Question

a convex mirror used in a bus has radius of curvature 3.5 metre if the driver of the first locates a car at 10m behind 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,

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

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

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

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

$v = \frac{70}{47} = 1.5m$

Thus, the car appears to be at 1.5m from the convex mirror.

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

Using,

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

$m = \frac{ - 1.5}{( - 10)} = 0.15$

Thus, the size of the image of the car is 0.15 times the actual size of the car.

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

Answer 2

R=3.5 , f=R/2= 1.75m

u= -10cm, v=?

1/f =1/v + 1/u

1/1.75 = 1/v - 1/10

v= 1.5 m

magnification (m)= -v/u =0.15

Image is Real, Erect and Diminished...