Question

Two thin identical shape plano-convex lenses $L_1({\mu }_1=1.4)$ and $L_2({\mu }_2=1.5)$ with radius of curvature $R=20\text{cm}$ have been arranged as shown in figure.


A parallel beam of light is incident on the system of the lenses, the position of image formed with respect to the common optical center is:

• A. $\frac{218}{10}\text{cm}$
• B. $\frac{200}{9}\text{cm}$
• C. $\frac{100}{7}\text{cm}$
• D. $\frac{156}{5}\text{cm}$

Answer 1

The correct option is B $\frac{200}{9}\text{cm}$

Since, the rays are coming parallel to the principal axis, the image will form at the principal focus on the common principal axis.
So, we need to find the equivalent focal length of the given lens system.
The arrangement shown in the figure is the combination of two plano-convex lenses.
From lens maker's formula, we have
$\frac{1}{f_1}=(\frac{{\mu }_1}{{\mu }_{air}}-1)(\frac{1}{R_1}-\frac{1}{\infty })$
since, $R_1=+R=+20\text{cm},{\mu }_1=1.4$
$\Rightarrow \frac{1}{f_1}=(\frac{1.4}{1}-1)(\frac{1}{20}-0)$
or, $f_1=50\text{cm}$
Similarly, for lens $L_2$
$\frac{1}{f_2}=(\frac{{\mu }_2}{{\mu }_{air}}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
since, $R_1=\infty ,R_2=-R=-20\text{cm},{\mu }_2=1.5$
$\Rightarrow \frac{1}{f_2}=(\frac{1.5}{1}-1)(\frac{1}{\infty }-\frac{1}{-20})$
or,$\frac{1}{f_2}=\frac{1}{40}$
therefore, $f_2=40\text{cm}$
the equivalent focal length of the system is,
$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}$
$\frac{1}{f_{eq}}=\frac{1}{50}+\frac{1}{40}$
$\Rightarrow$$f_{eq}=\frac{+200}{9}\text{cm}$

Why this question? Bottom line: Two thin lenses are placed together in close contact hence, visualize the system as a single lens with focal length '$f_{eq}$'.