A scuba diver shines a flashlight from beneath the surface of water (n = 1.33 such that the light strikes the water-air boundary with an angle of incidence of 37°. at what angle is the beam refracted?

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23-05-2017
Physics

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A scuba diver shines a flashlight from beneath the surface of water (n = 1.33 such that the light strikes the water-air boundary with an angle of incidence of 37°. at what angle is the beam refracted?

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andriansp andriansp · Answer 1 · 2017-05-30T15:05:57+02:00

The beam is refracted at 53.17-degree angle (asin 0.800414 = 53.17-degree). This problem can be solved by using the Snell's Law which described the refracted beam angle which traveled through a different media, in this case, through water and air. The formula of Snell's Law is stated as n1 sin θ1= n2 sin θ2. In this formula n1 is the refraction index of medium 1, θ1 is the normal light angle in the medium 1, n2 is the refraction index of medium 2, and θ2 is the normal light angle in the medium 2.