Question

A convex mirror used in a bus has radius of curvature 3.5 metre and if the driver of the bus locate a car at 10 metre behind the bus find the position nature and size of the image of the car.​

Answer 1

Solution:

Here,

$R = 3.5m$

$f = \frac{r}{2} = \frac{3.5}{2} = 1.75m$

$u = - 10.0m$

Step1: Determination of the position of the car.

Using,

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

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

$\frac{1}{v} = \frac{1}{1.75} - \frac{1}{( - 10)}$

$\frac{1}{v} = \frac{1}{1.75} + \frac{1}{10} \ = \frac{47}{70}$

$v = \frac{70}{47} = 1.49m$

Thus, the car appears to be at 1.49m from the convex mirror.

Step2: Determination of the size and nature of the image.

Using,

$m = - \frac{v}{u}$

$m = \frac{ - 1.49}{( - 10)} = 0.149$

Thus, the size of the image of the car is 0.149 times the actual size of the car.

Since, 'm' is positive, so image of the car is erect or upright.

Answer 2

ANSWER

Given,

Radius of Curvature, r = 3.5cm

Object distance i.e distance of mirror from car is -10m.(u)

Now we know that >

Focal length, f = $\dfrac{3.5}{2}$

=> f = 1.75 cm.

Image distance, v =?

Now, by using mirror formula we have >

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\implies \dfrac{1}{v} - \dfrac {1}{10} = \dfrac{1}{1.75}$

$\implies \dfrac{1}{v} =\dfrac {1}{1.75} + \dfrac {1}{10}$

$\implies \dfrac{1}{v} =+0.67 cm$

$\implies v =+ 1.49cm$

Hence the image is formed at a distance of 1.49cm from focal length.

Now,

magnification = $\dfrac{ - height\:of\:image}{height\:of\:Object}$

$\implies m =\dfrac{-v}{u}$

$\implies m = \dfrac{-(-1.49)} {10}$

$\implies m = +0.149$

Hence the height of image is 0.149 times more than object height.

Since, magnification is positive and image distance is also positive,hence the image formed is virtual,erect and smaller in size than the object.