Re: What equation relates angle of incidence to % reflected and refracted.

There are many sources which provide the answer to your question, such as the book "Optics" by Eugene Hecht, or the book "Fundamentals of Optics" by Jenkins and White, but the book I like the most is "The Feynman Lectures on Physics". The appropriate chapter is 33, and the section is 33.6.

See the figure here.

The left part of the figure shows EM radiation that is polarized perpendicular to the plane of the page (thus the dots), while the right side shows EM polarized parallel to the plane of the page. The letters I, a, b, A, and B are the amplitudes of the EM radiation. As usual the letters i and r represent the angles of incidence (and reflection, of course!) and refraction. The material on the left of the interface has refractive index n1 and on the right of the interface has refractive index n2. As usual,
n1 sin( i ) = n2 sin( r )

I will not go through the derivation, as that is given in the reference. The results are
b = - sin( i - r ) / sin( i + r )
B = - tan( i - r) / tan( i + r )

where, I say again, the b and B are amplitudes. To obtain reflection coefficients (which are based on energy, not amplitude of EM) these amplitudes need to be squared. Remember that the intensity is the square of the amplitude of EM.

You will have to determine the refractive index of water at the wavelength of interest, and use the first equation to determine r, and then use the other equations to determine a, b, A, and B.

If we assume the incident light has amplitude 1 (which we can do by normalizing everything to the incident light!) then
a = sqrt( 1 - b^2 )
A = sqrt( 1 - B^2 )

where "sqrt" is the square-root function and " ^ " denotes exponent, and, again, the intensities on which the reflection and refraction coefficients are calculated are the square of the amplitudes.

In your question you ask specifically about light impinging on water at 53 degrees from the normal. If we assume n2 = 1.33 (this is approximate) then
i = 53
r = 36.90
b = -0.277246 b^2 = 0.076865
a = 0.960799 a^2 = 0.923135
B = -0.00048284 B^2 = 0.000000233
A = 0.9999999 A^2 = 0.999999767

The example you asked for is obviously very close to Brewster's angle, at which the part of the beam polarized parallel to the plane of the page has no reflected amplitude. If n2 = 1.33 the actual angles for the Brewster condition to hold are
i = arctan ( n ) = 53.0612 degrees
which gives r = 36.9388 degrees

Have fun!!

John Link
MadSci Scientist