Question:
A spherical mirror is obtained as shown in the figure from a hollow glass sphere. If an object is positioned in front of the mirror, what will be the nature and magnification of the image of the object?
(Figure drawn as schematic and not to scale)
A spherical mirror is obtained as shown in the figure from a hollow glass sphere. If an object is positioned in front of the mirror, what will be the nature and magnification of the image of the object?
(Figure drawn as schematic and not to scale)
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Read the sphere diameter and object position from the scale: the mirror is concave with \(R\) about 8 cm; compare the object distance with \(2f\).
Updated On: Nov 5, 2026
- Inverted, real and unmagnified
- Inverted, real and magnified
- Erect, virtual and magnified
- Erect, virtual and unmagnified
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The Correct Option is A
Solution and Explanation
From the scale (mirror pole at 0 cm), the sphere extends to the 16 cm mark, so its diameter is about 16 cm and radius of curvature \(R\approx 8\) cm. The reflecting (silvered) side faces the object inside the sphere, so it is a concave mirror with \(f=R/2\approx 4\) cm.
The object is at about 10 cm from the pole, i.e. \(u\approx -10\) cm, which is beyond \(2f\) (8 cm).
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\), so \(\frac{1}{v}=\frac{1}{-4}+\frac{1}{10}=-0.15\), giving \(v\approx -6.7\) cm (real, in front of the mirror).
\(m=-\frac{v}{u}=-\frac{-6.7}{-10}=-0.67\): inverted and diminished.
Answer: (1) Inverted, real and unmagnified
The object is at about 10 cm from the pole, i.e. \(u\approx -10\) cm, which is beyond \(2f\) (8 cm).
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\), so \(\frac{1}{v}=\frac{1}{-4}+\frac{1}{10}=-0.15\), giving \(v\approx -6.7\) cm (real, in front of the mirror).
\(m=-\frac{v}{u}=-\frac{-6.7}{-10}=-0.67\): inverted and diminished.
Answer: (1) Inverted, real and unmagnified
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