2. Optics ISAT 241 Analytical Methods III Fall 2003 David J. Lawrence

3. Light as a Wave  Wave    motion (propagation) of a disturbance. u Waves transport energy. u Mechanical waves u sound, earthquakes, and ocean waves u require a “medium” that can be disturbed. u Light and radio waves are examples of “Electromagnetic waves”. u Disturbance is variations in electric (E) and magnetic (B) fields. u Since E and B can exist in a vacuum, electromagnetic waves do not require a medium.

4. For a “Harmonic Wave” (sinusoidal wave) Wave speed = frequency x wavelength v  f For light and other “Electromagnetic Waves”, in a vacuum: v = c = 2.9979  10 8 m/s In air: v air = 2.9970  10 8 m/s In glass: v glass = 2.0  10 8 m/s ~ 1 nanosecond = 1 ns = 1 x 10 -9 s = one billionth of a second In 1 ns, light travels a distance of: d = (3  10 8 m/s) (1  10 -9 s) = 0.3 m = 1 ft ~

5. Speed and Wavelength of Light in Several Media (for light of frequency = 5.09 x 10 14 Hz) ‘Yellow’ v = f  Medium Speed of Light Wavelength of Light vacuum* 2.9979 x 10 8  c 589 air 2.9970 x 10 8 589 water 2.25 x 10 8 442 glass (crown) 1.97 x 10 8 387 diamond 1.24 x 10 8 243 (m/s) (nm) * Light travels at its maximum speed in a vacuum

6. For the visible portion of the spectrum, the wavelength in vacuum (or in air) ranges from:  nm to  nm  Å to  Å Color f (Hz)   (nm) red 630-760 4.5 x 10 14 orange 590-630 4.9 x 10 14 yellow 560-590 5.2 x 10 14 green 500-560 5.7 x 10 14 blue 450-500 6.3 x 10 14 violet 380-450 7.1 x 10 14 These are average values!!

7. Light as a Wave Rays Wavefronts A “plane wave” propagating to the right.

8. The Ray Approximation Wavefronts A plane wave propagating to the right. Rays “Rays” are straight lines perpendicular to the wavefronts. They point in the direction of motion of the wave.

9. Serway & Jewett, Principles of Physics Figure 25.1

10. Serway & Jewett, Principles of Physics Figure 25.2

11. Serway & Jewett, Principles of Physics Figure 25.3

12. Specular Reflection = Reflection from a Smooth Surface Law of Reflection “Incident Ray”   “Reflected Ray” “normal” The incident ray, the reflected ray, and the normal, all lie in the same plane Angle of reflection = Angle of incidence     

13. Serway & Jewett, Principles of Physics Figure 25.5

14. Consider MRDs marching from a football field into a swamp... Swamp v1v1 Field v2v2

15. The Index of Refraction, n, of a medium is defined as the ratio: speed of light in a vacuum c n  speed of light in the medium = v (m/s) vacuum 2.9979 x 10 8 589 1 air 2.9970 x 10 8 589 1.000293 water 2.25 x 10 8 4421.333 glass (crown) 1.97 x 10 8 387 1.52 diamond 1.24 x 10 8 243 2.419 Medium Speed of Light Wavelength of Light (nm) Index of Refraction, n n > 1 n is dimensionless These values apply for light of frequency = 5.09 x 10 14 Hz ( ‘Yellow’)

16. What happens when light travels from one medium into another? The frequency does not change The speed and wavelength do change E.g., when light goes from air into glass rays Air = #1 Glass = #2 wavefronts speed in vacuum wavelength in vacuum

17. Serway & Jewett, Principles of Physics Figure 25.11

18. Field Swamp What if columns of MRD’s enter the swamp at an angle? v1v1 v2v2

19. Refraction of Light “Reflected Ray” Air Glass “Incident Ray” “Refracted Ray” “normal” n1n1 n2n2 All rays and the normal lie in the same plane. The “Refracted Ray” is bent towards the normal. Law of Refraction - Snell’s Law n 2 > n 1      n 1 sin   = n 2 sin  

