ScienceCBSE Class 10 · Sample paper, session 2026-27 · 3 marks · Short answer

CBSE Class 10 Science sample paper 2026-27: question 35

From the official CBSE sample question paper for 2026-27, section C, question 35, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 3 marks · Short answerLight – Reflection and Refraction

A concave lens forms an image that is 1/3rd the size of the object, as shown in the diagram. If the object is placed 24 cm in front of the lens, determine: A. the type of lens. B. the image distance. C. the focal length of the lens.

Figure in the paper: Ray diagram of a concave (diverging) lens with principal axis, optical centre O, focal points F1 and F2 and 2F1, 2F2 marked. Object AB stands on the principal axis to the left of F1, with image A'B' (virtual, erect, smaller) between F1 and O. Rays from B: one parallel to axis bent away as if from F1, one directed toward the centre, and a dashed line showing the virtual image path. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

An image that is erect and smaller than the object is a clue to the lens type. What sign does m take for an erect image?

Hint 2

Use m = v/u with u = −24 cm to find v.

Hint 3

Put u and v into 1/v − 1/u = 1/f, keeping the signs carefully.

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.

A. Type of lens: The image is virtual, erect and diminished (1/3rd the size of the object), and it is formed between $F_{1}$ and $O$ on the same side as the object. Only a concave (diverging) lens forms such an image for every object position. Object distance $u = - 24$ cm (the object is on the left of the lens, against the direction of incident light).
1 mark
B. Image distance: Magnification of a lens is $m = \frac{h _{i}}{h _{o}} = \frac{v}{u}$ . The image is erect and one-third the size of the object, so $m = + \frac{1}{3}$ .
$v = m \times u = \frac{1}{3} \times (- 24) = - 8$ cm. The image is 8 cm from the lens, on the same side as the object. It is virtual.
1 mark
C. Focal length: Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .
$\frac{1}{f} = \frac{1}{- 8} - \frac{1}{- 24} = - \frac{3}{24} + \frac{1}{24} = - \frac{2}{24} = - \frac{1}{12}$ , so $f = - 12$ cm. The negative sign confirms a diverging (concave) lens.
1 mark

Answer: A. Concave (diverging) lens, u = −24 cm; B. v = −8 cm (virtual, same side as object); C. f = −12 cm

Where the marks are

Identifies the lens as concave/diverging and takes u = −24 cm (1 mark)
Uses m = v/u with m = +1/3 to get v = −8 cm (1 mark)
Applies the lens formula correctly to get f = −12 cm (1 mark)

Revise the chapter

Chapter: Light – Reflection and Refraction (notes, formulas, where marks are lost and every past question)
Topic: Lens Formula, Magnification and Power of a Lens

Practise this chapterBoard-style questions from Light – Reflection and Refraction, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: every question, by section.

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CBSE Class 10 Science sample paper 2026-27: question 35

Hints

Step-by-step solution

Revise the chapter

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