c = ° (a) Find the critical angle for a water-air boundary if the index of refra

Question

c =  °

(a) Find the critical angle for a water-air boundary if the index of refraction of water is 1.33.

(b) Use the result of part (a) to predict what a fish will see (Fig 22.28) if it looks up toward the water surface at angles of 43.7°, 48.8°, and 57.3°.

Strategy After finding the critical angle by substitution, use the fact that the path of a light ray is reversible: at a given angle, wherever a light beam can go is also where a beam of light can come from, along the same path.
Figure 22.28 A fish looks upward toward the water's surface. Solution (a) Find the critical angle for a water-air boundary. Substitute into Equation 22.9 to find the critical angle.
c =  ° (b) Predict what the fish will see if it looks up toward the surface at angles of 44°, 49°, and 57°. 43.7° Direction I II III IV


48.8° Direction I II III IV


57.3° Direction I II III IV I A beam of light shone toward the surface will be completely reflected down toward the bottom of the pool. Reversing the path, the fish sees a reflection of some object on the bottom.

II A beam of light from underwater will be refracted at the surface and enter the air above. Because the path of a light ray is reversible (Snell's law works both going and coming), light from above can follow the same path and be perceived by the fish.

III A beam of light will travel through the air-water boundary without changing direction. The fish will see the object exactly as it would in the air.

IV Light from underwater is bent so that it travels along the surface. This means that light following the same path in reverse can reach the fish only by skimming along the water surface before being refracted towards the fish's eye.

Explanation / Answer

refractive index of water = 1.33

Using Snell law for critical angle,
(for critical angle, refraction angle is 90 deg )

ni sini = nr sinr

1.53 sin(iC) = 1.33 sin90

iC = 60.4 deg

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