PhysicsCBSE Class 12 · Board exam 2026, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2026: question 32(a)

Set in the CBSE Class 12 board exam of 2026, paper 55/1/1, section E, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(a) · 5 marks · Long answerRay Optics and Optical Instruments

(a) Using the relation for refraction at a curved spherical surface, derive the expression for lens maker’s formula.
(b) Three lenses $L_{1}$ , $L_{2}$ and $L_{3}$ , each of focal length 40 cm, are placed coaxially. The distance between $L_{1}$ and $L_{2}$ and between $L_{2}$ and $L_{3}$ are 120 cm and 20 cm respectively. An object is kept at a distance of 80 cm to the left of lens $L_{1}$ .
Find the distance of the final image formed from the object.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/1/1.

Hints

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

Hint 1

Apply $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ at each surface, using the first image as the object for the second.

Hint 2

In (b), see where the image from L₁ falls relative to the focus of L₂.

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) Take a thin lens of refractive index $n_{2}$ in a medium of index $n_{1}$ , with surface radii $R_{1}$ and $R_{2}$ . An object is at distance $u$ . Refraction at the first surface would form an image $I_{1}$ at $v_{1}$ :
$\frac{n _{2}}{v _{1}} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R _{1}}$ ...(1)
1 mark
$I_{1}$ acts as a virtual object for the second surface, since the lens is thin. The final image at $v$ satisfies
$\frac{n _{1}}{v} - \frac{n _{2}}{v _{1}} = \frac{n _{1} - n _{2}}{R _{2}}$ ...(2)
1 mark
Adding (1) and (2): $n_{1} (\frac{1}{v} - \frac{1}{u}) = (n_{2} - n_{1}) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ . For $u = \infty$ , $v = f$ , so
$\frac{1}{f} = (n_{21} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ , where $n_{21} = \frac{n _{2}}{n _{1}}$ .
1 mark
(b) Lens $L_{1}$ : $u = - 80$ cm, $f = + 40$ cm. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{40} - \frac{1}{80} = \frac{1}{80}$ , so $v = 80$ cm to the right of $L_{1}$ .
½ mark
Lens $L_{2}$ is 120 cm from $L_{1}$ , so the image lies 40 cm to its left: $u = - 40$ cm $= - f$ . The image is at infinity, i.e. the rays leaving $L_{2}$ are parallel.
½ mark
Lens $L_{3}$ focuses these parallel rays at its focus, 40 cm to its right. $L_{3}$ is $120 + 20 = 140$ cm from $L_{1}$ , so the final image is $180$ cm right of $L_{1}$ .
½ mark
Distance from the object: $80 + 180 = 260$ cm.
½ mark

Answer: $\frac{1}{f} = (n_{21} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ ; final image is 260 cm from the object (40 cm to the right of L₃)

Where the marks are

Refraction equation at the first surface (1 mark)
Refraction equation at the second surface with I₁ as object (1 mark)
Adds the equations and puts u = ∞ to get the lens maker's formula (1 mark)
Image by L₁ at 80 cm (½ mark)
Image by L₂ at infinity (½ mark)
Final image at the focus of L₃ (½ mark)
Distance from object 260 cm (½ mark)

Revise the chapter

Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)
Topic: Refraction at Spherical Surfaces and Lenses

Practise this chapterBoard-style questions from Ray Optics and Optical Instruments, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 55/1/1: 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 12 Physics 2026: question 32(a)

Hints

Step-by-step solution

Revise the chapter

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