CK-12-Physics - Intermediate

20.3. Index of Refraction http://www.ck12.org

Total internal reflection

So far we have discussed light passing into media with a greater index of refraction. Let us consider what happens
if we have a light source under water with the light passing into the air, as inFigure20.16.

FIGURE 20.16

We see there is a point at which the light is refracted at 90◦from the normal (the red normal and black rays). The
angle of incidence at which this occurs is called the critical angle and it is labeledθcin theFigure20.16. Rays
incident to the normal that are larger than the critical angle do not pass into the air. They are reflected back into the
water as shown by the ray labeled TIR in the figure. We call this condition total internal reflection (TIR).

A critical angle will always exist when light passes from a medium with a greater index of refraction to one with a
smaller index of refraction.

We can use Snell’s Law to determine the critical angle in terms of the indices of refraction of the two media involved

n 1 sinθ 1 =n 2 sinθ 2 →n 1 sinθc=n 2 sin 90◦=n 2 →

n 1 sinθc=n 2 →sinθc=

n 2

n 1

→θc=sin−^1

n 2

n 1

;where n 1 >n 2

Illustrative Example 21.3.1

a. What is the critical angle for the light source inFigure20.14?

Solution:

Using the equation derived above, we have

θc=sin−^1 nn^21 withn 1 = 1 .33 andn 2 = 1. 00

θc=sin−^111 ..^0033 = 48. 8 ◦

b. Any light that is incident to the normal with an angle greater than 48. 8 ◦will be internally reflected back into the
water as demonstrated inFigure20.16. An observer above the water will, therefore, see a circle of light.

If the light source inFigure20.16 is positioned 3 meters below the water surface, what is the radius of the circle of
light seen?

Solution:

A diagram of the situation is shown below.

We can see from the diagram that

tanθc=r 3 →r=3 tanθc=3 tan 48. 8 ◦= 3. 43 m