PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 4 marks · Case-based

CBSE Class 12 Physics 2025: question 30

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, section D, question 30, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 4 marks · Case-basedCompetency-basedRay Optics and Optical Instruments

A thin lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, one can obtain the 'lens maker formula' and then the 'lens formula'. A lens has two foci – called 'first focal point' and 'second focal point' of the lens, one on each side.
Consider the arrangement shown in figure. A black vertical arrow and a horizontal thick line with a ball are painted on a glass plate. It serves as the object. When the plate is illuminated, its real image is formed on the screen.

(i)
Which of the following correctly represents the image formed on the screen ?
1. (A)Arrow pointing up, horizontal line with ball on the right
2. (B)Arrow pointing up, line slanting with ball on the left
3. (C)Arrow pointing down, line with ball on the left
4. (D)Arrow pointing down, slanted line with ball on the upper right
1 mark
(ii)
Which of the following statements is incorrect ?
1. (A)For a convex mirror magnification is always negative.
2. (B)For all virtual images formed by a mirror magnification is positive.
3. (C)For a concave lens magnification is always positive.
4. (D)For real and inverted images, magnification is always negative.
1 mark
(iii)
A convex lens of focal length 'f' is cut into two equal parts perpendicular to the principal axis. The focal length of each part will be :
1. (A)f
2. (B)2 f
3. (C) $\frac{f}{2}$
4. (D) $\frac{f}{4}$
1 mark
OR
(iii)
If an object in case (i) above is 20 cm from the lens and the screen is 50 cm away from the object, the focal length of the lens used is
1. (A)10 cm
2. (B)12 cm
3. (C)16 cm
4. (D)20 cm
1 mark
(iv)
The distance of an object from first focal point of a biconvex lens is $X_{1}$ and distance of the image from second focal point is $X_{2}$ . The focal length of the lens is
1. (A) $X_{1} X_{2}$
2. (B) $X_{1} + X_{2}$
3. (C) $X_{1} X_{2}$
4. (D) $\frac{X _{2}}{X _{1}}$
1 mark

Figure in the paper: Optical bench: a light box with a plate (black vertical arrow and horizontal thick line with a ball painted on the plate on the hidden side of box), a convex lens in a holder, and a screen at the far end, all on a graduated rail. Four options show arrow/line images: (A) arrow up with line and ball on right; (B) arrow up with slanted line, ball at lower left; (C) arrow down with line, ball at left; (D) arrow down with slanted line, ball at upper right. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/1/1.

Hints

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

Hint 1

A real image from a single lens is rotated by 180° in its own plane. Which sign of m goes with virtual erect images?

Hint 2

For the cut lens, apply the lens-maker formula with one surface flat. For the focal length, take v = 50 − 20 cm. For (iv), substitute u = −(f + X₁) and v = f + X₂ in the lens formula.

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) A convex lens forms a real image that is inverted both up–down and left–right. So the arrow points down and the ball appears on the opposite side (left). Option (C).
1 mark
(ii) A convex mirror always forms virtual, erect images, so its magnification is always positive. Statement (A) is therefore incorrect. Option (A).
1 mark
(iii)(a) Cutting perpendicular to the principal axis gives two plano-convex halves, each with one curved surface: $\frac{1}{f ^{'}} = (n - 1) \frac{1}{R} = \frac{1}{2} \cdot \frac{1}{f}$ , so $f^{'} = 2 f$ . Option (B).
1 mark
(iii)(b) [alternative] $u = - 20$ cm, $v = 50 - 20 = + 30$ cm. $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{30} + \frac{1}{20} = \frac{5}{60}$ , so $f = 12$ cm. Option (B).
(iv) From Newton's formula, with object distance $f + X_{1}$ and image distance $f + X_{2}$ : $\frac{1}{f + X _{2}} + \frac{1}{f + X _{1}} = \frac{1}{f}$ simplifies to $X_{1} X_{2} = f^{2}$ , so $f = X_{1} X_{2}$ . Option (C).
1 mark

Answer: (i) C; (ii) A; (iii)(a) B, 2f; (iii)(b) B, 12 cm; (iv) C, √(X₁X₂)

Where the marks are

(i) Real image inverted in both directions, option C (1 mark)
(ii) Convex mirror magnification is positive, so A is incorrect (1 mark)
(iii) $f^{'} = 2 f$ , option B; or OR part f = 12 cm, option B (1 mark)
(iv) $X_{1} X_{2} = f^{2}$ , option C (1 mark)

Revise the chapter

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

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 paper2025 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2025: question 30

Hints

Step-by-step solution

Revise the chapter

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