91Ó°ÊÓ

The object distance is, $+40\text{cm}$

The magnification,$m=-0.7$

The object distance and the magnification ratio are given which find the image distance and the object distance. Then, using the image and object distance find the focal length. From the focal length, decide the type of mirror, and from the image distance, decide whether the image is real or virtual.

Formulae:

The radius of curvature is,

$r=2f$

The spherical mirror expression is,

role="math" localid="1663129482481" $\frac{1}{i}+\frac{1}{p}=\frac{1}{f}$

The magnification is,

$m=\frac{-i}{p}$

Where,$m$ is the magnification, $p$ is the pole, $f$is the focal length.

Write the formula for the magnification below.

$\begin{array}{rcl}M& =& \frac{-i}{p}\\ i& =& -Mp\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}i& =& -\left(-0.7\right)×40\text{cm}\\ & =& 28\text{cm}\end{array}$

Now the focal length is defined as follows.

role="math" localid="1663129856144" $\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{i}+\frac{1}{p}\\ & =& \frac{1}{28\text{cm}}+\frac{1}{40\text{cm}}\\ & =& \left(0.0357+0.025\right){\text{cm}}^{-1}\\ & =& 0.0607{\text{cm}}^{-1}\end{array}$

$\begin{array}{rcl}f& =& \frac{1}{0.0607{\text{cm}}^{-1}}\\ & =& 16.5\text{cm}\\ & ≈& +16\text{cm}\end{array}$

Hence, the mirror is concave because the focal length is positive.

The focal length is$+16\text{cm}$

Use the following formula to find the radius of curvature,

$\begin{array}{rcl}r& =& 2×f\\ & =& 2×16.5\text{cm}\\ & =& 33\text{cm}\end{array}$

Hence, the radius of curvature is$33\text{cm}$

The object distance is $p=+40\text{cm}$as given in the problem.

From the part (a), you can say that the image distance is,

$i=28\text{cm}$

As given in the problem, the magnification ratio is$M=-0.70$

Since image distance is positive, the image is real.

As the magnification is negative and image is real, so the image is inverted.

A real image is formed on the same side of the mirror as the object.