Optics 3 Question 42
42. 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)
Show Answer
Solution:
- The given system behaves like a mirror of power given by $P=2 P _L+P _M$
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 $11 cm$.
