SOLUTION In this example, we will apply Snell's law to the refraction of light SET UP (gm) shows our sketch. We take the water as medium a and the air as medium b; "a : 1.33 and m, : 1.00. Note that 0,, is 300 , not 60.0"! across a waterair interface. Suppose you kneel beside the We use Snell's law to find 91,. fishpond in your backyard and look at one of the fish, You see it by sunlight that reflects off the fish and retracts at the waterair SOLVE From Snell's law, interface. If the light from the fish to your eye strikes the waterair ' . interface at an angle of 60.00 to the interface, what is the angle of "a 51319\" = \"b 5111 9% refraction of the ray in the air? . 0 , 9!; Sinil "a 5111 a sin'1 1.33 51:: 30 41.70. m, 1.00 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. V Part A - Practlce Problem: You are Spearfishing from a boat and eye a large bass swimming below. It is apparentlyat an angle of 38.0" from the normal. At what angle should you aim your spear? Express your answer in degrees. Figure 1 of 1 931 = measured from the normal Muest Answer Provide Feedback Afr (m, - 1.00) Wafer (ita - 1.33) Next > Light with a frequency of 5.70x1014 Hz travels in a block of glass that has an index of refraction of 1.52. v PartA What is the wavelength of the light in vacuum '2 Express your answer in nanometers. muss! Answer V PartB What is the wavelength of the light in the glass? Express your answer in nanometers. Muest Answer Provide Feedback Next )