Question

Two cars $\mathrm{A}$ and $\mathrm{B}$ moving on a straight road. Car $\mathrm{B}$ passes car $\mathrm{A}$ by a relative speed of $45\mathrm{m}/\mathrm{s}$. At what speed does the driver of car $\mathrm{B}$ observe car $\mathrm{A}$ in the side mirror of focal length $10\mathrm{cm}$ when car A is at a distance of $1.9\mathrm{m}$ from car $\mathrm{B}$?

• A. $\frac{9}{80}$
• B. $\frac{-9}{80}$
• C. $\frac{-7}{80}$
• D. $\frac{7}{80}$

Answer 1

The correct option is B

$\frac{-9}{80}$

Step 1. Given data

Relative speed is $45\mathrm{m}/\mathrm{s}$ and focal length $\mathrm{f}$ is $10\mathrm{cm}$ and distance is $1.9\mathrm{m}$.

We have to find the out speed with respect to car $\mathrm{A}$

Step 2. Formula to be used

The expression of velocity of image with respect to mirror is,

${\mathrm{V}}_{\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}=-{\mathrm{m}}^2\times {\mathrm{V}}_{\raisebox{1ex}{$\mathrm{o}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}$

Here, ${\mathrm{V}}_{\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}$ is the image speed with respect to mirror, ${\mathrm{V}}_{\raisebox{1ex}{$\mathrm{o}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}$ is the velocity of an object with respect to the mirror.

The mirror formula is the relationship between the focal length $\mathrm{f}$ of the mirror, the distance $\mathrm{u}$ of the object from the pole of the mirror, and the distance $\mathrm{v}$ of the image from the pole.

So, the mirror formula is,

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

The derivation of mirror formula is,

$\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}$

$\frac{1}{\mathrm{v}}=\frac{\mathrm{u}-\mathrm{f}}{\mathrm{uf}}$

So,

$\mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}}$ -------- $\left(1\right)$

Now, the linear magnification for spherical mirrors is,

$\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$ -------- $\left(2\right)$

As size of image differs from the size of object, magnification exists.

Substitute equation $\left(1\right)$ in $\left(2\right)$, we get,

$\mathrm{m}=\frac{\left({\displaystyle \frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}}}\right)}{\mathrm{u}}$

$=\frac{-\mathrm{uf}}{\mathrm{u}\left(\mathrm{u}-\mathrm{f}\right)}$

$=\frac{\mathrm{f}}{-\mathrm{u}+\mathrm{f}}$

$=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}$

Here, $\mathrm{f}$ is focal length, $\mathrm{u}$ is object distance, $\mathrm{v}$ is image distance and $\mathrm{m}$ is magnification.

Step 3. Find the magnification.

From the given,

${\mathrm{V}}_{\raisebox{1ex}{$\mathrm{o}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}=45\mathrm{m}/\mathrm{s}$

$\mathrm{u}=1.9\mathrm{m}$

$=190\mathrm{cm}$

So,

$\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}$

Substitute the given values in the above formula, we get,

$=\frac{10}{10-\left(-190\right)}$

We take $190$ to be in negative, because the object distance is taken as negative, because from the above figure, the light is approaching from opposite end and the another car is moving against it.

So,

$=\frac{10}{10+190}$

$=\frac{10}{200}$

Step 4. Find the speed with respect to first car.

So, the expression of velocity of image with respect to mirror is,

${\mathrm{V}}_{\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}=-{\mathrm{m}}^2\times {\mathrm{V}}_{\raisebox{1ex}{$\mathrm{o}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}$

Substitute the above calculated values in the above formula, we get,

${\mathrm{V}}_{\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}=-{\left(\frac{10}{200}\right)}^2\times 45\mathrm{m}/\mathrm{s}$

$=-\frac{1}{400}\times 45\mathrm{m}/\mathrm{s}$

$=-\frac{9}{80}\mathrm{m}/\mathrm{s}$

$=-0.1\mathrm{m}/\mathrm{s}$

So, car will appear to move with speed is $0.1\mathrm{m}/\mathrm{s}$.

Hence, option $\mathrm{B}$ is correct answer.