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ScienceCBSE Class 10 · Board exam 2025, set 31/3/1 · 5 marks · Long answer

CBSE Class 10 Science 2025: question 36(a)

Set in the CBSE Class 10 board exam of 2025, paper 31/3/1, section D, question 36(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q36(a) · 5 marks · Long answerLight – Reflection and Refraction

(i) "In refraction of light through a rectangular glass slab, the emergent ray is always parallel to the direction of the incident ray." Why? Explain with the help of a ray diagram. What happens when a ray of light falls normally on one of the faces of a rectangular glass prism? Draw diagram.

(ii) An object is placed at a distance of 30 cm from the optical centre of a concave lens of focal length 20 cm. Use Lens formula to determine the position of the image formed in this case.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 31/3/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Compare what happens at the first face (air to glass) with what happens at the second face (glass to air).

Hint 2

If the angle of incidence is 0°, what does Snell's law give for the angle of refraction?

Hint 3

For a concave lens f is negative; use 1/v − 1/u = 1/f with u = −30 cm.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (i) Why the emergent ray is parallel to the incident ray: At the first face, light bends towards the normal as it goes from air into glass. At the second face, it bends away from the normal as it goes from glass into air. The two faces are parallel and the same pair of media is involved, so the bending at the second face is equal and opposite to the bending at the first face. The angle of emergence therefore equals the angle of incidence (). The emergent ray is parallel to the incident ray, only shifted sideways (lateral displacement).

    1 mark
  2. Ray diagram (glass slab): Draw a rectangle ABCD (the slab).

    • Incident ray PE meets face AB at E; draw normal and mark .
    • Refracted ray EF inside the glass bends towards the normal; mark .
    • At face CD, draw normal . The emergent ray FS bends away from the normal; mark .
    • Show FS parallel to PE: extend PE as a dotted line and mark the lateral displacement between it and FS.
    • Put arrows on all rays.
    1 mark
  3. Normal incidence: If a ray falls normally on a face (angle of incidence ), it enters the glass without bending, because . Diagram: a ray drawn perpendicular to one face passes straight through the glass with no deviation at that face, with an arrow showing its direction.

    1 mark
  4. (ii) For a concave lens, cm. The object is on the left, so cm.

    Lens formula: , so

    1 mark
  5. , so cm

    ½ mark
  6. The image is formed 12 cm from the optical centre, on the same side of the lens as the object. It is virtual and erect.

    ½ mark

Answer: (i) Bending at the two parallel faces is equal and opposite, so the emergent ray is parallel to the incident ray; a ray incident normally passes undeviated. (ii) v = −12 cm, i.e. 12 cm from the lens on the same side as the object (virtual, erect).

Where the marks are

  • Explains equal and opposite bending at the two parallel faces, so ∠e = ∠i (1 mark)
  • Correct slab ray diagram with normals, angles and arrows (1 mark)
  • States and draws that a normally incident ray passes undeviated (1 mark)
  • Correct sign convention and lens formula with substitution (1 mark)
  • v = −12 cm (½ mark)
  • States image is on the same side as the object (½ mark)

Revise the chapter

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.