A dedicated sports-car enthusiast polishes the inside and outsidesurfaces of a h

Question

A dedicated sports-car enthusiast polishes the inside and outsidesurfaces of a hubcap that is a section of a sphere. When he looksinto one side of the hubcap, he sees an image of his face29.1 cm behind the hubcap. He thenturns the hubcap over and sees another image of his face8.0 cm behind the hubcap.
(a) How far is his face from the hubcap?
cm
(b) What is the radius of curvature for the hubcap?
cm
(c) What is the magnification for each image?
First Image = Second Image = (a) How far is his face from the hubcap?
cm
(b) What is the radius of curvature for the hubcap?
cm
(c) What is the magnification for each image?
First Image = Second Image = First Image = Second Image =

Explanation / Answer

   let us take that the magnitude ofthe radius of curvature |R| is same for both sides of thehubcap
   for the convex side R = - |R|
   for the concave side R = + |R|
   the object distance p is positive (real object) andhas the same value in both the cases
   the virtual image distance as q = - |q|
   so from the mirror equation we get
   for the convex side as
   (1 / - |q|) = (2 / - |R|) - (1 / p)
   |q| = |R| p / |R| + 2 p .......... (1)
   for the concave side as
   (1 / - |q|) = (2 / |R|) - (1 / p)
   |q| = |R| p / |R| - 2 p .......... (2)
   on comparing equations (1) and (2) we can see that thesmaller magnitude image distance
   |q| = 11.0 cm occurs with the convex side of themirror so we get
   (1 / - 11.0 cm) = (2 / - |R|) - (1 / p)........(3)
   for the concave side |q| = 28.0 cm gives
   (1 / - 29.0 cm) = (2 / |R|) - (1 / p) ........(4)
(a)
   adding equations (3) and (4) we get
   p = ....... cm
(b)
   subtracting (3) from (4) we get
   |R| = ........ cm

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