The convex mirror in the park is an illustration of lens magnification
The man must focus his eyes at 4.50 meters to see his image
<h3>How to determine the object distance?</h3>
The given parameter is:
Radius of curvature, r = 3.00 m
The magnification (m) of the mirror is 1/2, because the image (v) is half as tall as the actual height (u).
So, we have:
m = u/v
So, we have:
u/v = 1/2
Make v the subject
v = 2u
The focal length is calculated as:
1/f = 1/u + 1/v
The focal length is calculated as:
f = r/2
f = 3/2
Substitute f = 3/2 in 1/f = 1/u + 1/v
2/3 = 1/u + 1/v
Substitute v = 2u in 2/3 = 1/u + 1/v
2/3 = 1/u + 1/2u
Take the LCM
2/3 = (2 + 1)/2u
This gives
2/3 = 3/2u
Take the inverse of both sides
3/2 = 2u/3
Cross multiply
2u = 3 * 3
4u = 9
Divide both sides by 4
u = 9/4
This gives
u = 2.25
Recall that:
v = 2u
So,we have:
v = 2 * 2.25
v = 4.50
Hence, the man must focus his eyes at 4.50 meters to see his image
Read more about magnification at:
brainly.com/question/2759705
