Question

A convex mirror used for rear view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror. Find the position, nature and size of the image.

Answer 1

We have:
Radius of curvature(r)= 3m
Object distance (u)= -5m
Position of image (v)= ?
Nature of image = ?
Size of the image (h')= ?

Since,
Radius of curvature (r) = 3m
.°.Focal length (f) = 3/2m

We know that,
By mirror formula,

1/u + 1/v = 1/f
=> 1/-5 + 1/v = 1/3/2
=> 1/v =2/3 + 1/5
=> 1/v = (10+3)/15
=> 1/v = 13/15
=> v = 15/13
=> v = 1.153 m
=> v = 1.15 m (approx.)

.°. Position of image = 1.15m,, behind the mirror.

Since, v= +ve
.°. Nature of image = Virtual and erect.

Again,
Let h be the height of the object and h' is the given height of image so formed.

.°. h'/h = -v/u
=> h'/h = -(15/13) / -(5)
=> h'/h = (15/13) × (1/5)
=> h'/h = 3/13
=> h' = (3/13)h

.°. Size of image is 3/13 times the size of the object.

Answer 2

Given:

Radius of curvature = 3.00 m

Bus located from this mirror = 5.00 m

Mirror formula:

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

Formula of magnification of a mirror:

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

Here,

v is the distance of the image and u is the distance of the object from the mirror.

Solution:

u = 5m

r = 3m

f = r/2

f = 3/2

f = 1.5m

Now in the mirror formula:

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \frac{6.5}{7.5}$

v = 1.15m

Magnification:

$m = - \frac{v}{u} \\ m = - \frac{ - 1.15}{ - 5} \\ m = 0.23$

Therefore, the size of the image on the convex mirror will be 0.23 times more than the bus (object).

The image formed will be:

• Erect
• Diminished