Optics Problem-II

The most obvious use of mirror is to see our reflection

Mirrors can be classified into different types

Plane mirrors show the reflection of any object that is in front of it

Spherical mirrors is useful when a large field of view is needed

For example, the rear view mirror on a car, which is used to see objects behind the car

The rear view mirror is mainly convex spherical mirror

For mirrors there is a relation between image and object distance

In spherical mirrors, the relation between image and object size along with the radius of curvature or focal length of mirror is known as Mirror Formula

The Mirror Formula is given as, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Where, $v$ = image distance $u$ = object distance $f$ = focal length of the mirror

While putting the values of the variables we also need to use proper sign convention

Let us solve a mirror problem using this equation

A car is fitted with a convex side view mirror of focal length 20 cm. A second car 2.8m behind the first car is overtaking the first car at a relative speed of 15 m/s

Find the speed of the image of the second car as seen in the mirror of the first one

Values that are supplied to us is, object distance $u$ = 2.8 m = 280 cm, and focal length $f$ = 20 cm

So, we have to first find out the image distance using mirror formula

According to sign convention, the focal length of a convex mirror is always positive and object distance is always negative

Therefore, $u$ = -280 cm ,and $f$ = +20 cm

Putting the values of u and f in mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-280}$

Solving the equation, the image distance is found to be, $v$ = $\frac{280}{15}$ cm

As we know the distance, speed can be calculated by differentiating $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ with respect to time t

Differentiating with respect to t, we get, $\frac{-1}{v^2}\frac{dv}{dt}=\frac{1}{u^2}\frac{du}{dt}$

The speed of the image of second car is, $\frac{dv}{dt}=\frac{{-v}^2}{u^2}\frac{du}{dt}$

Putting the values of variables, we get, $\frac{dv}{dt}=\frac{-\frac{{280}^2}{{15}^2}}{{280}^2}\times 15$

Where, $\frac{du}{dt}$ = 15 $u$ is positive as the second car overtakes the first car

Solving the equation, we get, speed of image of second car as $\frac{dv}{dt}=-\frac{1}{15}$

Therefore, the speed of image of second car is $\frac{1}{15}$ m/s

Revision

Mirror Formula, $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

Where, $v$ = image distance $u$ = object distance $f$ = focal length of the mirror

Proper sign convention should be used while solving mirror problems

The End