Read the following passage and answer the questions given: Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image. (a) A convex lens of focal length 20 cm is used to form an image. If an object is placed at 40 cm from the lens, what will be the position and nature of image? (b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens. (c)(i) A lens combination consists of a convex lens of focal length 30 cm and a concave lens of focal length 15 cm placed together. Find the equivalent focal length and power of this lens combination. (c)(ii) Two lenses are placed in contact. One is a concave lens with focal length 2 m and the other is a convex lens with focal length 1.5 m. What type of lens will the combination behave as (convex or concave)? Give reason.

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

CBSE Class 10 Science 2026: question 38

Set in the CBSE Class 10 board exam of 2026, paper 31/1/1, 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

Read the following passage and answer the questions given:
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.

(a)
A convex lens of focal length 20 cm is used to form an image. If an object is placed at 40 cm from the lens, what will be the position and nature of image?
1 mark
(b)
Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
1 mark
(c)(i)
A lens combination consists of a convex lens of focal length 30 cm and a concave lens of focal length 15 cm placed together. Find the equivalent focal length and power of this lens combination.
2 marks
OR
(c)(ii)
Two lenses are placed in contact. One is a concave lens with focal length 2 m and the other is a convex lens with focal length 1.5 m. What type of lens will the combination behave as (convex or concave)? Give reason.
2 marks

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

Hints

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

Hint 1

Use the sign convention: convex f is positive, concave f is negative, and object distance u is negative.

Hint 2

For a combination, convert each focal length to metres, find each power with P = 1/f, and add the powers.

Hint 3

For (a), notice that 40 cm is exactly twice the focal length.

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) Given $f = + 20$ cm and $u = - 40$ cm. Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ : $\frac{1}{v} = \frac{1}{20} + \frac{1}{- 40} = \frac{2 - 1}{40} = \frac{1}{40}$ , so $v = + 40$ cm. The image is 40 cm on the other side of the lens, at $2 F_{2}$ .
½ mark
(a) Nature: the image is real and inverted, and the same size as the object. This is because the object is at $2 F_{1}$ .
½ mark
(b) Ray diagram: draw a concave lens with $F_{1}$ and $F_{2}$ marked, and object AB between O and $F_{1}$ . Ray 1 from B goes parallel to the axis. After refraction it diverges, and its dashed back-extension passes through $F_{1}$ . Ray 2 from B passes undeviated through O. The extension and ray 2 meet at B′. The image A′B′ lies between O and $F_{1}$ on the same side as the object, and is virtual, erect and diminished.
1 mark
(c)(i) $f_{1} = + 30$ cm $= + 0.3$ m and $f_{2} = - 15$ cm $= - 0.15$ m. Using $P = \frac{1}{f}$ (f in metres): $P_{1} = \frac{1}{0.3} = + 3.33$ D.
½ mark
(c)(i) $P_{2} = \frac{1}{- 0.15} = - 6.67$ D.
½ mark
(c)(i) Equivalent power: $P = P_{1} + P_{2} = 3.33 - 6.67 = - 3.33$ D.
½ mark
(c)(i) Equivalent focal length: $f = \frac{1}{P} = \frac{1}{- 3.33} = - 0.3$ m $= - 30$ cm. The combination acts as a diverging lens.
½ mark
(c)(ii) $f_{1} = - 2$ m (concave) gives $P_{1} = - \frac{1}{2} = - 0.5$ D. $f_{2} = + 1.5$ m (convex) gives $P_{2} = + \frac{1}{1.5} = + 0.67$ D. Then $P = P_{1} + P_{2} = - \frac{1}{2} + \frac{2}{3} = + \frac{1}{6}$ D, so $f = + 6$ m.
1 mark
(c)(ii) The combined power (and focal length) is positive, so the combination behaves as a convex (converging) lens. Reason: the convex lens has the smaller focal length, so its power is greater and outweighs the concave lens.
1 mark

Answer: (a) 40 cm behind the lens (at 2F₂), real and inverted, same size; (b) virtual, erect, diminished image between O and F₁ on the object side; (c)(i) f = −30 cm, P = −3.33 D; (c)(ii) behaves as a convex lens (P = +1/6 D, f = +6 m)

Where the marks are

(a) Position: 40 cm on other side / at 2F (½ mark)
(a) Nature: real and inverted (½ mark)
(b) Correct ray diagram for concave lens with object between O and F (1 mark)
(c)(i) Uses $P = 1/ f$ with powers of each lens (½ + ½ mark)
(c)(i) Combined power $- 3.33$ D (½ mark)
(c)(i) Equivalent focal length $- 30$ cm (½ mark)
(c)(ii) States combination behaves as convex lens (1 mark)
(c)(ii) Reason: net power $+ 1/6$ D / focal length +6 m positive (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 paper2026 board paper, set 31/1/1: 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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