Q82 / 90 · JEE Main 2022 (24 Jun Shift 1) · Physics

Two identical thin biconvex lenses of focal length $15 cm$ and refractive index $1.5$ are in contact with each other. The space between the lenses is filled with a liquid of refractive index $1.25$ . The focal length of the combination is _____ $cm .$

Correct answer: 10

Solution

For convex lenses, let radius of curvature is $R$ . Then,

$\frac{1}{15 cm} = (1.5 - 1) (\frac{1}{R} - \frac{1}{- R})$

$\Rightarrow \frac{1}{R} = \frac{1}{15} \Rightarrow R = 15 cm$

Now for concave lens,

$\frac{1}{f_{c o n c a v e}} = (1.25 - 1) (\frac{1}{- R} - \frac{1}{R})$

$= \frac{1}{4} \times - \frac{2}{R} = \frac{- 1}{30}$

$\frac{1}{f_{e f f}} = \frac{1}{15} + \frac{1}{15} - \frac{1}{30} = - \frac{1}{10}$

$\Rightarrow f_{e f f} = - 10 cm$

Hence, the correct answer is 10.