ScienceCBSE Class 10 · Board exam 2024, set 31/4/1 · 5 marks · Long answer

CBSE Class 10 Science 2024: question 36(b)

Set in the CBSE Class 10 board exam of 2024, paper 31/4/1, 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 answerLight – Reflection and Refraction

(i) Define principal axis of a lens. Draw a ray diagram to show what happens when a ray of light parallel to the principal axis of a concave lens passes through it.
(ii) The focal length of a concave lens is 20 cm. At what distance from the lens should a 5 cm tall object be placed so that its image is formed at a distance of 15 cm from the lens ? Also calculate the size of the image formed.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 31/4/1.

Hints

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

Hint 1

Recall the lens formula 1/v − 1/u = 1/f. Which sign do the focal length and the image distance take for a concave lens?

Hint 2

Make 1/u the subject: 1/u = 1/v − 1/f, then substitute the values with their signs.

Hint 3

For the height, use h′/h = v/u.

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.

(i) Principal axis: the imaginary straight line passing through the two centres of curvature of the lens. It passes through the optical centre.
1 mark
(i) Ray diagram, described in words:
- Draw a concave lens with its principal axis, and mark the optical centre $O$ and the principal focus $F_{1}$ on the same side as the incoming light.
- A ray parallel to the principal axis strikes the lens and, after refraction, bends away from the axis (diverges).
- Produce the refracted ray backwards as a dotted line; it appears to come from $F_{1}$ .
- Show arrows on the rays.
1 mark
(ii) For a concave lens, $f = - 20$ cm. The image is virtual and on the object side, so $v = - 15$ cm. Object height $h = + 5$ cm.
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{u} = \frac{1}{v} - \frac{1}{f}$
½ mark
$\frac{1}{u} = \frac{1}{- 15} - \frac{1}{- 20} = \frac{- 4 + 3}{60}$
½ mark
$\frac{1}{u} = - \frac{1}{60}$ , so $u = - 60$ cm. The object must be placed 60 cm in front of the lens.
1 mark
Magnification: $m = \frac{h ^{'}}{h} = \frac{v}{u} \Rightarrow h^{'} = \frac{v}{u} \times h = \frac{- 15}{- 60} \times 5 = + 1.25$ cm. The image is 1.25 cm tall, erect and diminished.
1 mark

Answer: (ii) Object at 60 cm from the lens (u = −60 cm); image height = +1.25 cm (erect, diminished)

Where the marks are

Correct definition of principal axis (1 mark)
Ray diagram: a parallel ray diverges and appears to come from F on the same side (1 mark)
Writes the lens formula with correct signs f = −20 cm, v = −15 cm (½ mark)
Correct substitution (½ mark)
u = −60 cm (1 mark)
Image height h′ = (v/u)h = 1.25 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 paper2024 board paper, set 31/4/1: 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 2024: question 36(b)

Hints

Step-by-step solution

Revise the chapter

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