Question

The side-mirror of a truck has a convex section that enables the driver to see a wide field of view. Assume the convex portion of the mirror is spherical. A car is 30m behind the truck. If the virtual image of the car is 4% of the actual size of the car, then:

(a) Where is the virtual image, relative to the surface of the mirror?

(b) What is the radius of curvature of the mirror?

Answer 1

We know that magnification M=-v/u......(1)

Where v is image distance and u is object distance from the pole of the mirror

Given that image found 4% of object . If h_o is the object size and h_i is the image size then

Magnification M=h_i/h_{o, .} According to question h_i = 0.04h_{o .}

So magnification M = 0.04h_o/h_o = 0.04.

Here object distance is -30m . So using eq (1)

V=-M*u= (-0.04)*(-30) =1.2 m

So image distance is 1.2m. Ans

part (2) using mirror formula 1/v +1/u =1/f. f=focal length of the mirror

Substituting values. 1/1.2 +1/(-30) =1/f

f= 1.25 m. We also know that radius of curvature R =2f

So R =2*1.25 =2.50m. Ans