Question

A circular disc of diameter $1\text{cm}$ has been placed on the principal axis of a concave mirror $(R=20\text{cm})$ with its plane perpendicular to the principal axis at a distance of $15\text{cm}$ from the pole of the mirror. If the radius of disc starts increasing according to equation $r=(0.5+0.1t)\text{cm}$, where $t$ is time in $\text{s}$, then what will be the rate at which radius of its image is increasing?

• A. $0.2\text{cm/s}$
• B. $0.3\text{cm/s}$
• C. $0.4\text{cm/s}$
• D. $0.5\text{cm/s}$

Answer 1

The correct option is A $0.2\text{cm/s}$

Considering the radius of the disc as an extended object placed in front of the concave mirror.


For the given concave mirror,
$u=-15\text{cm},f=\frac{-20}{2}=-10\text{cm}$
Using mirror formula,
$\frac{1}{v}-\frac{1}{15}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-1}{30}$
$\Rightarrow v=-30\text{cm}$
So, $m=\frac{-v}{u}=\frac{-(-30)}{-15}=-2$
Now, radius of image,
$r_i=|m|r_o$
$\Rightarrow r_i=2(0.5+0.1t)=1+0.2t$
Rate at which radius of image is increasing,
$\frac{d{r}_i}{dt}=0+0.2=+0.2\text{cm/s}$
Hence, the radius of the image is increasing at the rate of $0.2\text{cm/s}$.
$\begin{array}{c}\text{Why this question ?}\\ \text{As the plane of the disc is perpendicular to the principal axis,}\\ \text{we can consider it as a linear extended object}\\ \text{and can apply mirror formula.}\end{array}$