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Analyse the following… (Class 10 Science 2025, Q36B)

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ScienceCBSE Class 10 · Board exam 2025, set 31/5/1 · 5 marks · Long answer

CBSE Class 10 Science 2025: question 36(B)

Set in the CBSE Class 10 board exam of 2025, paper 31/5/1, section D, question 36(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q36(B) · 5 marks · Long answerCompetency-basedLight – Reflection and Refraction

Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow without doing any calculations:

  1. (a)

    Determine the focal length of the lens. Give reason for your answer.

    0 marks
  2. (b)

    Find magnification of the image formed in Observation No. 3.

    0 marks
  3. (c)

    The numerical value of magnifications in cases of observation 1 and 2 is same. List two differences in the images formed in these two cases.

    0 marks

Figure in the paper: Observation table (object distance u in cm; image distance v in cm): 1: u = −15, v = −60; 2: u = −25, v = +100; 3: u = −30, v = +60; 4: u = −40, v = +40; 5: u = −60, v = +30; 6: u = −100, v = +25. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 31/5/1.

Hints

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

Hint 1

Look for the observation where the object and image distances are numerically equal.

Hint 2

For a convex lens, where must the object be for the image to be the same distance away on the other side?

Hint 3

For (c), check the sign of v in each case: a negative v means a virtual 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.

  1. (a) Focal length: In Observation 4, cm and cm. The object and image distances are equal, so the object is at and the image is at . Therefore cm, which gives cm.

    1 mark

  2. Reason: For a convex lens, an object placed at forms a real, inverted image of the same size at on the other side of the lens. Only then are and equal.

    1 mark

  3. (b) Magnification in Observation 3: . The image is real, inverted and twice the size of the object.

    1 mark

  4. (c) Difference 1: In Observation 1, cm (negative), so the image is virtual and erect (). In Observation 2, cm (positive), so the image is real and inverted ().

    1 mark

  5. Difference 2: In Observation 1 the image forms on the same side as the object and cannot be obtained on a screen. In Observation 2 it forms on the other side of the lens and can be obtained on a screen.

    1 mark

Answer: (a) f = 20 cm, since cm gives cm (object at 2F gives image at 2F); (b) m = −2; (c) Obs. 1: virtual, erect, same side as the object; Obs. 2: real, inverted, other side of the lens.

Where the marks are

  • Identifies cm, so cm (1 mark)
  • Reason: object at 2F forms image at 2F for a convex lens (1 mark)
  • Magnification (1 mark)
  • Image in Obs. 1 is virtual and erect (1 mark)
  • Image in Obs. 2 is real and inverted (1 mark)

Revise the chapter

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 paper2025 board paper, set 31/5/1: 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.