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  • A biconvex lens of focal length is in front of a plane mirror. The distance between the

A biconvex lens of focal length \mathrm{15\ cm} is in front of a plane mirror. The distance between the lens and the mirror is \mathrm{10\ cm}. A small object is kept at a distance of \mathrm{30\ cm} from the lens. The final image is

Option: 1

virtual and at a distance of \mathrm{16\ cm } from the mirror


Option: 2

real and at a distance of \mathrm{16\ cm} from the mirror


Option: 3

virtual and at a distance of \mathrm{20\ cm} from the mirror


Option: 4

real and at a distance of \mathrm{20\ cm} from the mirror


Answers (1)

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Focal length of the biconvex lens is \mathrm{15\ cm}. A small object is placed at a distance of \mathrm{30\ cm} from the lens i.e. at a distance of \mathrm{2\ f}. Therefore the image should form at \mathrm{30\ cm} from the lens at \mathrm{I_{1}}

But since the ray strike the plane mirror before reaching \mathrm{I_{1}}, the image \mathrm{I_{1}} acts as the virtual object for reflection on plane mirror kept at a distance of \mathrm{20\ cm} from it. It should produce an image \mathrm{I_{2}} but as the ray encounters the lens, it gets refracted and the final image is formed at \mathrm{I_{3}} . For the last refraction from the biconvex lens, \mathrm{u=10\ cm}
Applying lens formula \mathrm{\frac{1}{v}-\frac{1}{u}=\frac{1}{f}}
\mathrm{\Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{10}=\frac{1}{15} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{15}+\frac{1}{10}=\frac{25}{150} \Rightarrow v=6 \mathrm{~cm}}
Therefore a real image is formed at a distance of \mathrm{16\ cm} from the plane mirror.

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