# A scuba diver uses his waterproof flashlight to shine a beam of light so that it strikes the surface of the water at an angle of incidence θi. Use Snell’s law to find the angle of incidence that...

A scuba diver uses his waterproof flashlight to shine a beam of light so that it strikes the surface of the water at an angle of incidence θi. Use Snell’s law to find the angle of incidence that would give an angle of refraction for the refracted ray to be directed right along the surface, and show that θi is the same as the critical angle for total internal reflection.

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Snell's Law states

n_1sin\theta_1=n_2sin\theta_2

where n is the index of refraction And \theta is the angle of incidence or refraction.

For total internal reflection to occur, the beam must go from a higher index to a lower index of refraction.

If a refracted ray is directed along the surface, the angle of refraction is 90. Let n_1 be the water and n_2 be air. So the equation becomes

n_1sin\theta_i=n_2sin(90)

n_1sin\theta_i=n_2

:. \theta_i = sin^(-1)(n_2/n_1)

n_1 gt n_2

If we look at this, we see the two materials dictate this maximum angle of incidence.

Now let's use some numbers to see if this is true. n for air is 1.0. n for water is 1.3. thus this angle becomes 50.3.

But what if we use a larger angle of incidence? If this is the critical angle , it should be apparant in the result. Let's use 60.

n_1sin\theta_1=n_2sin(\theta_r)

:.\theta_r=sin^(-1)(1.3/1.0sin(60))

Trying to solve that produces no solution. Which means there is not a refracted ray. So the maximum angle of refraction is 90 and the angle of incidence that causes this is the largest angle possible before total internal reflection.