A dedicated sports car enthusiast polishes the inside and outside surfaces of a

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 10.8 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 31.8 cm in back of the hubcap.

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 10.8 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 31.8 cm in back of the hubcap. How far is his face from the hubcap?

Explanation / Answer

first situation: object distance is "p" image distance is q = -30 so

use 1/f = 1/o + 1/q

Or

1/f = 1/o - 1/10.8

Now... do the same thing for the second situation. The hubcap is convex, so f is the same number but negative and...

-1/f = 1/o - 1/31.8

Add the two equations:

0 = 2/o - 1/10.8- 1/31.8

0 = 2/o -0.12403913347

2/o = 0.12403913347

o = 16.123943662 cm is the first answer

Then...

1/f = 1/16.123943662 - 1/10.8

f = 32.7085714286cm

so radius of curvature = 2f = 65.4171428571 cm

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