Question

A concave mirror and a convex mirror of focal lengths $10\text{cm}$ and $15\text{cm}$ are placed at a distance of $70\text{cm}$. An object $\text{AB}$ of height $2\text{cm}$ is placed at a distance of $30\text{cm}$ from the concave mirror as shown in the figure. If at first, ray is incident on the concave mirror and then on the convex mirror.Find the size and position of the image,

• 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 $0.2cm,\frac{165}{14}cm$

The diagram of the given problem:


$\text{For concave mirror}:$

Object distance, $u=-30\text{cm}$
Focal length, $f=-10\text{cm}$
Object height, $AB=2\text{cm}$

Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

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

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

Magnification,

$m=\frac{\text{Image height}}{\text{Object height}}=\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{-v}{u}$

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

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

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


Image formed by first reflection will be real inverted and diminished as shown in the figure.

$\text{Convex mirror:}$
For convex mirror, the image formed by concave mirror will act as object for convex mirror.
Now, the object distance for convex mirror, $u^{\prime }=O_2{A}^{\prime }=(70-15)=-55\text{cm}$

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


Using the mirror formula,

$\frac{1}{v^{\prime }}+\frac{1}{u^{\prime }}=\frac{1}{f^{\prime }}$

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

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

Magnification:

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

$\Rightarrow \frac{+A^{\prime \prime }{B}^{\prime \prime }}{A^{\prime }{B}^{\prime }}=\frac{-\left(\frac{165}{14}\right)}{-55}$

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

$\Rightarrow A^{\prime \prime }{B}^{\prime \prime }=0.214\text{cm}\approx 0.2\text{cm}$

So, the size of the final image is $A^{\prime \prime }{B}^{\prime \prime }=0.2\text{cm}$ and the position of image is $v^{\prime \prime }=\frac{165}{14}\text{cm}$.

Hence, option (b) is the correct answer.

Why this question: To understand the image formation by combination of spherical mirrors.