20. Serway & Jewett, Principles of Physics Figure 25.8

22. Refraction of Light Air “Incident Ray” “Reflected Ray” “Refracted Ray” Glass “normal” All rays and the normal lie in the same plane. The “Refracted Ray” is bent away from the normal. Law of Refraction - Snell’s Law       n 1 sin   = n 2 sin   n 2 < n 1

25. Dispersion The speed of light in any material depends on the frequency (or wavelength, or color) of the light. n depends on wavelength. Light of different wavelengths will be refracted or bent at different angles when it enters a material. Therefore: The material is said to exhibit “Dispersion” “Normal” “White light”

31. Total Internal Reflection air n 2 = 1.00 n 1 = 1.50 glass 0 %       c the “critical angle” 100 % Snell’s Law : n 1 sin  1 = n 2 sin  2 For n 1 > n 2, when    c,  2 , and Snell’s Law gives n 1 sin  c = n 2 sin  n 2

32. Total Internal Reflection air n 2 = 1.00 n 1 = 1.50 glass 0 %       c the “critical angle” 100 %

33. Total Internal Reflection (cont.) Total internal reflection occurs only when light attempts to go from a medium of higher index of refraction to a medium of lower index of refraction. For this example, when light attempts to go from glass with n 1 = 1.50 into air with n 2 = 1,  c = sin -1 ( ) = sin -1 ( )  c = 41.8° n2n2 n1n1 1 1.50 For  1 >  c, the beam is entirely reflected at the boundary.

37. Fiber Optics end view n1n1 n2n2 Cladding Core cc n1n1 n2n2 typically  < 8 ° Light Ray

38. Fiber Optics (cont.) Light must fall inside this angle to be guided in the fiber core. 2 × Acceptance Angle 2  accept Acceptance Angle,  accept air n air = 1 Core Cladding n2n2 n2n2 n1n1 cc  accept = sin -1 [ n 1 2 - n 2 2 ]Numerical Aperture = NA = sin  accept = n 1 2 - n 2 2

39. Not all of the light that gets into one end of an optical fiber gets out the other end. P in P out Total Fiber Transmission Loss (dB) = Total Fiber Attenuation (dB) = = - 10 log 10 [ ] light power exiting the fiber light power coupled into the fiber a = - 10 log 10 [ ] P out P in a = 10 log 10 [ ] P in P out What if the length of the fiber is l ?

40. Fiber Loss per Unit Length (dB/km) = Fiber Attenuation per Unit Length (dB/km) fiber length in km 10 l = log 10 [ ] P out P in 10 l = log 10 [ ] P in P out a l Total Fiber Loss (dB) = Fiber Loss per Unit Length (dB/km) × l (km) Accounts for transmission loss in the fiber, NOT the losses encountered in getting the light into or out of the fiber (coupling losses).

41. Coupling Loss (in decibels) Not all of the source light output gets into the fiber. Coupling Loss (dB) = a coupling [ ] power coupled into the fiber power produced by the source =  10 log 10 [ ] power produced by the source power coupled into the fiber = 10 log 10

42. Interference of Light Waves u “Superposition” u In order for us to be able to observe sustained interference between two or more light waves - (1) The sources of the waves must be “monochromatic”, which means that they must have a single wavelength (or a single frequency, or a single “color”). (2) The sources of the waves must be “coherent”, which means that the phase difference between the waveforms must remain constant, e.g.,

43. Huygens’ Principle All points on a given wave front are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate outward with speeds characteristic of waves in that medium. After some time has elapsed, the new position of the wave front is the surface tangent to the wavelets. Page 1119 of the text

44. Huygens’ Principle

49. Interference of Waves Huygen’s Principle p. 1119 of the text

51. Young’s Double-Slit Experiment viewing screen Source d two slits in metal plate u Light exiting the two slits interferes.  In Phase  Constructive Interference

