Question:
Object at 20 cm from plano-convex lens (f = 15 cm), plane surface silvered. Find image position.
Object at 20 cm from plano-convex lens (f = 15 cm), plane surface silvered. Find image position.
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Silvered lens $\Rightarrow$ apply lens formula twice.
Updated On: Apr 23, 2026
- 60 cm left of AB
- 30 cm left of AB
- 12 cm left of AB
- 60 cm right of AB
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The Correct Option is C
Solution and Explanation
Concept:
Light passes twice through lens due to reflection.
Step 1: First refraction
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{20} = \frac{1}{60} \] \[ v = 60\,cm \]
Step 2: Reflection at plane mirror
Image forms at same distance: \[ 60\,cm \]
Step 3: Second refraction
Now object for lens is at \(+60\): \[ \frac{1}{15} = \frac{1}{v'} - \frac{1}{60} \Rightarrow \frac{1}{v'} = \frac{1}{15} + \frac{1}{60} = \frac{5}{60} \] \[ v' = 12\,cm \] Image on left side. Conclusion: \[ 12\,cm\ \text{left of AB} \]
Step 1: First refraction
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{20} = \frac{1}{60} \] \[ v = 60\,cm \]
Step 2: Reflection at plane mirror
Image forms at same distance: \[ 60\,cm \]
Step 3: Second refraction
Now object for lens is at \(+60\): \[ \frac{1}{15} = \frac{1}{v'} - \frac{1}{60} \Rightarrow \frac{1}{v'} = \frac{1}{15} + \frac{1}{60} = \frac{5}{60} \] \[ v' = 12\,cm \] Image on left side. Conclusion: \[ 12\,cm\ \text{left of AB} \]
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