1. Total Internal Reflection
When light passes from a medium of larger refractive index into one of smaller refractive index (i.e., water to air), the refracted ray bends away from the normal

2. Total Internal Reflection
As the angle of incidence increases, so does the angle of refraction.

3. Total Internal Reflection
At the critical angle, Ɵc, the angle of refraction is 90°
The refracted ray points along the surface

4. Total Internal Reflection
When the angle of incidence is greater than the critical angle, there is no refracted light
All the light is reflected back into the medium from which it came.

5. Total Internal Reflection
TIR only occurs when light is traveling from a higher-index medium into a lower index medium. (Never in the opposite direction.)

6. Total Internal Reflection – Critical Angle
If n1 > n2
n1sinƟc = n2sinƟ2
sinƟc = n2sin90 n1 = n2 n1

7. Total Internal Reflection – Critical Angle
Example: find the critical angle for light traveling from water (n1=1.33) to air (n2=1.00)
sinƟc = n2 n1 =1/1.33
Ɵc = sin-1(1/1.33) = 48.8°

8. Total Internal Reflection – Critical Angle for water/air
Ɵc = sin-1(1/1.33) = 48.8°
For angles greater than 48.8°, the air water interface acts like a mirror

9. Example 5 p. 803
A beam of light is propagating through diamond (n1=2.42) and strikes a diamond-air interface at an angle of incidence of 28°.
A) will part of the beam enter the air (n2=1.00) or will the beam be totally reflected at the interface?
B) repeat part A, but assume
that the diamond is totally
surrounded by water (n2=1.33)
instead of air.

10. Example 5 p. 803
A beam of light is propagating through diamond (n1=2.42) and strikes a diamond-air interface at an angle of incidence of 28°.
B) Find the critical angle for diamond/air.
sinƟc = n2 n1 =1.00/2.42
Ɵc = sin-1(1.00/2.42) = 24.4°
28° > Ɵc so no light is refracted
Only TIR at the interface.

11. Example 5 p. 803
A beam of light is propagating through diamond (n1=2.42) and strikes a diamond-air interface at an angle of incidence of 28°.
B) If the angle is greater than the critical angle, there will be total internal reflection.
sinƟc = n2 n1 =1.33/2.42
Ɵc = sin-1(1.33/2.42) = 33.3°
28° < Ɵc so light is refracted
and reflected at the interface.

12. Applications
Many optical instruments such as binoculars, periscopes, telescopes, etc. use glass prisms and TIR to turn a beam of light through 90° or 180°

13. Applications – Fiber optics
Optical fibers (hair-thin threads of glass or plastic) “pipe” light from one place to another.
Much less loss of signal and much higher capacity than metal wires, so it is very useful in telecommunications.

14. Example 7: optical fiber
An optical fiber consists of a core made of flint glass (nflint=1.667) surrounded by cladding made of crown glass (ncrown=1.523). A ray of light in air enters the fiber at angle Ɵ1. What is Ɵ1 if this light also strikes the core-cladding interface at an angle that just barely exceeds the critical angle?

15. Assignment
Focus p. 834 #6,7 Problems p. 836 #26-31