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A lens is 5 \mathrm{~cm} thick and its radius of curvature of its surfaces are 10 \mathrm{~cm} and 25 \mathrm{~cm} respectively. A point object is placed at a distance 12 \mathrm{~cm} from the surface whose radius of curvature is 10 \mathrm{~cm}. How far beyond the other surface image is formed? 

Option: 1

90 \mathrm{~cm}


Option: 2

95 \mathrm{~cm}


Option: 3

100 \mathrm{~cm}


Option: 4

105 \mathrm{~cm}


Answers (1)

best_answer

We know that \frac{\mu_{2}}{\mathrm{~V}}-\frac{\mu_{1}}{\mathrm{u}}=\frac{\mu_{2}-\mu_{1}}{\mathrm{R}}

\mathrm{u=-12 \mathrm{~cm}, \mathrm{R}=10 \mathrm{~cm}, \quad \mu_{1}=1, \mu_{2}=1.5}
\mathrm{\frac{1.5}{v}-\frac{1}{-12}=\frac{1.5-1}{10} \Rightarrow v=-45 \mathrm{~cm}}

This image will service as an object for the second surface.

For the second surface = object distance

\mathrm{u}=5+45=50 \mathrm{~cm}

For the second surface again
\mathrm{u}=-50 \mathrm{~cm}, \mathrm{R}=-25 \mathrm{~cm}, \mu_{1}=1.5, \mu_{2}=1
\mathrm{\frac{\mu_{2}}{\mathrm{~V}}-\frac{\mu_{1}}{\mathrm{u}}=\frac{\mu_{2}-\mu_{1}}{\mathrm{R}} \Rightarrow \frac{1}{\mathrm{~V}}-\frac{1.5}{-50}=\frac{1-1.5}{-25}}

\mathrm{or \: v=-100 \mathrm{~cm}}
Final image will be at a distance -95 \mathrm{~cm}  from the first surface on the same side as the objects.

 

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