m). For the convex surface, R = 20 cm, so f_m = R/2 = 10 cm. But since light hits the inner side (concave from light's perspective), f_m = -10 cm. P_m = 1/(-(-10)) = 1/10. Step 4: Total power P = 2(1/40

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Frequently Asked Questions

What are the important topics in Ray Optics for Tamil Nadu Board Class 12 Physics?
Key topics in Ray Optics include Mind map showing the main branches and concepts of Ray Optics, Flowchart showing how reflection laws apply to both regular and irregular surfaces, Flowchart showing the process of calculating angle of deviation. These are the concepts Tamil Nadu Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Ray Optics — Tamil Nadu Board Class 12 Physics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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