Optics 7 Question 37
38. In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is $R$. Determine the focal length of this system.
$(2003,2 M)$
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Solution:
- For refraction at first surface,
$$ \frac{\mu _2}{v _1}-\frac{\mu _1}{-\infty}=\frac{\mu _2-\mu _1}{+R} $$
For refraction at second surface,
$$ \frac{\mu _3}{v _2}-\frac{\mu _2}{v _1}=\frac{\mu _3-\mu _2}{+R} $$
Adding Eqs. (i) and (ii), we get
$$ \begin{aligned} \frac{\mu _3}{v _2} & =\frac{\mu _3-\mu _1}{R} \\ v _2 & =\frac{\mu _3 R}{\mu _3-\mu _1} \end{aligned} $$
Therefore, focal length of the given lens system is
$$ \frac{\mu _3 R}{\mu _3-\mu _1} $$