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  • An object is placed at a distance of to the left on the axis of a convex lens A of foca

An object is placed at a distance of 10 \mathrm{~cm} to the left on the axis of a convex lens A of focal length 20 \mathrm{~cm}. A second convex lens of focal length 10 \mathrm{~cm} is placed co-axially to the right of the lens A at a distance of 5 \mathrm{~cm} from A. Find the position of the final image and its magnification. Trace the path of the rays.   

Option: 1

1.2


Option: 2

1.33


Option: 3

1.4


Option: 4

1.5


Answers (1)

best_answer

Here for 1 \mathrm{st}$ lens, $\mathrm{u}_{1}=-10 \mathrm{~cm}


\mathrm{f}_{1}=20 \mathrm{~cm}.
\frac{1}{\mathrm{v}_{1}}-\frac{1}{\mathrm{u}_{1}}=\frac{1}{\mathrm{f}_{1}}
\Rightarrow \frac{1}{\mathrm{v}_{1}}=\frac{1}{20}-\frac{1}{10}
\Rightarrow \mathrm{v}_{1}=-20 \mathrm{~cm}

i.e. the image is virtual and hence lies on the same side of the the object.

This will behave as an object for the second lens.

For 2nd lens
\mathrm{\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}}
\mathrm{\text { Here } u_{2}=-(20+5)}
\mathrm{ f_{2}=10 \mathrm{~cm} }
\mathrm{ \frac{1}{v_{2}}+\frac{1}{25}=\frac{1}{10} \Rightarrow v_{2}=\frac{50}{3}=16 \frac{2}{3} \mathrm{~cm}}

i.e. the final image is at a distance of \mathrm{ 16 \frac{2}{3} \mathrm{~cm}}on the right of the second lens.
The magnification of the image is given by,

\mathrm{m}=\frac{\mathrm{v}_{1}}{\mathrm{u}_{1}} \cdot \frac{\mathrm{v}_{2}}{\mathrm{u}_{2}}=\frac{20}{10} \cdot \frac{50}{3 \times 25}=\frac{4}{3}=1.33

Posted by

Ritika Harsh

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