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 :
| Case | Mirror | Focal Length (cm) | Object Distance (cm) |
|---|---|---|---|
| 1 | A | 20 | 45 |
| 2 | B | 15 | 30 |
| 3 | C | 30 | 20 |
- (i)1 mark
In which one of the above cases the mirror will form a diminished image of the object? Justify your answer.
- (ii)1 mark
List two properties of the image formed in case 2.
- (iii)(A)2 marks
What is the nature and size of the image formed by mirror C? Draw ray diagram to justify your answer.
OR
(iii)(B)2 marksAn 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.
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.
- ½ mark
(i) Case 1 (mirror A) forms a diminished image.
- ½ mark
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.
- 1 mark
(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).
- ½ mark
(iii)(A) Mirror C: cm, object at 20 cm, i.e. between P and F. The image is virtual and erect,
- ½ mark
and magnified (larger than the object), formed behind the mirror.
- 1 mark
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.
- ½ mark
(iii)(B) Using sign convention: cm, cm. Mirror formula: .
- ½ mark
- ½ mark
cm
- ½ mark
The image is formed 36 cm in front of the mirror (on the same side as the object), so it is real.
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
- Chapter: Light – Reflection and Refraction (notes, formulas, where marks are lost and every past question)
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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.