Geometrical Optics – Page 3

JEE Advanced 2019 Paper 1, Question 10

A thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$ as shown in figure. The radius of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_1 = n_2 = n$ . The focal length is $f + \Delta f$ when $n_1 = n$ and $n_2 = n + \Delta n$ . Assuming $\Delta n \ll (n-1)$ and $1 < n < 2$ , the correct statement(s) is/are,

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$\bigg| \frac{\Delta f}{f} \bigg| < \bigg| \frac{\Delta n}{n} \bigg|$
For $n=1.5, \Delta n = 10^{-3}$ and $f = 20$ cm, the value of $|\Delta f|$ will be 0.02 cm (round off to $2^{\rm nd}$ decimal place).
If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$
The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

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Category: Geometrical Optics

Convex lens with two materials