Question

Two half lenses $L_1({\mu }_1=1.4)$ & $L_2({\mu }_2=1.5)$ having radius of the curvature $R=20\text{cm}$ are placed together as shown. Find the position of the image of the parallel beam of light with respect to the common principal axis.

• A. $100/7\text{cm}$
• B. $200/9\text{cm}$
• C. $31.2\text{cm}$
• D. $21.8\text{cm}$

Answer 1

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

By using lens maker's formula,

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

For lens $L_1:$

Here, ${\mu }_2=\mu =1.4,{\mu }_1=1$
$R_1=R=20\text{cm},R_2=\infty$

$\frac{1}{f_1}=(\frac{1.4}{1}-1)(\frac{1}{20}-\frac{1}{\infty })$

$\therefore f_1=50\text{cm}$

For lens $L_2:$

Here, ${\mu }_2=\mu =1.5,{\mu }_1=1$
$R_1=\infty ,R_2=R=-20\text{cm}$

$\frac{1}{f_2}=(\frac{1.5}{1}-1)(\frac{1}{\infty }-\frac{1}{-20})$

$\therefore f_2=40\text{cm}$

Now,
$\therefore \frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{50}+\frac{1}{40}$

$\Rightarrow f_{eq}=\frac{200}{9}\text{cm}$

Hence, option $\left(b\right)$ is correct.

Why this question? To understand the effect of cutting of lenses along a direction perpendicular to the principal axis on the focal length of the lens.