Question

A dedicated sports car enthusiast polishes the inside and outside surfaces of a hubcap that is a section of a sphere. When he looks into one side of the hubcap, he sees an image of his face 10.3 cm in back of it. He then turns the hubcap over, keeping it the same distance from his face. He now sees an image of his face 29.0 cm in back of the hubcap.

(a) How far is his face from the hubcap?
cm
(b) What is the magnitude of the radius of curvature of the hubcap?
cm

Answer 1

for convex side, Suppose

f = -|f|

u = object distance

v = -10.3 cm = image distance

Now using the equation

1/u + 1/v = 1/f

1/u - 1/10.3 = -1/f

1/f = (u - 10.3)/(u*10.3)

for concave side,

f = +|f|

u = object distance

v = -29.0 cm = image distance

Now using the equation

1/u + 1/v = 1/f

1/u - 1/29.0 = 1/f

1/f = (29.0 - u)/(u*29.0)

Now from both equation

(u - 10.3)/(u*10.3) = (29.0 - u)/(u*29)

29*(u - 10.3) = 10.3*(29.0 - u)

39.3u = 29*10.3 + 29*10.3

u = 2*29*10.3/(39.3)

u = 15.2 cm = distance of face from hubcap

Part B

we know that radius is curvature is double of focal length

1/f = 2/R

Also

1/f = (u - 10.3)/(u*10.3)

2/R = (u - 10.3)/(u*10.3)

Using value of u = 15.2

R = 2*15.2*10.3/(15.2 - 10.3)

R = 63.9 cm