Index of Refraction and Snell's Law in Optics

1. 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 = 1.0003  1 • (because density is low); • nwater = 1.333  4/3; nglass = 1.4-1.7; • 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. 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. 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.  = 30o What are values of  and ?  nair = 1   nglass = 1.5 • 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?  ndiamond = 2.4   

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 horizon atmosphere sun

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