PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2025-26: question 32(I)

From the official CBSE sample question paper for 2025-26, section E, question 32(I), for 5 marks. Try it first; the hints and our step-by-step solution are below.

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

(A) Derive lens maker’s formula.
(B) Equi-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 10cm?

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

Treat the lens as two spherical refracting surfaces taken one after the other; the image from the first surface becomes the object for the second.

Hint 2

Use $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ for each surface, add the two equations, then put $u = - \infty$ .

Hint 3

For an equi-convex lens, $R_{1} = R$ and $R_{2} = - R$ , so the bracket becomes $2/ R$ .

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) Set-up. Consider a thin lens of refractive index $n_{2}$ placed in a medium of refractive index $n_{1}$ , with poles B and D of its two surfaces and centres of curvature $C_{1}$ , $C_{2}$ . A point object O on the principal axis forms an image I. Since the lens is thin, BI $_{1} \approx$ DI $_{1}$ , and so on.
Refraction at the first surface (ABC). The first surface alone would form an image I $_{1}$ inside the glass. Using the single-surface formula:
$\frac{n _{1}}{O B} + \frac{n _{2}}{B I _{1}} = \frac{n _{2} - n _{1}}{B C _{1}} \dots (1)$
1 mark
Refraction at the second surface (ADC). I $_{1}$ acts as a virtual object for the second surface, which forms the final image I in the rarer medium:
$- \frac{n _{2}}{D I _{1}} + \frac{n _{1}}{D I} = \frac{n _{2} - n _{1}}{D C _{2}} \dots (2)$
1 mark
Adding (1) and (2) with $B I_{1} \approx D I_{1}$ (thin lens):
$\frac{n _{1}}{O B} + \frac{n _{1}}{D I} = (n_{2} - n_{1}) (\frac{1}{B C _{1}} + \frac{1}{D C _{2}})$
Now apply the Cartesian sign convention: $O B = - u$ , $D I = + v$ , $B C_{1} = + R_{1}$ and $D C_{2} = - R_{2}$ . This gives
$\frac{1}{v} - \frac{1}{u} = (\frac{n _{2}}{n _{1}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
For an object at infinity ( $u = - \infty$ ), the image forms at the focus ( $v = f$ ). With $n_{21} = μ$ :
$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
This is the lens maker's formula.
1 mark
(B) For an equi-convex lens, $R_{1} = + R$ and $R_{2} = - R$ , with $μ = 1.55$ and $f = 10$ cm. Then
$\frac{1}{10} = (1.55 - 1) (\frac{1}{R} + \frac{1}{R}) = 0.55 \times \frac{2}{R}$
1 mark
Solving: $R = 2 \times 0.55 \times 10 = 11$ cm. Each face needs a radius of curvature of 11 cm.
1 mark

Answer: (A) $\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ ; (B) R = 11 cm

Where the marks are

Writes the refraction equation at the first surface (1 mark)
Writes the refraction equation at the second surface, treating I $_{1}$ as the object (1 mark)
Adds the equations, applies the sign convention and puts $u = - \infty$ , $v = f$ to get the lens maker's formula (1 mark)
Substitutes $R_{1} = R$ , $R_{2} = - R$ , $μ = 1.55$ , $f = 10$ cm (1 mark)
Obtains R = 11 cm (1 mark)

Revise the chapter

Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)

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CBSE Class 12 Physics sample paper 2025-26: question 32(I)

Hints

Step-by-step solution

Revise the chapter

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