Goal Apply the concept of total internalreflection. Problem (a) Find the critica

Question

Goal Apply the concept of total internalreflection.

Problem (a) Find the criticalangle for a water-air boundary if the index of refraction of wateris 1.33.

(b) Use the result of part (a) to predict what afish will see (Fig 22.28) if it looks up toward the water surfaceat angles of 42.8°, 48.8°, and 55.8°.

Strategy After finding the critical angle bysubstitution, use the fact that the path of a light ray isreversible: at a given angle, wherever a light beam can go is alsowhere a beam of light can come from, along the same path.

Solution

(a)   Find the critical angle for awater-air boundary.

(b)   Substitute into Equation 22.9 tofind the critical angle.

(c)   

I A beam of light shone toward the surface will becompletely reflected down toward the bottom of the pool. Reversingthe path, the fish sees a reflection of some object on thebottom.

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


IV Light from underwater is bent so that ittravels along the surface. This means that light following the samepath in reverse can reach the fish only by skimming along the watersurface before being refracted towards the fish's eye.

Suppose alayer of oil with n = 1.52 coats the surface of the water.What is the critical angle for total internal reflection for lighttraveling in the oil layer and encountering the oil-waterboundary?

Goal Apply the concept of total internalreflection.

Problem (a) Find the criticalangle for a water-air boundary if the index of refraction of wateris 1.33.

(b) Use the result of part (a) to predict what afish will see (Fig 22.28) if it looks up toward the water surfaceat angles of 42.8°, 48.8°, and 55.8°.

Strategy After finding the critical angle bysubstitution, use the fact that the path of a light ray isreversible: at a given angle, wherever a light beam can go is alsowhere a beam of light can come from, along the same path.

Solution

Figure 22.28 A fish looks upward toward thewater's surface.

(a)   Find the critical angle for awater-air boundary.

(b)   Substitute into Equation 22.9 tofind the critical angle.

(c)   

42.8° 2 DirectionIIIIIIIV


48.8° 3 DirectionIIIIIIIV


55.8° 4 DirectionIIIIIIIV

c = °

I A beam of light shone toward the surface will becompletely reflected down toward the bottom of the pool. Reversingthe path, the fish sees a reflection of some object on thebottom.

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


IV Light from underwater is bent so that ittravels along the surface. This means that light following the samepath in reverse can reach the fish only by skimming along the watersurface before being refracted towards the fish's eye.

Explanation / Answer

The angle of refraction for the side with lower index of refraction (water) will be 90 degrees.

The sin of 90 is 1.

So the equation to find the critical angle is sin(theta) = 1.33/1.52

sin^-1 (1.33/1.52)

= 61.045 degrees

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