# The index of refraction of a transparent liquid

(similar to water but with a different index of refraction) is 1.39. A flashlight held under the
transparent liquid shines out of the transparent liquid in a swimming pool. This beam of light exiting the surface of the transparent liquid makes an angle of �a = 39 degrees with respect
to the vertical.

At what angle (with respect to the vertical)is the flashlight being held under transparent liquid?

The flashlight is slowly turned away from the vertical direction.
At what angle will the beam no longer be visible above the surface of the pool? Answer in units of degrees.

Question ID
538971

Created
April 29, 2011 2:27am UTC

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1. We can use Snell's Law: n1 * sin(theta1) = n2 * sin(theta2), where n1 and n2 are indices of refraction, and theta1 and theta2 are the incident and refracted angles, respectively.

For the first part:
n1 = 1.39 (from the problem statement)
n2 = 1 (assuming the air's index of refraction)
theta2 = 39 degrees

Solving for theta1:
1.39 * sin(theta1) = 1 * sin(39) = 0.629
theta1 = arcsin(0.629 / 1.39) = 27.35 degrees (approximately)
So the flashlight is being held at approximately 27.35 degrees with respect to the vertical under the transparent liquid.

For the second part, we want to find the critical angle, the angle at which light will no longer be visible above the surface of the pool. This occurs when the light exits at an angle of 90 degrees with respect to the vertical. So:
n1 * sin(theta1_critical) = n2 * sin(90)
1.39 * sin(theta1_critical) = 1
theta1_critical = arcsin(1 / 1.39) = 46.91 degrees (approximately)

So the beam of light will no longer be visible above the surface of the pool if the flashlight is held at an angle greater than 46.91 degrees (approximately) with respect to the vertical.