The variation of image distance (v) with the object distance (u) for a convex lens is given in the following observation table. Analyse it and answer the questions given below : S.No. Object distance (u) cm Image distance (v) cm --- --- --- 1 – 150 + 30 2 – 75 + 37·5 3 – 50 + 50 4 – 37·5 + 75 5 – 30 + 150 6 – 15 + 37·5 (a) Without calculation, find out the focal length of the given convex lens. Justify your answer. (b) Which one of the observations given in the table is not correct and why ? (c)(i) Find out the value of magnification for u = -30 cm. Write the nature of the image formed. Give reason for your answer. (c)(ii) Analyse the data given in the table above and draw the labelled ray diagram for u = -30 cm and v = 150 cm.

ScienceCBSE Class 10 · Board exam 2026, set 31/3/1 · 4 marks · Case-based

CBSE Class 10 Science 2026: question 38

Set in the CBSE Class 10 board exam of 2026, paper 31/3/1, section C, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedLight – Reflection and Refraction

The variation of image distance ( $v$ ) with the object distance ( $u$ ) for a convex lens is given in the following observation table. Analyse it and answer the questions given below :

S.No.	Object distance ( $u$ ) cm	Image distance ( $v$ ) cm
1	– 150	+ 30
2	– 75	+ 37·5
3	– 50	+ 50
4	– 37·5	+ 75
5	– 30	+ 150
6	– 15	+ 37·5

(a)
Without calculation, find out the focal length of the given convex lens. Justify your answer.
1 mark
(b)
Which one of the observations given in the table is not correct and why ?
1 mark
(c)(i)
Find out the value of magnification for $u = - 30$ cm. Write the nature of the image formed. Give reason for your answer.
2 marks
OR
(c)(ii)
Analyse the data given in the table above and draw the labelled ray diagram for $u = - 30$ cm and $v = 150$ cm.
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 31/3/1.

Hints

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

Hint 1

For a convex lens, where must the object be for the image distance to equal the object distance?

Hint 2

Once you know f, check which object distance lies inside the focal length. What should the sign of v be there?

Hint 3

Magnification for a lens is v/u. A negative sign tells you about the orientation of the image.

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) In observation 3, $∣ u ∣ = ∣ v ∣ = 50$ cm. For a convex lens the image distance equals the object distance only when the object is at $2 F_{1}$ , so $2 f = 50$ cm and $f = 25$ cm.
1 mark
(b) Observation 6 is not correct.
½ mark
(b) Reason: with $u = - 15$ cm the object lies between the optical centre O and $F_{1}$ (15 cm < 25 cm). The image is then virtual and on the same side as the object, so $v$ must be negative. The table gives $+ 37.5$ cm.
½ mark
(c)(i) Magnification is $m = \frac{v}{u} = \frac{+ 150}{- 30}$ .
½ mark
$m = - 5$
½ mark
Nature of image: real and inverted (and 5 times magnified).
½ mark
Reason: the magnification is negative, which means the image is inverted, and an inverted image formed by a lens is real.
½ mark
(c)(ii) [alternative] Ray diagram (described):
- Draw the principal axis and a convex lens at O. Mark $F_{1}$ and $2 F_{1}$ (25 cm and 50 cm) on the left, and $F_{2}$ and $2 F_{2}$ on the right.
- Place an upright object AB at 30 cm, between $F_{1}$ and $2 F_{1}$ .
- Ray 1: from A parallel to the axis, refracting through $F_{2}$ .
- Ray 2: from A through O, going straight undeviated.
- The rays meet beyond $2 F_{2}$ at 150 cm, forming the image A′B′, which is real, inverted and enlarged.
- Arrows must show the direction of the rays.
2 marks

Answer: (a) f = 25 cm (object at 2F, u = v = 50 cm). (b) Observation 6; object within focus, so v should be negative. (c)(i) m = −5; real and inverted, since m is negative. (c)(ii) Object between F₁ and 2F₁ gives a real, inverted, enlarged image beyond 2F₂.

Where the marks are

f = 25 cm with the reason that u = v at 2F (1 mark)
Identifies observation 6 (½ mark)
Reason: object between F and O gives a virtual image, so v should be negative (½ mark)
Uses m = v/u (½ mark)
m = −5 (½ mark)
Image is real and inverted (½ mark)
Reason: m is negative (½ mark)
OR (c)(ii): correct labelled ray diagram (2 marks; ½ mark deducted if ray directions are missing or wrong)

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 paper2026 board paper, set 31/3/1: every question, by section.

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.

CBSE Class 10 Science 2026: question 38

Hints

Step-by-step solution

Revise the chapter

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