Question

A convex mirror used for rear view on a car has radius of curvature $3m$. If a bus is located at $5m$ from this mirror. Find position, nature and size of the image.

Answer 1

Hint: We will use a mirror formula to solve the given problem. We should also use the fact that convex mirrors always produce diminished and virtual images. Image is always produced on the opposite side of the object.
Formula Used:
We are using the following formula to find the correct answer:-
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$.

Complete step by step solution:
From the given problem we have the following parameters with us:-
Object distance, $u=-5m$.
Radius of curvature, $R=3m$.
We have to find the image distance, $v$.
We have mentioned the signs for parameters using sign conventions for convex mirrors.
We know that focal length is half of the radius of curvature.
Therefore, focal length, $f=\dfrac{R}{2}$
$\Rightarrow f=\dfrac{3}{2}m$(Using sign convention focal length of a convex mirror is always positive).
Now using mirror formula we have
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$……………….. $(i)$
Putting values of the given parameters in equation $(i)$ we get
$\Rightarrow \dfrac{1}{v}+\dfrac{\left( -1 \right)}{5}=\dfrac{1}{\dfrac{3}{2}}$
$\Rightarrow \dfrac{1}{v}+\dfrac{\left( -1 \right)}{5}=\dfrac{2}{3}$
$\Rightarrow \dfrac{1}{v}=\dfrac{1}{5}+\dfrac{2}{3}$
$\Rightarrow \dfrac{1}{v}=\dfrac{3+10}{15}$
$\Rightarrow \dfrac{1}{v}=\dfrac{13}{15}$
$\Rightarrow v=0.87m$
Therefore, the position of the image is on the opposite side of the mirror with image distance of $0.87m$ .
From the properties of convex mirrors we know that the image produced is always virtual and behind the mirror.
Now using the relation of magnification, $m$ for the given mirror we can find the size of the image. Formula of magnification is given as follows:-
$m=-\dfrac{v}{u}$……………. $(ii)$
Using the parameters in equation $(ii)$ we get
$m=-\dfrac{0.87}{\left( -5 \right)}$
$\Rightarrow m=0.17$
Hence the size of image will be $0.17$ times the size of object and erect in nature as magnification has positive value.

Note:
We should use the sign conventions carefully for the given mirrors. Regardless of the nature of the mirror, object distance is always taken as negative. Sign of image distance depends on the side at which it forms. We should also be careful about the use of mirror formula and not lens formula. Formula for magnification should also be taken care of.