Question

Figure shows a concave mirror of focal length $20\text{cm}$. A point object on the principal axis at a distance of $30\text{cm}$ from the pole starts moving perpendicular to the principal axis and its position time relation is given by $y=(3t^2+4t)\text{cm/s}$. The speed of the image at $t=1\text{s}$ will be :

• A. $15\text{cm/s}$
• B. $18\text{cm/s}$
• C. $20\text{cm/s}$
• D. $12\text{cm/s}$

Answer 1

The correct option is C $20\text{cm/s}$

Given,
Focal length, $f=-20\text{cm}$
Object distance, $u=-30\text{cm}$
From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{30}=\frac{-1}{20}$
$\Rightarrow \frac{1}{v}=\frac{-1}{60}$
$\Rightarrow v=-60\text{cm}$

Now, lateral magnification,
$m=\frac{h_2}{h_1}=\frac{-v}{u}$
$\Rightarrow h_2=\frac{-(-60)}{-30}{h}_1$
$\Rightarrow h_2=-2h_1$

Now, object height can be assumed to be the distance of the point object from the P.A at time $t$
$\Rightarrow h_2=-2(3t^2+4t)$
$\Rightarrow h_2=-6t^2-8t$
On differentiating both sides w.r.t time
$\Rightarrow \frac{d{h}_2}{dt}=-12t-8$
$\Rightarrow v_i=-12t-8$ is the velocity of image perpendicular to P.A at time $t$.

At $t=1\text{s}$,
$v_i=-12\times 1-8$
$\Rightarrow v_i=-20\text{cm/s}$