Asked by Chirayu
A point source of light is 96.7 cm below the surface of a body of water. Find the diameter of the circle at the surface through which light emerges from the water. Water has an index of refraction of 1.33.
Answers
Answered by
drwls
There is a maximum angle that a ray from the source can have with the vertical, and not be totally reflected. That angle is called the critical angle, theta. It is giben by the equation
sin theta = 1/N
where N is the refractive index of water
Therefore theta = 48.7 degrees.
The radius of the circle through which light can escape the water surface, from a depth d below, is
R = d tan theta = 1.14 d
The diameter of the circle is 2R.
sin theta = 1/N
where N is the refractive index of water
Therefore theta = 48.7 degrees.
The radius of the circle through which light can escape the water surface, from a depth d below, is
R = d tan theta = 1.14 d
The diameter of the circle is 2R.
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