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 10.2 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 27.5 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

Mirror equation is given by:

1/u + 1/v = -1/f

for first case when

v = image distance = -10.2 cm

1/u - 1/10.2 = -1/f

So if we flip the hubcap, now mirror equation will be

1/u + 1/v = 1/f

in this case v = -27.5 cm

1/u - 1/27.5 = 1/f

1/27.5 - 1/u = -1/f

from both bold equations

1/u - 1/10.2 = 1/27.5 - 1/u

2/u = 1/27.5 + 1/10.2

2/u = 0.1344

u = 2/0.1344 = 14.88 cm

His face is 14.88 cm from the hubcap

Now Part B

using above values:

1/u - 1/10.2 = -1/f

1/14.88 - 1/10.2 = -1/f

-f = 14.88*10.2/(10.2 - 14.88)

-f = -32.43 cm

f = 32.43 cm

Radius of curvature = 2*f

R = 2*32.43 = 64.86 cm

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