Question

A convex mirror used on a moving automobile has a radius of curvature of 3.0m.If a truck is following it at a constant distance of 3.5m,calculate
(a) the position of the image
(b) the nature of the image
(c) the magnification of the image

Answer 1

refer to this attachment

Answer 2

Position of the image, v = +1.05 m

Image is virtual and erect

Magnification, m = 0.3

Explanation:

It is given that,

Radius of curvature of the convex mirror, R = 3 m

Focal length of the convex mirror, f = +1.5 m (R = 2f)

Distance between convex mirror and the truck, u = -3.5 m

(a) Let v is the position of the image wrt to the object. Using mirror's formula to find it as :

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

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{(1.5)}-\dfrac{1}{(-3.5)}$

v = +1.05 m

So, the image is located at a distance of 1.05 m from the mirror.

(b) The image formed by the convex mirror is virtual and erect.

(c) Let m is the magnification of mirror. It is given by :

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

$m=\dfrac{-(1.05)}{-3.5}$

m = 0.3

Hence, this is the required solution.

Learn more,

Mirror's formula

https://brainly.in/question/10321690