Question 13 Marks
A convex mirror used for rear-view on an automobile has a radius of curvature of $3.00 m$. If a bus is located at $5.00 m$ from this mirror, find the position, nature and size of the image.
Answer
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$R=+3.00 m \quad(\because$ given mirror is convex $)$
$u=-5.00 m(\because$ object distance is always negative)
$v=?$
$h^{\prime}=?$
(i) For spherical mirror, $f=\frac{R}{2}=\frac{+3.00 m}{2}$
$\therefore f=+1.50 m$
* Now, according to mirror formula,
$\frac{1}{f} =\frac{1}{v}+\frac{1}{u}$
$\therefore \quad \frac{1}{v} =\frac{1}{f}-\frac{1}{u}=\frac{1}{(+1.50)}-\frac{1}{(-5.00)}$
$\therefore \frac{1}{v} =\frac{1}{1.50}+\frac{1}{5.00}=\frac{5.00+1.50}{(1.50)(5.00)}$
$\therefore \frac{1}{v} =\frac{6.50}{7.50} \Rightarrow v=\frac{7.50}{6.50}=+1.15 m$
$\Rightarrow$ Image will be formed on the back side of mirror, at $1.15 m$ from the pole of mirror.
$(ii)$ Now, magnification is,
$m=-\frac{v}{u}=-\frac{(+1.15 m)}{(-5.00 m)}$
$\therefore m=+0.23 \Rightarrow \frac{h^{\prime}}{h}=0.23$
$\Rightarrow h^{\prime}=0.23 h$
$\Rightarrow$ Image will be
$(i)$ virtual and erect
$(ii)$ diminished ( $\because m$ is positive)
$(\because|m|<1)$
$(iii)$ of height 0.23 times the height of object ( $\because$ $h^{+}=0.23 h$ )
(iv) formed on the back side of mirror, at $1.15 m$ from the mirror
$(\because v=+1.15 m )$
$R=+3.00 m \quad(\because$ given mirror is convex $)$
$u=-5.00 m(\because$ object distance is always negative)
$v=?$
$h^{\prime}=?$
(i) For spherical mirror, $f=\frac{R}{2}=\frac{+3.00 m}{2}$
$\therefore f=+1.50 m$
* Now, according to mirror formula,
$\frac{1}{f} =\frac{1}{v}+\frac{1}{u}$
$\therefore \quad \frac{1}{v} =\frac{1}{f}-\frac{1}{u}=\frac{1}{(+1.50)}-\frac{1}{(-5.00)}$
$\therefore \frac{1}{v} =\frac{1}{1.50}+\frac{1}{5.00}=\frac{5.00+1.50}{(1.50)(5.00)}$
$\therefore \frac{1}{v} =\frac{6.50}{7.50} \Rightarrow v=\frac{7.50}{6.50}=+1.15 m$
$\Rightarrow$ Image will be formed on the back side of mirror, at $1.15 m$ from the pole of mirror.
$(ii)$ Now, magnification is,
$m=-\frac{v}{u}=-\frac{(+1.15 m)}{(-5.00 m)}$
$\therefore m=+0.23 \Rightarrow \frac{h^{\prime}}{h}=0.23$
$\Rightarrow h^{\prime}=0.23 h$
$\Rightarrow$ Image will be
$(i)$ virtual and erect
$(ii)$ diminished ( $\because m$ is positive)
$(\because|m|<1)$
$(iii)$ of height 0.23 times the height of object ( $\because$ $h^{+}=0.23 h$ )
(iv) formed on the back side of mirror, at $1.15 m$ from the mirror
$(\because v=+1.15 m )$











