Question:
An object is placed at a distance \(20\ \text{cm}\) from the pole of a convex mirror of focal length \(20\ \text{cm}\). The image is produced at
An object is placed at a distance \(20\ \text{cm}\) from the pole of a convex mirror of focal length \(20\ \text{cm}\). The image is produced at
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For convex mirror, image is always virtual, erect, and diminished.
Updated On: Apr 23, 2026
- \(13.3\ \text{cm}\)
- \(20\ \text{cm}\)
- \(25\ \text{cm}\)
- \(10\ \text{cm}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). For convex mirror, \(f\) is positive, \(u\) is negative.
Step 2: Detailed Explanation:
Given \(f = +20\ \text{cm}\), \(u = -20\ \text{cm}\).
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} - \frac{1}{(-20)} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}\)
\(\Rightarrow v = 10\ \text{cm}\). Image is virtual, behind the mirror.
Step 3: Final Answer:
Thus, image distance = \(10\ \text{cm}\).
Mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). For convex mirror, \(f\) is positive, \(u\) is negative.
Step 2: Detailed Explanation:
Given \(f = +20\ \text{cm}\), \(u = -20\ \text{cm}\).
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} - \frac{1}{(-20)} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}\)
\(\Rightarrow v = 10\ \text{cm}\). Image is virtual, behind the mirror.
Step 3: Final Answer:
Thus, image distance = \(10\ \text{cm}\).
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