You are watching: A ray of light emitted beneath the surface of an unknown liquid with air above it

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This is College Physics Answers through Shaun Dychko. Light is being emitted by some light resource inside this liquid and light travels up and hits this interconfront between the liquid and also air and then has actually full internal reflection and also this is the smallest angle such that total internal reflection occurs and that makes it the critical angle. So we need to figure out what is the index of refraction of this product, what is n 1 in other words and also then looking up in the table <25.1>, we"ll figure out what this material can be. So we can number out what this angle α is making use of trigonomeattempt based upon the dimensions provided to us below and also then understanding that this dotted black line is a perpendicular to the interface, we deserve to say that Θ c then is 90 minus this angle α that we"ll uncover and also then understanding the crucial angle, we"ll figure out what n 1 is. Okay! So the critical angle is 90 minus α so let"s number out α. The tangent of α is the opposite leg of this best triangle separated by the adjacent leg so that"s the inverse tan then of 15.0 centimeters divided by 13.4 centimeters and also that provides an angle of 48.22452 degrees, maintaining several digits since this is an intermediate calculation. So I took the inverse tan of both sides here by the method to number out α. Okay! So that"s α and so then Θ c — the important angle— is 90 minus that α and that works out to 41.77548 levels. And then we usage this formula with the important angle and also we have the initially index of refraction the index of refraction of the initially material that the light is traveling with which is this oselection liquid multiplied by sin of whatever the crucial angle is amounts to the index of refractivity of the second tool, which is air in this case. So we settle for n 1 by dividing both sides by sin Θ c and also then n 1 then is n 2 over sin Θ c. So that"s 1.000 divided by sin of this angle that we uncovered and also this functions out to 1.501. And once we look up in table <25.1>, we look at liquids with an index of refraction of 1.501... we find it here and the material need to be Benzene. So this liquid is Benzene.

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