Question

A convex mirror with a radius of curvature of 0.450 m is placed above theaisles in a store. Determine the image distance and magnification of acardboard box on the floor 2.75 m below the mirror. Is the image virtual orreal? Is the image inverted or upright?

Answer 1

Given:

The radius of the curvature of the convex mirror is

$R=0.450\text{ m}$

The distance of the object is

$u=-2.75\text{ m}$

Required: image distance and the magnification

Explanation:

first, we find the focal length of the mirror by the relation is

$f=\frac{R}{2}$

plugging the values in the above, we get

$\begin{gathered} f=\frac{0.450\text{ m}}{2} \\ f=0.22\text{5 m} \end{gathered}$

the mirror formula is given by

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

Plugging all the values in the above relation, we get

$\begin{gathered} \frac{1}{0.225\text{ m}}=\frac{1}{v}+\frac{1}{-2.75\text{ m}} \\ \frac{1}{v}=\frac{1}{0.225}+\frac{1}{2.75} \\ \frac{1}{v}=\frac{2.75+0.225}{2.75\times0.225} \\ \frac{1}{v}=\frac{2.975}{0.618} \\ v=0.21\text{ m} \end{gathered}$

the distance of the image is 0.21 m. it will be formed behind the mirror.

now calculate the magnification. magnification is given by

$m=-\frac{v}{u}$

Plugging all the values in the above, we get

$\begin{gathered} m=-\frac{0.21m}{-2.75\text{ m}} \\ m=0.076\text{ } \end{gathered}$

The magnification is 0.076.

the image formed by the convex mirror is virtual, inverted, and erect.