Question

a convex mirror used in a bus has radius of curvature 3.5 metre is the driver of the bus locates a car at 10 metres behind the bus then 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

Hey mate,..

R= 3.5

U= 10.0

f= R/2   =3.5/2

3.5*10/2*10

35/20

7/4

1.75

1/f= 1/v +1/u    1/v =  1/f- 1/u       1/v= 1/1.75-1/-10

= 1/7/4-1/-10   =4/7- 1/-10

= -40-7\ 70 = 47/70

v = 70\47

v = 1.5

m= -v\u

m = -1.5\[-10]

= 0.15

Hope it will help you