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Q. Two plano-concave lenses $(1$ and $2)$ of glass of refractive index $1.5$ have radii of curvature $25 \,cm$ and $20 \,cm$. They are placed in contact with their curved surface towards each other and the space between them is filled with liquid of refractive index $4/3$. Then the combination is

BITSATBITSAT 2017

Solution:

$\frac{1}{f_{1}}=\left(\frac{3}{2}-1\right)\left(\frac{1}{\infty}-\frac{1}{25}\right)=-\frac{1}{50}$
$\frac{1}{f_{2}}=\left(\frac{4}{3}-1\right)\left(\frac{1}{25}+\frac{1}{20}\right)=\frac{3}{100}$
and $\frac{1}{f_{3}}=\left(\frac{3}{2}-1\right)\left(\frac{1}{-20}-\frac{1}{\infty}\right)=-\frac{1}{40}$
Now $ \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}$
$=-\frac{1}{50}+\frac{3}{100}-\frac{1}{40}$
$\therefore f=-66.6 cm$