A search light is being used at the edge of a tank to find a lost key at the bottom of the tank. The tank is filled to the top with a fluid for which the index of refraction is 1.68. The key is 8.176 m below the surface and 6.84 m from the edge of the tank. At what angle to the normal will the searchlight have to point in order to locate the key if the light enters the water 1 m from the edge of the tank?
n1 sin theta1 = n1 sin theta2
The Attempt at a Solution
inverse sin of 6.84/8.176
inverse sign of 7.84/8.176
and then i added those two together..
Im totally lost and dont understand what to do . Any help will be much appreciated =]