Question

Example 10.1
A convex mirror used for rear-view on an automobile
curvature of 3.00 m. If a bus is located at 5.00 m
find the position, nature and size of the image.​

Answer 1

Answer:

Position = 1.15 m behind the mirror.

Nature = Virtual and erect.

Magnification = 0.23

Explanation:

Step 1:- Write what's given.

Radius of curvature, R = 3.00m

Object distance , u = -5.00 m

Step 2 :- Write we have to find out

Image distance, v = ?

Height of image, h' = ?

Solution:-

We know that , focal length , f = R/2 = 3.00/2 = 1.50 m

By Mirror's formula :-

$\sf{\large{\boxed{1/f = 1/v + 1/u}}}$

${\implies{\sf{}}}$ 1/v = 1/f - 1/u

${\implies{\sf{}}}$ 1/v = 1/1.50 - 1/(-5.00)

${\implies{\sf{}}}$ 1/v = 1/1.50 + 1/500

${\implies{\sf{}}}$ 1/v = 5.00 + 1.50 / 7.50

${\implies{\sf{}}}$ 1/v = 6.50 / 7.50

${\implies{\sf{}}}$ v = 7.50 /6.50

${\implies{\sf{}}}$ v = + 1.15 m

The positive sign indicates the image formed behind the convex mirror.

$\huge{\boxed{\textsf{\Large{\red{Magnification\:formula:-}}}}}$

m = h'/ h = - v/u = - 1.15 m/ (- 5.00)m = +0.23

$\sf{\large{\boxed{Refer\: attachment}}}$