The above image is that of a Digital Single Lense Reflector (DSLR) Camera which are used to take high resolution photographs by professional photographers. The second image of the above two is a schematic diagram of how an image is formed on the sensor of the camera. Based on your understanding of the lenses, answer the following questions. A What type of lens is used in the DSLR camera shown in the image? B What type of image is formed on the sensor? C A photographer is using a DSLR camera with a lens of focal length f = 50 mm to take a close-up photograph of a small object. The lens projects an image onto the camera sensor that is located 60 mm behind the lens. Calculate the object distance (i.e., the distance between the object and the lens). D A photographer is using a DSLR camera to take a picture of a flower. The flower is positioned 150 mm away from the camera lens. The actual height of the flower is 80 mm, and the image height formed on the camera's sensor is measured to be 20 mm. Calculate the focal length of the camera lens.

ScienceCBSE Class 10 · Sample paper, session 2025-26 · 4 marks · Case-based

CBSE Class 10 Science sample paper 2025-26: question 38

From the official CBSE sample question paper for 2025-26, 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 above image is that of a Digital Single Lense Reflector (DSLR) Camera which are used to take high resolution photographs by professional photographers. The second image of the above two is a schematic diagram of how an image is formed on the sensor of the camera. Based on your understanding of the lenses, answer the following questions.

A
What type of lens is used in the DSLR camera shown in the image?
1 mark
B
What type of image is formed on the sensor?
1 mark
C
A photographer is using a DSLR camera with a lens of focal length f = 50 mm to take a close-up photograph of a small object. The lens projects an image onto the camera sensor that is located 60 mm behind the lens. Calculate the object distance (i.e., the distance between the object and the lens).
2 marks
OR
D
A photographer is using a DSLR camera to take a picture of a flower. The flower is positioned 150 mm away from the camera lens. The actual height of the flower is 80 mm, and the image height formed on the camera's sensor is measured to be 20 mm. Calculate the focal length of the camera lens.
2 marks

Figure in the paper: Top: photograph of a DSLR camera with a large lens barrel on the left and the camera body with dials and viewfinder. Bottom: schematic ray diagram showing an object (a palm tree on the left) with a lens drawn as a rectangle with a converging (convex) lens symbol; rays from the top and bottom of the object pass through the lens and converge to form an inverted image of the tree on the camera sensor, labelled 'image', inside the camera body at right. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Think about which kind of lens forms an image that can be caught on a screen such as the sensor.

Hint 2

Use 1/f = 1/v − 1/u with the New Cartesian sign convention: the object distance is negative and a real image distance is positive.

Hint 3

For D, find m from the heights (the image is inverted, so m is negative), then v = m × u, then use the lens 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.

A. The rays from the object converge to form an image on the sensor, so the camera uses a convex (converging) lens.
1 mark
B. The image falls on a screen (the sensor) and is upside down, so it is real and inverted (and diminished).
1 mark
C. Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ , which gives $\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ .
With the sign convention: $f = + 50$ mm and $v = + 60$ mm (the image is real, behind the lens).
½ mark
$\frac{1}{u} = \frac{1}{60} - \frac{1}{50} = \frac{5 - 6}{300} = - \frac{1}{300}$
1 mark
$u = - 300$ mm. The object is 300 mm (30 cm) in front of the lens.
½ mark
C/D are alternatives. D (OR, worth 2 marks).
Given: $u = - 150$ mm, $h = + 80$ mm, and $h^{'} = - 20$ mm (the image is inverted).
Magnification: $m = \frac{h ^{'}}{h} = \frac{- 20}{80} = - 0.25$ [½ mark]
D (continued): from $m = \frac{v}{u}$ , we get $v = m \times u = (- 0.25) \times (- 150) = + 37.5$ mm [½ mark]
D (continued): $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{37.5} + \frac{1}{150} = \frac{4 + 1}{150} = \frac{1}{30}$
So $f = + 30$ mm [1 mark]
For visually impaired students (alternative).
VI A: a concave lens is used in the flashlight and a convex lens in the solar-cooker concentrator [1 mark].
VI B: a concave lens diverges light, so the flashlight beam spreads over a wider area. A convex lens converges sunlight to a point, which raises the temperature there for cooking [1 mark].
VI C: the solar cooker uses a convex lens, and the image is real, inverted and twice the size, so $m = - 2$ . With $u = - 40$ cm, $v = m u = (- 2) (- 40) = + 80$ cm.
$\frac{1}{f} = \frac{1}{80} - \frac{1}{- 40} = \frac{1 + 2}{80} = \frac{3}{80}$ , so $f = \frac{80}{3} \approx 26.67$ cm [2 marks].
(If a virtual image with $m = + 2$ were taken instead, $v = - 80$ cm and $f = 80$ cm.)
VI D (OR): $u = - 20$ cm and $v = - 10$ cm (the image is on the same side as the object).
$\frac{1}{f} = \frac{1}{- 10} - \frac{1}{- 20} = \frac{- 2 + 1}{20} = - \frac{1}{20}$
So $f = - 20$ cm, which means a concave (diverging) lens [2 marks].

Answer: A: convex lens; B: real and inverted (diminished); C: u = −300 mm (object 300 mm in front of the lens); D: f = 30 mm. VI: A concave (flashlight), convex (solar cooker); C f ≈ 26.67 cm; D f = −20 cm

Where the marks are

A: convex lens (1 mark)
B: real and inverted image (1 mark)
C: lens formula with correct signs, f = +50 mm and v = +60 mm (½ mark)
C: correct substitution and working to 1/u = −1/300 (1 mark)
C: u = −300 mm, i.e. 300 mm in front of the lens (½ mark)
D (alternative): m = −0.25 (½ mark), v = 37.5 mm (½ mark), f = 30 mm via the lens formula (1 mark)
VI: lens choices (1 mark), reasons (1 mark), C f ≈ 26.67 cm or D f = −20 cm (2 marks)

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 2025-26: every question, by section.

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CBSE Class 10 Science sample paper 2025-26: question 38

Hints

Step-by-step solution

Revise the chapter

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