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  • A pin is placed in front of a convex lens of focal length <img alt="20 \mathrm{~cm}" sr

A pin is placed 10 \mathrm{~cm} in front of a convex lens of focal length 20 \mathrm{~cm} made of a material having refractive index 1.5. The surface of the lens is farther away from the pin is silvered and has a radius of curvature 22 \mathrm{~cm}.Determine the position of the final image. State the nature of the image.

Option: 1

9 \mathrm{~cm}


Option: 2

10 \mathrm{~cm}


Option: 3

11 \mathrm{~cm}


Option: 4

12 \mathrm{~cm}


Answers (1)

best_answer

The curved silvered surface will behave as a concave lens of focal length


\mathrm{f_{m}=\frac{R}{2}=-\frac{22}{2}=-11 \mathrm{~cm}=-0.11 \mathrm{~m}}
And hence
\mathrm{P_{M}= \text{the power of the mirror }=-\frac{1}{f_{M}}=-\frac{1}{-0.11}=+\frac{1}{0.11} D}

Further as the focal length of lens is 20 \mathrm{~cm}  i.e., 0.20 \mathrm{~m} its power will be
\mathrm{P_{L}=\frac{1}{f_{L}}=\frac{1}{20} D}

Now as in image formation, light after passing through the lens will be reflected back by the curved mirror through the lens again.

\mathrm{P=P_{L}+P_{M}+P_{L}=2 P_{L}+P_{M}}
\mathrm{P=\frac{2}{0.20}+\frac{1}{0.11}=\frac{210}{11} D}

So the focal length of equivalent mirror
\mathrm{F}=-\frac{1}{\mathrm{P}}=\frac{11}{210} \mathrm{~m}=-\frac{110}{21} \mathrm{~cm}
i.e. the silvered lens behaves as a concave mirror of focal length (110/21) cm. So for object at a distance 10 cm in front of it

\mathrm{\frac{1}{v}+\frac{1}{-10}=\frac{21}{110} \quad i.e., \quad v=-11 \mathrm{~cm}}

i.e. image will be 11 \mathrm{~cm} infront of the silvered lens and will be real as shown in figure.

Posted by

Ritika Jonwal

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