Question

#3 Suppose you have a fish tank made of a plastic polymer that has a mirror on the bottom. The fish tank is exactly half full of water. The fish tank is lit by a light coming from the upper, left hand corner, which you can approximate as a ray of light traveling at 45 degrees from the normal towards the water. For fun you put a mirror on the top of the tank as well. A) Draw the transmitted light ray as it reflects and refracts towards the right hand side of the tank. Show the angles and exactly where it strikes the other side of the tank. B) If you could control the index of refraction of the plastic polymer, what must it be so that there is Total Internal Reflection of the ray on the far side of the tank? Minns 45.0 0.500 m Water 13.5 m

Answer 1

Refractive index of water is 1.33. From Snell's law we get theta_r = sin^-1[eta_t sin theta_t/eta_r] = sin^-1[sin 45/1.33] theta_r = 32.12 degrees The horizontal distance travelled by light before first refraction is d_45 degrees = h/2 cos theta_t = 0.177 m The horizontal distance travelled by light between first refraction and reflection is d_32.12 degrees = h/2 sin theta_r = 0.133 m Now, the total distance travelled is 13.5 m thus we have 13.5 > 0.177x + 0.133y and the difference between x and y cannot be more than 1. So we get for x = 43 and y = 44 the distance covered as 13.463 m implies delta t = 0.037 m As the light is emerging out of the air we have the vertical distance travelled from the surface of water as h = delta d cos 32.12 degrees = 0.031 m The above diagram does not contain the exact number of reflections and refractions