Light - Reflection & Refraction CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Light - Reflection & Refraction CBQs

1. Question 1MCQ1 mark
   When an incident ray of light enters a medium from air, it bends towards the normal. Which of the following is TRUE about the refractive index of the medium ($n_m$) as compared to the refractive index of air ($n_a$)?
   • $n_m$ is equal to $n_a$.
   • $n_m$ is less than $n_a$.
   • $n_m$ is more than $n_a$.
   • The refractive indices cannot be compared based on the given information.
   Show step-wise solution

   1. [1 mark] Bending towards the normal means the light slows down: the medium is optically denser than air, so $n_m$ is more than $n_a$.

   Answer: $n_m$ is more than $n_a$.

2. Question 2Assertion-Reason1 mark
   Assertion (A): When Neha shines a torch beam straight at a plane mirror, perpendicular to its surface, the beam returns along the same path into the torch. Reason (R): For a ray falling along the normal, the angle of incidence is 0°, so the angle of reflection is also 0°.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A) Both are true and R explains A: a ray along the normal has i = 0°; since the angle of reflection equals the angle of incidence, r = 0°, so the ray goes back along the normal, retracing its path.

   Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

3. Question 3Short answer3 marks
   Smriti is looking at herself in a convex mirror in a science museum, standing 2 m away from the mirror. Her image appears to be around half her actual height. Estimate the focal length of the mirror.
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   1. [½ mark] Identify the object distance: u = -2 m (object in front of the mirror).
   2. [1 mark] The image is half her height, virtual and erect, so m = +1/2. Using m = -v/u: v = -m × u = -(1/2) × (-2 m) = +1 m (image behind the mirror).
   3. [½ mark] Use the correct mirror formula: 1/v + 1/u = 1/f.
   4. [1 mark] 1/f = 1/(+1) + 1/(-2) = 1/2, so f = +2 m. The mirror is convex with focal length 2 m.

   Answer: f = +2 m

4. Question 4Case study4 marks
   Students of a school in Rajkot built a simple solar cooker using a concave mirror. They pointed the mirror towards the Sun so that sunlight fell on it parallel to its principal axis. They found that the reflected rays met at a point on the principal axis 30 cm in front of the pole of the mirror, and placed a small black vessel of water at that point. The water became hot within a few minutes. Later, one of the students noticed that when she held her face close to the same mirror, she saw an enlarged image of her face, just like in a make-up mirror. (i) What is the radius of curvature of this mirror? (ii) Why was the vessel placed exactly at the point where the reflected rays met? (iii) The student's face is 20 cm in front of the pole of the same mirror. Using the mirror formula, find the position of the image and its magnification. State the nature of the image.
   Show step-wise solution

   1. [1 mark] (i) Parallel rays meet at the principal focus, so f = 30 cm; R = 2f = 2 × 30 = 60 cm.
   2. [1 mark] (ii) That point is the principal focus of the concave mirror: the parallel rays of sunlight converge there, so the heat is concentrated on the vessel and the water heats up quickly.
   3. [1 mark] (iii) u = −20 cm, f = −30 cm. 1/v + 1/u = 1/f ⇒ 1/v = 1/f − 1/u = −1/30 + 1/20 = (−2 + 3)/60 = 1/60 ⇒ v = +60 cm, i.e. the image is 60 cm behind the mirror.
   4. [1 mark] (iii) m = −v/u = −(+60)/(−20) = +3; the image is virtual, erect and magnified three times.

   Answer: (i) 60 cm; (ii) it is the principal focus, where sunlight converges; (iii) v = +60 cm (behind the mirror), m = +3 — virtual, erect, magnified.

5. Question 5Long answer5 marks
   In a school laboratory in Lucknow, Farhan wants to obtain a sharp image of a candle flame on a screen using a concave mirror of focal length 20 cm. (a) State the two laws of reflection of light. (b) The flame, 3 cm tall, is placed 30 cm in front of the mirror. At what distance from the mirror should the screen be placed to get a sharp image? Use the mirror formula. (c) Find the magnification and the height of the image, and state its nature. (d) Farhan then moves the flame to 10 cm from the mirror. Can he now obtain its image on the screen? Give a reason.
   Show step-wise solution

   1. [1 mark] (a) Laws: (i) angle of incidence = angle of reflection; (ii) incident ray, normal at the point of incidence and reflected ray lie in the same plane.
   2. [0.5 mark] (b) u = −30 cm, f = −20 cm; mirror formula 1/v + 1/u = 1/f.
   3. [1 mark] (b) 1/v = 1/f − 1/u = −1/20 + 1/30 = −1/60, so v = −60 cm: screen 60 cm in front of the mirror.
   4. [0.5 mark] (c) m = −v/u = −(−60)/(−30) = −2.
   5. [1 mark] (c) h' = m × h = −2 × 3 = −6 cm; image is real, inverted and magnified (6 cm tall).
   6. [1 mark] (d) No: object between P and F gives 1/v = −1/20 + 1/10 = 1/20, v = +20 cm, a virtual, erect image behind the mirror, which cannot be taken on a screen.

   Answer: (a) i = r; incident ray, normal and reflected ray in one plane; (b) screen 60 cm in front of the mirror (v = −60 cm); (c) m = −2, image 6 cm tall, real, inverted, magnified; (d) no — the image is virtual (object within focus).

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