1. (n depends on wavelength, type of atom and inversely with
Index of refraction:
n  c/v
(n depends on wavelength, type
of atom and inversely with
density of atoms)
Snell’s law:
n2 sin 2 = n1 sin1
V1 = c/n1
Notes:
v  c, so n  1.
[nvacuum = 1; nair =  1
(because density is low);
nwater =  4/3; nglass = ;
ndiamond = 2.42]
Angles defined with respect to normal and are between 0o and 90o.
If normal incidence: 1 = 0, so 2 = 0.
If 1 increases, so does 2.
If n2 increases, 2 decreases !
V2 = c/n2
Ray passing from small n to large n bends toward normal (away from surface).
Ray passing from large n to small n bends away from normal (toward surface).

2. (n depends on wavelength, type of atom and inversely with
Index of refraction:
n  c/v
(n depends on wavelength, type
of atom and inversely with
density of atoms)
Snell’s law:
n2 sin 2 = n1 sin1
Note: frequency of light (# cycles/second) doesn’t change in passing from 12. Since  = v/f, the wavelength changes:
2 / 1 = v2 / v1 = n1/ n2

3. Since normals are parallel, the incoming and outgoing rays will be
If two surfaces are parallel (e.g. window), the incoming and outgoing rays will be parallel.
Proof:
Since normals are parallel, the incoming and outgoing rays will be
parallel if 1 = 3.
Because normals are parallel, 2’ = 2 (they are alternate angles).
n1 sin 1 = n2 sin 2 = n2 sin 2’
n2 sin 2’ = n1 sin 3
n1 sin1 = n1 sin 3
3 = 1
qed
d will increase if the thickness of n2 material increases.
d will increase if 1 increases (e.g. d = 0 if normal incidence (1 = 0)).
2’

4. nair = 1   
nglass = 1.5
ndiamond = 2.4
What are values of  and ?
 = 30o
Note:
Both angles labeled  are equal – why?
Both angles labeled  are equal – why?
Both angles labeled  are equal, i.e. the outgoing beam will be parallel to the incoming beam – why?  

5. nd sin = ng sin
sin = ng sin / nd
sin = (1.5) (0.333)/2.4 = 0.208
 = 12.0o 

6. 
n, , and 1 will be given.
Find change in direction = deviation = 

7. 180o - 
1 - 2
4 - 3 
Deviation :
180o -  = 180o - (1 - 2) - (4 - 3)
 = 1 - 2 + 4 - 3

8. 90o - 2
90o - 3
 = 1 + 4 - 2 - 3; n, , and 1 will be given.
n sin(2) = sin(1)
sin(2) = sin(1) / n [use to find 2]
+ (90o- 2) + (90o- 3) = 180o
3 =  - 2 [use to find 3]
sin(4) = n sin(3) [use to find 4]

9. Example: n = 1.4,  = 45o,
1 = 60o
sin(2) = 0.866/1.4 = 0.619
2 = 38.2o
3 = 45o – 38.2o = 6.8o
sin(4) = 1.4*0.12 = 0.166
4 = 9.5o
 = 1 + 4 - 2 - 3
 = 1 + 4 - 2 - 3 = 24.5o

10. At sunrise & sunset, can see the sun slightly below the horizon:
continuous variation of n:
earth
atmosphere
sun
horizon

11. “Green Flash” at sunrise/sunset is due to refraction, dispersion and scattering:
Dispersion: Short wavelengths (green, blue, violet) refract (bend) more than long wavelengths (red, orange, yellow).
Scattering: Shortest wavelengths (blue, violet) scatter (away from sun) more than green.

12. Dispersion n1 sin 1 = n2 sin 2 n1 n2
Typically, index of refraction varies (slightly) with frequency (and wavelength) of light, with n decreasing with decreasing f (increasing vac = c/f).
Therefore, shorter wavelengths (e.g. violet) will change direction more than longer wavelengths (e.g. red) when passing from vacuum (or air) into/out of material.
vac

17. TIR: Basis of transmission in optical fibers (communication cables, laparascopy, lamps): can feed signal around bends (if not too sharp)
communication fiber bundle