Question

A convex mirror used in a bus has radius 3.5 m. If the driver of the bus locates a car at 10:00 behind bus find the position nature and size of the image of the car.

Answer 1

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...

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Answer 2

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.