Question

A convex mirror used on automobile has 3 m radius of curvature. If a bus is located at 5m from this mirror, find the position and size of the image

Answer 1

Step 1: Given that:

The radius of curvature(R) of the convex mirror = $3m$

The distance of the bus from the mirror = object distance(u) = $-5m$

Step 2: Calculation of focal length of the mirror:

The focal length of a mirror is half of the radius of curvature of the mirror.
Therefore;

$f=\frac{3m}{2}$

$f=1.5m$

Step 2: Calculation of position of image:

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

Where; u = object distance, v= image distance and f is the focal length of the mirror.

Putting the given values we get;

$\frac{1}{v}+\frac{1}{-5m}=\frac{1}{1.5m}$

$\frac{1}{v}=\frac{1}{1.5}+\frac{1}{5}$

$\frac{1}{v}=\frac{10}{15}+\frac{1}{5}$

$\frac{1}{v}=\frac{10+3}{15}$

$\frac{1}{v}=\frac{13}{15}$

$v=\frac{15}{13}$

$v=1.15cm$

Thus, the position of the image is

$1.15cm$

behind the mirror.

Step 4: Finding the nature of the image:

Magnification produced by the mirror is given by;

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

Now,

$m=-\frac{\frac{15}{13}cm}{-5cm}$

$m=\frac{15}{13\times 5}$

$m=+\frac{3}{13}$

$m=+0.23$

Positive magnification represents that the image is virtual and erect and since the value of the magnification is less than 1, the image is smaller than the object.

Hence,

The image is formed behind the mirror at $1.15cm$ and the image is smaller than the object, virtual and erect.