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ScienceCBSE Class 10 · Board exam 2024, set 31/1/1 · 4 marks · Case-based

CBSE Class 10 Science 2024: question 39

Set in the CBSE Class 10 board exam of 2024, paper 31/1/1, section E, question 39, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q39 · 4 marks · Case-basedCompetency-basedLight – Reflection and Refraction

Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors :

CaseMirrorFocal Length (cm)Object Distance (cm)
1A2045
2B1530
3C3020
  1. (i)

    In which one of the above cases the mirror will form a diminished image of the object? Justify your answer.

    1 mark
  2. (ii)

    List two properties of the image formed in case 2.

    1 mark
  3. (iii)(A)

    What is the nature and size of the image formed by mirror C? Draw ray diagram to justify your answer.

    2 marks
  4. OR

    (iii)(B)

    An object is placed at a distance of 18 cm from the pole of a concave mirror of focal length 12 cm. Find the position of the image formed in this case.

    2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 31/1/1.

Hints

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

Hint 1

Find the centre of curvature (2f) for each mirror and see where each object lies relative to F and C.

Hint 2

Recall the image table for a concave mirror: beyond C, at C, between P and F.

Hint 3

For (iii)(B), take both u and f negative in the mirror formula.

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) Case 1 (mirror A) forms a diminished image.

    ½ mark
  2. Justification: for A, cm, so cm; the object at 45 cm lies beyond C, and a concave mirror forms a diminished image between F and C.

    ½ mark
  3. (ii) Case 2: cm, object at 30 cm , i.e. at C. The image is formed at C and is real and inverted, and of the same size as the object (any two).

    1 mark
  4. (iii)(A) Mirror C: cm, object at 20 cm, i.e. between P and F. The image is virtual and erect,

    ½ mark
  5. and magnified (larger than the object), formed behind the mirror.

    ½ mark
  6. Ray diagram (described): draw the principal axis with P, F (30 cm) and C (60 cm); place object AB at 20 cm. Ray 1 from A parallel to the axis reflects passing through F. Ray 2 from A directed towards the pole reflects symmetrically about the axis (or a ray along the line from C through A reflects back on itself). The reflected rays diverge; extend them backward with dotted lines behind the mirror — they meet at A′, giving an enlarged, erect image A′B′ behind the mirror. Mark arrows on all rays.

    1 mark
  7. (iii)(B) Using sign convention: cm, cm. Mirror formula: .

    ½ mark
  8. ½ mark
  9. cm

    ½ mark
  10. The image is formed 36 cm in front of the mirror (on the same side as the object), so it is real.

    ½ mark

Answer: (i) Case 1 (mirror A), object beyond C; (ii) real, inverted, same size; (iii)(A) virtual, erect, magnified; (iii)(B) v = −36 cm, 36 cm in front of the mirror

Where the marks are

  • Identifies mirror A / case 1 (½ mark)
  • Justifies: object beyond C (½ mark)
  • Any two properties of image in case 2 (½ + ½ mark)
  • Nature virtual and erect (½ mark)
  • Size magnified (½ mark)
  • Correct ray diagram with arrows (1 mark; ½ deducted if ray directions not marked)
  • Mirror formula with correct signs (½ mark)
  • Substitution and simplification (½ mark)
  • v = −36 cm (½ mark)
  • Image position stated: 36 cm in front of mirror (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.