A 10 cm long pencil is placed 5 cm in front of a concave mirror having a radius of curvature of 40 cm. (i) Determine the position of the image formed by this mirror. (ii) What is the size of the image? (iii) Draw a ray diagram to show the formation of the image as mentioned in the part (i).

(i) v = +6.67 cm, i.e. 6.67 cm behind the mirror (virtual, erect). (ii) m = 4/3, image size 13.3 cm (enlarged, erect). (iii) Ray diagram with object between P and F and a virtual image behind the mirror.

ScienceCBSE Class 10 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 10 Science sample paper 2023-24: question 36(b)

From the official CBSE sample question paper for 2023-24, 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

A 10 cm long pencil is placed 5 cm in front of a concave mirror having a radius of curvature of 40 cm.

(i)
Determine the position of the image formed by this mirror.
2 marks
(ii)
What is the size of the image?
1 mark
(iii)
Draw a ray diagram to show the formation of the image as mentioned in the part (i).
2 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

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

Hint 1

The focal length is half the radius of curvature. Signs for a concave mirror: $f$ and $u$ are negative.

Hint 2

Use $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ , then $m = - \frac{v}{u}$ and $h^{'} = m \times h$ .

Hint 3

The object is closer than F, so the reflected rays diverge. Extend them behind the mirror to find 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.

(i) Given $R = 40$ cm, so $f = \frac{R}{2} = 20$ cm. For a concave mirror $f = - 20$ cm. Object distance $u = - 5$ cm and object height $h = 10$ cm.
½ mark
Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ , so $- \frac{1}{20} = \frac{1}{v} - \frac{1}{5}$
½ mark
$\frac{1}{v} = - \frac{1}{20} + \frac{1}{5} = \frac{- 1 + 4}{20} = \frac{3}{20}$
½ mark
$v = \frac{20}{3} \approx + 6.67$ cm. The positive sign means the image is 6.67 cm behind the mirror, so it is virtual and erect.
½ mark
(ii) Magnification $m = \frac{h ^{'}}{h} = - \frac{v}{u} = - \frac{20/3}{- 5} = \frac{4}{3}$ .
Image size $h^{'} = m \times h = \frac{4}{3} \times 10 \approx 13.3$ cm. The image is erect and enlarged.
1 mark
(iii) Ray diagram (described): Draw the concave mirror with pole P, focus F (20 cm) and centre of curvature C (40 cm) on the principal axis. Place the pencil AB upright at 5 cm, between P and F.
- Ray 1 from A, parallel to the axis, reflects and passes through F.
- Ray 2 from A, directed as if coming from C (along the normal), reflects back along itself.
1 mark
The reflected rays diverge. Extend them backwards (dotted lines) behind the mirror; they meet at A′. Draw A′B′ behind the mirror. The image is virtual, erect and enlarged. Show arrows on all rays.
1 mark

Answer: (i) $v = + 6.67$ cm, i.e. 6.67 cm behind the mirror (virtual, erect). (ii) $m = 4/3$ , image size $\approx 13.3$ cm (enlarged, erect). (iii) Ray diagram with object between P and F and a virtual image behind the mirror.

Where the marks are

Takes $f = - 20$ cm and $u = - 5$ cm (0.5 mark)
Writes the mirror formula and substitutes (0.5 mark)
Obtains $1/ v = 3/20$ (0.5 mark)
States $v = + 6.67$ cm, behind the mirror (0.5 mark)
Finds $m = 4/3$ and image size $\approx 13.3$ cm (1 mark)
Ray diagram: object between P and F, two correct rays (1 mark)
Arrows, dotted extensions and virtual erect image behind the mirror (1 mark)

Revise the chapter

Chapter: Light – Reflection and Refraction (notes, formulas, where marks are lost and every past question)
Topic: Mirror Formula, Magnification and Sign Convention

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 2023-24: every question, by section.

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CBSE Class 10 Science sample paper 2023-24: question 36(b)

Hints

Step-by-step solution

Revise the chapter

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