Physics - Geometrical Optics Question with Solution | TestHub

Let $L_1$ be the first lens and $L_2$ be the second lens.

The first refraction is in $L_1$.

Using lens formula,

$$$$\begin{gathered} u_1=\infty , f_1=20 \mathrm{cm}, \\ \frac{1}{v_1}-0=\frac{1}{f_1} \\ \Rightarrow v_1=20 \mathrm{cm} \end{gathered}$$

Hence, the distance between the second image and the $L_2$ would be $\left(60-20\right)=40 \mathrm{cm}$.

For the second lens $L_2$,

$$\begin{gathered} u_2=-40 \mathrm{cm}, f_2=20 \mathrm{cm}, \\ \frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2} \\ \Rightarrow \frac{1}{v_2}+\frac{1}{40}=\frac{1}{20} \\ \Rightarrow \frac{1}{v_2}=\frac{1}{40} \\ \Rightarrow v_2=40 \mathrm{cm} \end{gathered}$$

So, the distance of image from first lens is

$x=40 \mathrm{cm}+60 \mathrm{cm}$

$=100 \mathrm{cm}$