JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Geometrical Optics

A thin convex lens is made of two materials with refractive indices ${\mathrm{n}}_1$ and ${\mathrm{n}}_2$, 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 ${\mathrm{n}}_1={\mathrm{n}}_2=\mathrm{n}$. The focal length is $\mathrm{f}+\Delta \mathrm{f}$ when ${\mathrm{n}}_1=\mathrm{n}$ and ${\mathrm{n}}_2=\mathrm{n}+\Delta \mathrm{n}$. Assuming $\Delta \mathrm{n}\ll$ (n-1) and 1 < n < 2, the correct statement(s) is/are :

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

Answer: (A), (C), (D)

Solution

When $n_1=n_2=n$, $\frac{1}{f}=(n-1)\frac{2}{R}$. Replacing the right half's index by $n+\Delta n$ adds $\frac{\Delta n}{R}$ to the power, so $\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$. This relation contains no $R$, so it is unchanged if the surfaces become concave of the same radius, and it shows that a negative $\frac{\Delta n}{n}$ makes $\frac{\Delta f}{f}$ positive. For $n=1.5$, $\Delta n=10^{-3}$ and $f=20$ cm, $|\Delta f|=\frac{f\Delta n}{2(n-1)}=\frac{20\times 10^{-3}}{2\times 0.5}=0.02$ cm. The inequality $\left|\frac{\Delta f}{f}\right|<\left|\frac{\Delta n}{n}\right|$ does not hold, since for $1<n<2$ one has $\frac{1}{2(n-1)}>\frac{1}{n}$.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options