Q142 / 160 · AP EAMCET 2022 (06 Jul Shift 1) · Physics
A concave lens and a convex lens are arranged as shown in the figure. The position of the final image.
- A$17 \mathrm{~cm}$ to the left of convex lens
- B$24.2 \mathrm{~cm}$ to the right of concave lens
- Correct answerC$29.2 \mathrm{~cm}$ to the right of concave lens
- D$24.2 \mathrm{~cm}$ to the left of convex lens
Solution
For concave lens
$$
\begin{aligned}
& \mathrm{u}=-30 \mathrm{~cm} \\
& \mathrm{f}=-20 \mathrm{~cm}
\end{aligned}
$$
$$
\begin{aligned}
& \text { So, } \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
& \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-30}=\frac{1}{-20} \Rightarrow \frac{1}{\mathrm{v}}=-\frac{1}{30}-\frac{1}{20} \\
& \Rightarrow \frac{1}{\mathrm{~V}}=\frac{-20-30}{600} \Rightarrow \mathrm{v}=-12 \mathrm{~cm}
\end{aligned}
$$
For convex lens
$$
\begin{aligned}
& \mathrm{u}=-(12+5)=-17 \mathrm{~cm} \\
& \mathrm{f}=+10 \mathrm{~cm}
\end{aligned}
$$
So, $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
$$
\begin{aligned}
& \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-17}=\frac{1}{10} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{17} \\
& \Rightarrow \frac{1}{\mathrm{v}}=\frac{7}{170} \Rightarrow \mathrm{v}=\frac{170}{7}=24.2 \mathrm{~m}
\end{aligned}
$$
So, final image will be $(24.2+5) \mathrm{cm}=29.2 \mathrm{~cm}$ to the right of concave lens.
Hence, the correct answer is option: 'C'.