Question

The rays of light incident at the spherical surface of the lens from two different directions as shown in the figure. The relation between focal length of the lens in two cases is:
($f_1$ and $f_2$ is focal length of lens for ray $1$ and $2$ respectively.)

• A. $f_I>f_{II}$
• B. $f_{II}>f_I$
• C. $f_I=f_{II}$
• D. Can't be compared

Answer 1

The correct option is C $f_I=f_{II}$

Lens maker formula,
$\frac{1}{f}=\underset{A}{\underset{}{(\frac{{\mu }_2}{{\mu }_1}-1)}}\underset{B}{\underset{}{(\frac{1}{R_1}-\frac{1}{R_2})}}$
For both cases the term $A$ will remain same.

Case I: When light is incident from left side.
$R_1=+R,R_2=-R$
or, $\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{R}-\frac{1}{-R}$
$\Rightarrow (\frac{1}{R_1}-\frac{1}{R_2})=\frac{2}{R}$
$\Rightarrow \frac{1}{f_1}=(\frac{{\mu }_2}{{\mu }_1}-1)\left(\frac{2}{R}\right)$

Case II: When light is incident from right side.
The surface at which it falls become $R_1$.
$\Rightarrow R_1=+R$ and $R_2=-R$
again, $(\frac{1}{R_1}-\frac{1}{R_2})=\frac{2}{R}$
$\Rightarrow \frac{1}{f_2}=(\frac{{\mu }_2}{{\mu }_1}-1)\left(\frac{2}{R}\right)$
Thus,
$\Rightarrow f_1=f_2$


$\begin{array}{c}\text{Why this question}?\\ \text{Caution:}{R}_1\text{will correspond to the curvature}\\ \text{of the surface on which light is incident}\\ \text{Tip: For analysis of focal length "always use}\\ \text{use the lens maker's formula as a tool" to}\\ \text{compare the various parameters.}\end{array}$