Question

centre of curvature of convex mirror used on a moving automobile is 2 metre a truck is coming behind him at a constant distance of 3.5 M calculate the position nature and size

Answer 1

Given :

R = 2 m

=> focal length, f = R/2

=> f = 1 m

Object distance, u = -3.5 m

To find :

Position of the image, magnification and nature of the image.

Solution :

1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = 1/1 - 1/(-3.5)

1/v = (3.5 + 1) / 3.5

1/v = 4.5 / 3.5

v = 3.5 / 4.5

v = 0.77 m

Therefore, the image will be formed at a distance of 0.77 m from the mirror.

Magnification, m = h'/h = - v/u

m = - (0.77) / (-3.5)

m = 0.77 / 3.5

m = 0.22 m

=> m < 1

Therefore, the image will be diminished, virtual and erect. 

Answer 2

Given :

R = 2 m

=> focal length, f = R/2

=> f = 1 m

Object distance, u = -3.5 m

To find :

Position of the image, magnification and nature of the image.

Solution :

1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = 1/1 - 1/(-3.5)

1/v = (3.5 + 1) / 3.5

1/v = 4.5 / 3.5

v = 3.5 / 4.5

v = 0.77 m

Therefore, the image will be formed at a distance of 0.77 m from the mirror.

Magnification, m = h'/h = - v/u

m = - (0.77) / (-3.5)

m = 0.77 / 3.5

m = 0.22 m

=> m < 1

Therefore, the image will be diminished, virtual and erect.

Hope this helps u............