Optics Ques 92
- A pin is placed $10 $ $cm$ in front of a convex lens of focal length $20 $ $cm$ and made of a material of refractive index $ 1.5$ . The convex surface of the lens farther away from the pin is silvered and has a radius of curvature of $22$ $ cm$. Determine the position of the final image. Is the image real or virtual ?
(1978)
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Answer:
Correct Answer: 92.(at a distance of $11$ $cm$, virtual)
Solution:
Formula:
- The given system behaves like a mirror of power given by
$ \quad P=2 P _L+P _M$
or $\quad-\frac{1}{F}=2 (\frac{1}{0.2})+(\frac{-2}{0.22})$
As $\quad P _L=\frac{1}{f(m)}$
and $\quad P _M=\frac{-1}{f(m)}=\frac{-2}{R(m)}$
Solving this equation, we get
$ F=-1.1$ $ m=-110$ $ cm $
i.e. the system behaves as a concave mirror of focal length $18.33$ $ cm$.
Using the mirror formula
or
$ \begin{aligned} \frac{1}{v}+\frac{1}{u} & =\frac{1}{f} \text { we have } \\ \frac{1}{v}-\frac{1}{10} & =-\frac{1}{110} \\ v & =11 cm \end{aligned} $
i.e. virtual image will be formed at a distance of $11 $ $cm$.
BETA
