Question

the radius of curvature of a convex mirror used as a rear view mirror in a moving car is '20' m a truck is coming from behind it at a distance of 3.5 metre calculate the a)position b)size of the image relative to the size of truck what will be the nature of the image.​

Answer 1

Answer :-

(a) Postion of the image is 2.59 m from the mirror.

(b) Realtive size [magnification] is 0.74 .

Nature of the image is erect and virtual .

Explanation :-

We have :-

→ Radius of curvature (R) = 20 m

→ Distance of the truck (u) = 3.5 m

______________________________

Firstly, let's calculate focal length of the mirror .

⇒ f = R/2

⇒ f = 20/2

⇒ f = 10 m

Since the mirror is convex :-

• f = + 10 m

• u = -3.5 m

According to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/10 - (-1)/3.5

⇒ 1/v = 1/10 + 10/35

⇒ 1/v = (7 + 20)/70

⇒ 1/v = 27/70

⇒ 27v = 70

⇒ v = 70/27

⇒ v = 2.59 m

______________________________

Now, from the formula of magnification of a spherical mirror :-

m = -(v/u)

⇒ m = -(2.59/-3.5)

⇒ m = -(-0.74)

⇒ m = + 0.74

Answer 2

Given :-

The radius of curvature of a convex mirror used as a rear view mirror in a moving car is '20' m a truck is coming from behind it at a distance of 3.5 metre

To Find :-

a)position b)size of the image relative to the size of truck what will be the nature of the image.​

Solution :-

We know that

Focal Length = Radius/2

Focal length = 20/2 = 10 m

1/u + 1/v = 1/f

1/-3.5 + 1/v = 1/10

-1/3.5 + 1/v = 1/10

1/v = 1/10 - (-1/35)

1/v = 1/10 + 1/3.5

1/v = 1/10 + 10/35

1/v = 7 + 20/70

1/v = 27/70

v =  70/27

Now

Size of image = m = -v/u

-(70)/27/-3.5

70/27/3.5

0.74