A trigonometric dog enjoys watching goldfish from the top of the fish tank. If the goldfish appears to be at an angle of 28.0deg from the normal to the surface of the water, what will its true...
A trigonometric dog enjoys watching goldfish from the top of the fish tank. If the goldfish appears to be at an angle of 28.0deg from the normal to the surface of the water, what will its true angle from the normal be?
According to Snell's Law, a wavefront of light (represented by a ray drawn in the direction perpendicular to the wavefront and in the direction of energy travel) will be bent as it moves from a media of one optical density into a media of differing optical density at any angle other than perpedicular to the surface. The relationship Snell developed is given by
N1Sin(theta1) = N2Sin(theta2)
Where N1 is the index of refraction for the media from which the ray originates, and theta1 is the angle it makes to the normal at the surface interface; N2 is the index of refraction for the media into which the ray travels, and theta2 is the angle the ray makes to the normal as it enters the new media.
For this example, the ray is originating at the fish in water which has an index of refraction of 1.33 and is striking the surface at some unknown angle theta1 to the normal. The ray leaves the water and enters air which has an index of refraction of 1.003 at an angle of 28.0 degrees to the normal.
Solving Snells equation for theta1 allows us to determine the unknown angle:
theta1 = invSin[(1.003/1.33)*sin(28.0)]
theta1 = invSin(0.3540) = 20.7deg.
The equation is n1(Sine of angle1)= n2(Sine of angle 2) so the index of refraction of air is 1 and of water is 1.333. So it would be this 1(Sin 28 degrees)= 1.333(Sin of angle 2) you would find sin of 28 divide it by 1.333 and find the inverse of sine. So it would come to equal 20.62 degrees.