0/5 points Previous Answers SerPSE10352OPO11 My Notes Ask Your Teacher A dedicat

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0/5 points Previous Answers SerPSE10352OPO11 My Notes Ask Your Teacher A dedicated sports car enthusiast polishes the inside and outside surfaces of a hubcap that is a thin section of a sphere. When she looks into one side of the hubcap, she sees an image of her face 50.0 cm in the back of the hubcap. She then flips the hubcap over and sees another image of her face 10.0 cm in the back of the hubcap (a) How far (in cm) is her face from the hubcap? 100 Does the near image (at 10.0 cm) form when using the concave or convex side of the hubcap? cm (b) What is the radius of curvature (in cm) of the hubcap? cm (c what If? If she wishes to project an image of herself on the garage all 2 60 m behind her, how far from her face mcm should she hold the ap cm (d) What is the magnification of the image on the wall? Need Help?Read It Watch It Submit Answer Save ProgressPractice Another Version

Explanation / Answer

a)

case 1 :

do = object distance

di = image distance = 50 cm

f = focal length = ?

using the equation

1/do + 1/di = 1/f

1/do - 1/50 = 1/f eq-1

case 2 :

do = object distance

di = image distance = 10 cm

f = focal length = ?

using the equation

1/do + 1/di = 1/f

1/do - 1/10 = - 1/f eq-2

adding eq-1 and eq-2

(1/do - 1/50) + (1/do - 1/10) = 0

do = 50/3

b)

using eq-1

1/do - 1/50 = 1/f  

3/50 - 1/50 = 1/f  

f = 25

c)

do = object distance

f = 25

di = image distance = - 2.60 m = 260 cm

using the equation

1/do - 1/di = 1/f  

1/do - 1/260 = 1/25  

do = 22.8 cm

d)

m = magnification = - di/do = - (- 260)/22.8 = 11.4

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