| n1 sin θ1= n2 sin θ2. |
where ni represents the index of refraction
in medium i, and θi
represents the angle the light makes with the normal in medium i.
| (a) | Since the light is traveling from a rarer region (lower n) to a denser region (higher n), it will bend toward the normal. |
| (b) | We will identify air as medium 1 and the fiber as medium
2. Thus, n1 = 1.00, n2
= 1.44, and θ/font>1
= 22o. Snell's Law then becomes
sin θ2 = (1.00/1.44) sin 22o = 0.260 θ2 = sin-1 (0.260) = 15o. |
| (c) | The path of the light is shown in the figure below.
![]() |
| (a) | Since the light is now traveling from the fiber into
air, we will call the fiber material 1 and air material 2. Thus,
n1
= 1.44, n2 = 1.00, and θ1
= 30o. Snell's Law then becomes
sin θ2 = (1.44/1.00) sin 30o = 1.44 (0.500) = 0.720 θ2 = sin-1 (0.720) = 46o. |
| (b) | Replacing the angle of incidence with 50o
gives
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The situation in part (b) is an
example of total internal reflection, discussed on the next content page.
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