Question

A concave mirror and a convex mirror of focal length $10\text{cm}$ and $15\text{cm}$ respectively are placed at a distance of $70\text{cm}$. An object $\text{AB}$ of height $2\text{cm}$ is placed in between both both the mirrors at a distance of $30\text{cm}$ from the concave mirror. The first ray is incident on the concave mirror then on the convex mirror. Then final size and position of the image is

• A. $2cm,\frac{165}{7}cm$
• B. $0.2cm,\frac{165}{14}cm$
• C. $0.2cm,\frac{165}{28}cm$
• D. $2cm,\frac{165}{14}cm$

Answer 1

The correct option is B $0.2cm,\frac{165}{14}cm$


Here, the first ray is incident on concave mirror.

So, for concave mirror

$u=-30\text{cm};f=-10\text{cm}$

Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}-\frac{1}{30}=\frac{-1}{10}$

$\Rightarrow v=-15\text{cm}$

Now applying magnification formula,

$m=\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{-v}{u}$

$\Rightarrow \frac{A^{\prime }{B}^{\prime }}{2}=\frac{-(-15)}{-30}$

$\Rightarrow A^{\prime }{B}^{\prime }=-1\text{cm}$

So, image formed by first reflection will be real inverted and diminished.


Now, object distance for convex mirror

$u^{\prime }=O_2{A}^{\prime }=(70-15)=-55\text{cm}$ and

$f^{\prime }=+15\text{cm}$

Again using mirror formula,
$\frac{1}{v^{\prime }}+\frac{1}{u^{\prime }}=\frac{1}{f^{\prime }}$

$\Rightarrow \frac{1}{v^{\prime }}=\frac{1}{15}+\frac{1}{55}$

$\Rightarrow v^{\prime }=\frac{165}{14}\text{cm}$

Again applying magnification formula,

$m^{\prime }=\frac{A^{\prime \prime }{B}^{\prime \prime }}{A^{\prime }{B}^{\prime }}=\frac{-v^{\prime }}{u^{\prime }}=\frac{-\left(\frac{165}{14}\right)}{-55}$

$\Rightarrow A^{\prime \prime }{B}^{\prime \prime }=\left(\frac{3}{14}\right)A^{\prime }{B}^{\prime }$

$\Rightarrow A^{\prime \prime }{B}^{\prime \prime }=\left(\frac{3}{14}\right)(-1)=-0.2\text{cm}$

$\begin{array}{c}{A}^{\prime \prime }{B}^{\prime \prime }=-0.2\text{cm};v^{\prime \prime }=\frac{165}{14}\text{cm}\end{array}$

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