3. Define and apply the concept of the index of refraction and discuss its effect on the velocity and wavelength of light.Define and apply the concept of the index of refraction and discuss its effect on the velocity and wavelength of light. Determine the changes in velocity and/or wavelength of light after refraction.Determine the changes in velocity and/or wavelength of light after refraction. Apply Snell’s law to the solution of problems involving the refraction of light.Apply Snell’s law to the solution of problems involving the refraction of light. Define and apply the concepts of total internal reflection and the critical angle of incidence.Define and apply the concepts of total internal reflection and the critical angle of incidence.

4. Water Air Refraction is the bending of light as it passes from one medium into another. refraction N ww AA Note: the angle of incidence  A in air and the angle of refraction  A in water are each measured with the normal N. The incident and refracted rays lie in the same plane and are reversible.

5. Water Air Water Air The eye, believing that light travels in straight lines, sees objects closer to the surface due to refraction. Such distortions are common.

6. The index of refraction for a material is the ratio of the velocity of light in a vacuum (3 x 10 8 m/s) to the velocity through the material. c v Index of refraction Examples: Air n= 1; glass n = 1.5; Water n = 1.33

7. v air = c v G = 2 x 10 8 m/s Glass Air For glass: n = 1.50 If the medium were water: n W = 1.33. Then you should show that the velocity in water would be reduced from c to 2.26 x 10 8 m/s.

8. Sand Pavement Air Glass Light bends into glass then returns along original path much as a rolling axle would when encountering a strip of mud. 3 x 10 8 m/s 2 x 10 8 m/s v s < v p

9. Medium 1 Medium 2      Consider two light rays. Velocities are v 1 in medium 1 and v 2 in med. 2. Segment R is common hypotenuse to two rgt. triangles. Verify shown angles from geometry. v1v1 v1tv1t v2v2 v2tv2t 1111 R

10. 11 22 Medium 1 Medium 2 The ratio of the sine of the angle of incidence  1 to the sine of the angle of refraction  2 is equal to the ratio of the incident velocity v 1 to the refracted velocity v 2. Snell’s Law: v1v1 v2v2

11. The incident angle is:  A = 90 0 – 30 0 = 60 0  W = 35.3 0 Air H2OH2O 30 0 WW AAAA

12. Another form of Snell’s law can be derived from the definition of the index of refraction: Snell’s law for velocities and indices: Medium 1 11 22 Medium 2

13. Since the indices of refraction for many common substances are usually available, Snell’s law is often written in the following manner: The product of the index of refraction and the sine of the angle is the same in the refracted medium as for the incident medium.

14. Glass Air Air n=1.5 First find  G inside glass: 50 0 GG   G = 30.7 0 From geometry, note angle  G same for next interface. GG Apply to each interface:  e = 50 0 Same as entrance angle!

15. The energy of light is determined by the frequency of the EM waves, which remains constant as light passes into and out of a medium. (Recall v = f.) Glass Air n=1 n=1.5 A G f A = f G G  A

16. Refraction is affected by the index of refraction, the velocity, and the wavelength. In general: All the ratios are equal. It is helpful to recognize that only the index n differs in the ratio order. Snell’s Law:

17. n G = 1.5; A = 632 nm Note that the light, if seen inside the glass, would be blue. Of course it still appears red because it returns to air before striking the eye. Glass Air Air n=1.5  GG  GG

18. Red Orange Yellow Green Blue Indigo Violet Dispersion is the separation of white light into its various spectral components. The colors are refracted at different angles due to the different indexes of refraction.

19. Water Air light The critical angle  c is the limiting angle of incidence in a denser medium that results in an angle of refraction equal to 90 0. When light passes at an angle from a medium of higher index to one of lower index, the emerging ray bends away from the normal. When the angle reaches a certain maximum, it will be reflected internally. i = r Critical angle cccc 90 0

20. For critical angle,  A = 90 0 n A = 1.0; n W = 1.33 Critical angle:  c = 48.8 0 Water Air cccc 90 0 Critical angle In general, for media where n 1 > n 2 we find that:

21. Snell’s Law: Refraction is affected by the index of refraction, the velocity, and the wavelength. In general: c = 3 x 10 8 m/s v Index of refraction Medium n

22. The critical angle  c is the limiting angle of incidence in a denser medium that results in an angle of refraction equal to 90 0. In general, for media where n 1 > n 2 we find that: n 1 > n 2 cccc 90 0 Critical angle n1n1 n2n2