Use the worked example above to help you solve this problem. A coin 1.87 cm in d

Question

Use the worked example above to help you solve this problem. A coin 1.87 cm in diameter is imbedded in a solid glass ball of radius r = 33.7 cm (see figure). The index of refraction of the ball is 1.50, and the coin is d = 22.5 cm from the surface. (a) Find the position of the image of the coin. (b) Find the height of the coin's image. A coin is embedded 22.5 cm from the surface of a similar ball of transparent substance having radius 33.7 cm and unknown composition. The coin's image is virtual and located 16.9 cm from the surface. (a) Find the index of refraction of the substance. (b) Find the magnification.

Explanation / Answer

practice question:

part a:

as the rays are moving from a medium of high index of refraction to a medium of low index of refraction, the image formed will be virtual

corresponding equation is given as:

(n1/u)+(n2/v)=(n2-n1)/R

where n1=index of refraction for the first medium=1.5

n2=index of refraction for the second medium=1

u=distance of the coin from the curved surface=22.5 cm

v=image distance

R=radius of curvature=-33.7 cm

solving the equation , we get v=-19.29 cm

part b:

magnification is given by M=-n1*v/(n2*u)=-1.5*(-19.29)/(1*22.5)

=1.286

so height of coin's image is 1.286*1.87=2.4048 cm

exercise:

part a:

let index of refraction of the substance consisting the ball is n1.

index of refraction of second medium=n2=1

radius=R=-33.7 cm

object distance=u=22.5 cm

image distance v=-16.9 cm

using the above formula:

(n1/22.5)-(1/16.9)=(1-n1)/(-33.7)

==>n1=1.997

part b:
magnification=-n1*v/(n2*u)=-1.997*(-16.9)/(1*22.5)

=1.5

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