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 29.0 cm in back of the hubcap. He then turns the hubcap over, keeping it the same distance from his face. He now sees an image of his face 10.4 cm in back of the hubcap.
(a) How far is his face from the hubcap?
cm
(b) What is the radius of curvature of the hubcap?
cm

I keep getting it wrong please help. Thank you

Explanation / Answer

image of his face back of the hubcap is at q1 = -29 cm then he turns the hubcap over, keeping it the same distance from his face. He now sees an image of his face at q2 = -10.6 cm in back of it. lens equation for first case 1/f =1/q+1/q1   where 'q' is the face distance from the hubcup                                         1/f = 1/q -1/29......................(1) lens equation for second case is -1/f =1/q+1/q2                                                    -1/f =1/q -1/10.6 ................(2)     (1)-(2) gives                              2/f = -1/29+1/10.6           = 0.05985686 focal length is f =33.41 cm _______________________________________________________________________ (a) from equation (1) 1/f =1/q+1/q1 1/33.41= 1/q -1/29 1/q =1/33.41+1/29 q =15.52 cm his face from the hubcap is q =15.52 cm ______________________________________________________________________ radius of curvature of the hubcap is R = 2*f ______________________________________________________________________ radius of curvature of the hubcap is R = 2*f                                                            = 2*(33.41)                                                             =68.82 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.