Question

a convex mirror used in a bus has radius of curvature 3.5 metre if the driver of the bus locates 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

❣Holla user❣

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...

Hope it helps u^_^