CBSE Class 12 Physics 2025: question 18(a)
Set in the CBSE Class 12 board exam of 2025, paper 55/7/1, section B, question 18(a), for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q18(a) · 2 marks · Very short answerRay Optics and Optical Instruments
Using the mirror equation and the formula of magnification, deduce that "the virtual image produced by a convex mirror is always diminished in size and is located between the pole and the focus."
Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/7/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
For a convex mirror f is positive and u is always negative in the Cartesian sign convention.
Hint 2
Get 1/v from the mirror formula and compare it with 1/f; then write m = −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.
- ½ mark
For a convex mirror, take (focus behind the mirror). For a real object, with . Mirror formula: .
- ½ mark
Both terms on the right are positive, so : the image is always behind the mirror, i.e. virtual. Also . Hence : the image lies between the pole and the focus.
- ½ mark
Solving, . Magnification: .
- ½ mark
Since , we get . The image is therefore always erect and diminished.
Answer: 0 < v < f (between pole and focus, virtual) and 0 < m = f/(x+f) < 1 (diminished, erect).
Where the marks are
- Applies the mirror formula with the correct signs for a convex mirror (½ mark)
- Shows v is positive and less than f, so the image lies between P and F (½ mark)
- Writes m = −v/u in terms of f and x (½ mark)
- Concludes m is positive and less than 1, so the image is diminished (½ mark)
Revise the chapter
- Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)
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