You sight along the rim of a glass with vertical sides so that the top rim is lined up with the opposite edge of the bottom (see the figure (a)). The glass is a thin-walled hollow cylinder 16.0 cm high with a top and bottom of the glass diameter of 8.0 cm. While you keep your eye in the same position, a friend fills the glass with a transparent liquid, and you then see a dime that is lying at the center of the bottom of the glass (see the figure(b)).
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What is the index of refraction of the liquid?
http://session.masteringphysics.com/prob…
What is the index of refraction of the liquid?
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Since we are not given any information about velocities, its fairly obvious that we have to use Snell's law which says
Refractive index = sin (angle of incidence) / sin (angle of refraction)
Recall that these two angles are measured from the normal at the point of incidence, which, in this case, is the point where the ray just passes over the rim of the glass.
The first diagram allows us to calulate the angle of incidence and the second, the angle of refraction.
So,
angle of incidence = tan^-1 (8/16) = 26.6 degrees
Sin (26.6) degrees = 0.447
angle of refraction = tan^-1 (4/16) = 14.0 degrees
Sin (14) degrees = 0.24
Refractive index of liquid = 0.447 / 0.24 = 1.86 to two decimal places.
Refractive index = sin (angle of incidence) / sin (angle of refraction)
Recall that these two angles are measured from the normal at the point of incidence, which, in this case, is the point where the ray just passes over the rim of the glass.
The first diagram allows us to calulate the angle of incidence and the second, the angle of refraction.
So,
angle of incidence = tan^-1 (8/16) = 26.6 degrees
Sin (26.6) degrees = 0.447
angle of refraction = tan^-1 (4/16) = 14.0 degrees
Sin (14) degrees = 0.24
Refractive index of liquid = 0.447 / 0.24 = 1.86 to two decimal places.