52. Young’s Double-Slit Experiment Source d two slits in metal plate viewing screen u Light exiting the two slits interferes.  Out of Phase  Destructive Interference

53. Young’s Double-Slit Experiment u Light exiting the two slits interferes.  Constructive interference  bright band (fringe)  Destructive interference  dark fringe Source d two slits in metal plate viewing screen

57. Source d  P d sin  path difference r1r1 r2r2  Young’s Double-Slit Experiment u In order to have a bright area at point P (called a “bright fringe” or “constructive interference”), we must have  d sin  = m  (m = 0, ±1, ±2,...)

58. Source d  P d sin  path difference r1r1 r2r2  u In order to have a dark area at point P (called a “dark fringe” or “destructive interference”), we must have  d sin  = (m + 1/2)  (m = 0, ±1, ±2, …) Young’s Double-Slit Experiment

59. The Diffraction Grating

61. A diffraction grating consists of a large number of equally spaced parallel slits, e.g., 5000 slits/cm. Light diffracts (spreads out) as it leaves each slit. Light from each slit interferes with light from the other slits. Each wavelength experiences constructive interference at a different angle .  blue  red The Diffraction Grating

62. Just as with the double slit, for the diffraction grating we get constructive interference (“maxima” in the interference pattern) when d sin  = m  (m = 0,  1,  2,  3,...) or sin  = m d or  = sin -1 ( ) m d e.g., for m=1 and  = blue  = 475 nm  blue = sin -1 ( ) 475 nm d and for m=1 and  = red  = 660 nm  red = sin -1 ( ) 660 nm d   red  >  blue

63. Interference in Thin Films u Interference effects are commonly observed in “thin films”. u Examples: u (1) thin layer of oil on water, u (2) soap bubbles, u (3) coatings on camera lenses and solar cells. u The varied colors observed when white light is incident on a thin film result from the interference of waves reflected from the two surfaces of the film.

65. Interference in Thin Films Ray 2 reflects off the bottom surface of the film, so it travels 2t farther than ray 1. 1 2 If the thickness of the film is constant, then for some color (wavelength), rays 1 and 2 will add  constructive interference.  For another color (wavelength), rays 1 and 2 will cancel  destructive interference. air n = 1 soap film n  1.3 t = thickness u In reality, a soap film varies in thickness, so different parts have different colors.

68. Nonreflective Coatings for Solar Cells u Solar cells u Convert sunlight directly into electricity. u Frequently made of silicon. t = thickness Air n = 1 SiO 2 n  1.45 Si n = 3.5 1 2 u Often coated with a thin, transparent film u Such as silicon dioxide (SiO 2 ). u Reduces the amount of reflected (and hence lost) sunlight. u Ray 2 travels 2t farther than Ray 1.

70. Solar Cell Example (cont.)  What must the film thickness, t, be so that we will get the least amount of green light ( o = 550 nm) reflected from the surface? u The light wavelength inside the SiO 2 is SiO 2 = = = 379 nm o n SiO 2 550 nm 1.45 ( n = ) o n

71. Air SiO 2 Si n = 1 n = 1.45 n = 3.5 0 1 2 0 = 550 nm These two waves tend to cancel.  The least reflection of green light SiO 2 = 379 nm t Solar Cell Example (cont.)

72. Ray 2 travels 2t farther than ray 1, and it does so in SiO 2, where SiO 2 = 0 n SiO 2 550 nm 1.45 = = 379 nm for green light. In order to make waves 1 and 2 cancel, the path difference, 2t, must equal SiO 2 2, so we have 2t = SiO 2 2 t = = = 94.8 nm SiO 2 4 379 nm 4 For an SiO 2 coating of this thickness, we get the minimum amount of green light reflected. However, other colors (wavelengths) are reflected. Solar Cell Example (cont.)