Question

A security mirror used in a big showroom has a radius of curvature $5{\rm{ m}}$. If a customer is standing at a distance of $20\,\;{\rm{m}}$ from the cash counter, find the position, nature and size of the image formed in the security mirror.

Answer 1

We will use the mirror formula for the given situation which gives us the relation between focal length, object distance and image distance. We will also write the expression for magnification of the customer for the image formed on the mirror.

Complete step by step answer:
Given:
The radius of curvature of the security mirror is $R = 5\;{\rm{m}}$.
The distance of the customer from the cash counter is called the image distance and it is given $u = 20\,\;{\rm{m}}$.

We are required to find the position, nature and size of the image of the customer formed on the security mirror when it is standing at a distance $20\,\;{\rm{m}}$ from the cash counter.

Let us write the expression for relation between object distance, image distance and the focal length of the given security mirror.
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\\\ \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$……(1)
Here f is the focal length of the mirror.

We know that the expression for focal length of the given mirror in terms of radius of curvature can be expressed as:
$f = \dfrac{R}{2}$

We will substitute $5\;{\rm{m}}$ for R in the above expression to find the value of focal length of the mirror.$f = \dfrac{{5{\rm{ m}}}}{2}\\\ = 2.5\;{\rm{m}}$

On observing the given situation closely and taking sign conventions for object distance, image distance and focal length, we will find that object distance and focal length have negative value because they on the other side of the image formed so we can write:
$f = - 2.5{\rm{ m}}$
And,
$u = - 20{\rm{ m}}$

We will substitute $- 2.5{\rm{ m}}$ for f and $- 20{\rm{ m}}$ for u in equation (1).
$\dfrac{1}{v} = \dfrac{1}{{ - 2.5{\rm{ m}}}} - \dfrac{1}{{ - 20{\rm{ m}}}}\\\ \implies \dfrac{1}{v} = - \dfrac{7}{{20}}{\rm{ m}}\\\ \implies v = - \dfrac{{20}}{7}{\rm{ m}}$

We can write the expression for magnification as below:
$m = - \dfrac{v}{u}$

We will substitute $- \dfrac{{20}}{7}{\rm{ m}}$ for v and $- 20{\rm{ m}}$ for u in the above expression.
$m = - \dfrac{{\left( { - \dfrac{{20}}{7}{\rm{ m}}} \right)}}{{ - 20{\rm{ m}}}}\\\ = - \dfrac{1}{7}$

Therefore, we can say that the image of the customer is formed at a distance $\dfrac{{20}}{7}$ in the direction opposite to which we considered earlier. The image of the object formed is smaller than the actual image by a factor of $\dfrac{1}{7}$.

Note:
We should not forget to consider the sign convention of focal length, distance of customer from the counter and the distance of image formed on the mirror for the given situation.