SOLUTION SET UP (Figure 1) shows our sketch. We take the water as medium a and the air as medium b; na = 1.33 and nb = 1.00. Note that a is 30.0 , not 60.0 ! We In this example, we will apply Snell's law to the refraction of light use Snell's law to find Ob. across a water-air interface. Suppose you kneel beside the fishpond in your backyard and look at one of the fish. You see it SOLVE From Snell's law, by sunlight that reflects off the fish and refracts at the water-air interface. If the light from the fish to your eye strikes the water-air na sin Ga = nb sin Ob, interface at an angle of 60.0 to the interface, what is the angle of efraction of the ray in the air? 06 = sin-1 (na sin a ) _ sin-1 1.33 sin 30 no 1.00 = 41.70. REFLECT Because the difference in refractive index between water and air is substantial, the actual position of the fish is quite different from its apparent position to you. Part A - Practice Problem: Figure 1 of 1 You are spearfishing from a boat and eye a large bass swimming below. It is apparently at an angle of 38.0 from the normal. At what angle should you aim your spear? Express your answer in degrees. ? Air (nb = 1.00) measured from the normal 60 Water (na = 1.33) Submit Previous Answers Request Answer 130 A X Incorrect; Try Again; 3 attempts remaining