Question

A convex mirror used for rear-view on an automobile has a radius of curvature of $3 \cdot 00{\text{m}}$. If a bus is located at $5 \cdot 00{\text{m}}$ from this mirror, find the position, nature and size of the image.

Answer 1

Hint:A convex mirror is a curved mirror whose reflecting surface bulges out towards the source of light. Here, the focal length of the mirror will be half of the given radius of curvature of the mirror. We can then easily obtain the image distance using the mirror equation. Determining the magnification of the mirror will help us to specify the nature and size of the image. A positive magnification implies that the image formed is erect and if the magnification is greater than one, then the image is a magnified one.

Formulas used:
-The mirror equation is given by, $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $u$ is the object distance from the mirror, $v$ is the image distance and $f$ is the focal length of the mirror.
-The magnification of a mirror is given by, $m = \dfrac{{ - v}}{u}$ where $u$ is the object distance from the mirror and $v$ is the image distance.

Complete step by step solution.

Step 1: List the known parameters involved in the given problem.
Here the bus is the object of the convex mirror whose image is formed. The position of the bus (or object) from the mirror is given to be $u = 5 \cdot 00{\text{m}}$.

The radius of curvature of the convex mirror is given to be $R = 3 \cdot 00{\text{m}}$, then its focal length will be $f = \dfrac{R}{2} = \dfrac{3}{2} = 1 \cdot 5{\text{m}}$.

Let $v$ be the position of the image from the mirror.

Step 2: Express the mirror equation to obtain the position of the image formed.
The mirror equation is given by, $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $u$ is the object distance from the mirror, $v$ is the image distance and $f$ is the focal length of the mirror.
$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$-------- (1)

Substituting for $u = - 5 \cdot 00{\text{m}}$ and $f = 1 \cdot 5{\text{m}}$ in equation (1) we get, $ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{1 \cdot 5}} + \dfrac{1}{5} = \dfrac{{5 + 1 \cdot 5}}{{5 \times 1 \cdot 5}} = \dfrac{{6 \cdot 5}}{{7 \cdot 5}}$

$ \Rightarrow v = 1 \cdot 15{\text{m}}$

Thus the image distance from the mirror is obtained to be $v = 1 \cdot 5{\text{m}}$. The positive value indicates that the image is formed behind the mirror and so it is a virtual image.

Step 3: Express the relation for the magnification of the mirror to determine the nature and size of the image.

The magnification of the mirror can be expressed as $m = \dfrac{{ - v}}{u}$ ------- (2)

Substituting for $v = 1 \cdot 5{\text{m}}$ and $u = - 5 \cdot 00{\text{m}}$ in equation (2)
we get, $m = \dfrac{{ - 1 \cdot 15}}{{ - 5}} = 0 \cdot 23$
The magnification obtained is $m = 0 \cdot 23$ and it is positive. This suggests that the image formed is erect and since the magnification is less than one, the image formed will be diminished.

So the size of the image will be $0 \cdot 23$ times the size of the object (bus).

Note:For problems related to ray optics, it is advised to use sign convention when substituting the values of the image distance or object distance or focal length of mirrors and lenses in different equations. Here the given mirror is convex and by sign convention, all distances in front of the mirror will be taken to be negative while those behind the mirror will be taken to be positive. So the focal length of the convex mirror is positive while the object distance is negative. Hence we substituted the object distance as $u = - 5 \cdot 00{\text{m}}$ in equations (1) and (2).