Question

The size of the image of an object, which is at $\infty ,$ is formed by a convex lens of focal length $30\text{cm}$ is $2\text{cm}.$ If a concave lens of focal length $20\text{cm}$ is placed between the convex lens and the image at a distance of $26\text{cm}$ from the convex lens. Find the new size of the image-

• A. $1.25\text{cm}$
• B. $2.5\text{cm}$
• C. $1.05\text{cm}$
• D. $2\text{cm}$

Answer 1

The correct option is B $2.5\text{cm}$



Let $f_1$ and $f_2$ be the focal lengths of convex and concave lens respectively and $I_1$ and $I_2$ be the size of images formed by the convex and concave lens respectively.

Given, $f_1=30\text{cm};f_2=20\text{cm}{I}_1=2\text{cm};I_2=?$

For the convex lens,

$u=\infty ;f_1=30\text{cm}$

From lens formula,

$\frac{1}{f_1}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f_1}=\frac{1}{v}-\frac{1}{\infty }$

$\therefore v=f_1=30\text{cm}$

Hence, the convex lens will form an image $I_1$ at its focus, which acts as a virtual object for the concave lens.

Hence, for the concave lens,

$u=+4\text{cm};f_2=-20\text{cm}$

From lens formula,

$\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{-20}=\frac{1}{v}-\frac{1}{4}$

$\Rightarrow v=5\text{cm}$

From the lateral magnification,

$m=\frac{v}{u}=\frac{h_i}{h_0}$

Here, $u=4\text{cm};v=5\text{cm}{h}_0=2\text{cm}$

$\Rightarrow \frac{h_i}{h_0}=\frac{v}{u}$

$\Rightarrow h_i=\frac{v}{u}\times h_0$

$\Rightarrow h_i=\frac{5}{4}\times 2=2.5\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.