1. The wave nature of light Interference Diffraction Polarization
Physical Optics
The wave nature of light
Interference
Diffraction
Polarization

2. Huygens’ Principle
Every point on a propagating wavefront serves as the source of spherical wavelets, such that the wavelets at sometime later is the envelope of these wavelets.
If a propagating wave has a particular frequency and speed, the secondary wavelets have that same frequency and speed.
“Isotropic”

3. Diffraction Diffraction – Bending of light into the shadow region
Grimaldi - 17th Century observation of diffraction
Diffraction vs. Refraction?

5. Explanation of Snell’s Law

6. Superposition of waves
Constructive
Interference
Destructive
Interference

7. Conditions for Interference
To observe interference in light waves, the following two conditions must be met:
1) The sources must be coherent
They must maintain a constant phase with respect to each other
2) The sources should be monochromatic
Monochromatic means they have a single wavelength

8. Young’s Experiment

9. Young’s Experiment

10. Young’s Experiment
Maxima occur when:
maxima
Minima occur when:

11. What the pattern looks like

12. Intensity Distribution, Electric Fields
The magnitude of each wave at point P can be found
E1 = Eo sin ωt
E2 = Eo sin (ωt + φ)
Both waves have the same amplitude, Eo

13. Intensity Distribution, Resultant Field
The magnitude of the resultant electric field comes from the superposition principle
EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)]
This can also be expressed as
EP has the same frequency as the light at the slits
The magnitude of the field is multiplied by the factor 2 cos (φ / 2)

14. Intensity Distribution, Equation
The expression for the intensity comes from the fact that the intensity of a wave is proportional to the square of the resultant electric field magnitude at that point
The intensity therefore is

15. Resulting Interference Pattern
The light from the two slits forms a visible pattern on a screen
The pattern consists of a series of bright and dark parallel bands called fringes
Constructive interference occurs where a bright fringe occurs
Destructive interference results in a dark fringe

16. Example
A He-Ne Laser has a wavelength of 633 nm. Two slits are placed immediately in front of the laser and an interference pattern is observed on a screen 10 m away. If the first bright band is observed 1 cm from the central bright fringe, how far apart are the two slits?
At what angle Q is the fourth dark band found?

17. Thin Films Constructive Interference (maxima)
Destructive Interference (minima)

18. Chapter 35 - Problem 45
Stealth aircraft are designed to not reflect radar whose wavelength is 2 cm, by using an anti-reflecting coating. Ignoring any change in wavelength in the coating, estimate its thickness.

19. Phase shift on reflection
External Reflection
Internal Reflection
Now, If one reflection is internal and one reflection is external half wavelength path differences will result in constructive interference

20. Phase Changes Due To Reflection
An electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was traveling
Analogous to a pulse on a string reflected from a rigid support

21. Lloyd’s Mirror
An arrangement for producing an interference pattern with a single light source
Waves reach point P either by a direct path or by reflection
The reflected ray can be treated as a ray from the source S’ behind the mirror

22. Lloyd’s Mirror

23. Interference in Thin Films Again
Assume the light rays are traveling in air nearly normal to the two surfaces of the film
Ray 1 undergoes a phase change of 180° with respect to the incident ray
Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave
For constructive interference
d=2t = (m + ½)λn (m = 0, 1, 2 …)
This takes into account both the difference in optical path length for the two rays and the 180° phase change
For destructive interference
d=2t = mλn (m = 0, 1, 2 …)

24. Newton’s Rings
Maxima - bright
Minima - dark

25. Chapter 37 – Problem 62
Show that the radius of the mth dark Newton’s ring as viewed from directly above is given by:
Where R is the radius of curvature of the curved glass surface and l is the wavelength of the light used. Assume that the thickness of the air gap is much less than R at all points and that x<<R

26. Newton’s Rings Minima Maxima Newtons rings are a
special case of Fizeau fringes.
They are useful for testing
surface accuracy of a lens.
Maxima

28. Wedges – Fringes of Equal Thickness
Minima (destructive)
Maxima (constructive)
Fringes of this type are also
known as Fizeau fringes

29. Example: thin film of air

30. Michelson Interferometer
A lens can be used to form fringes of equal inclination (rings)
Tilting the mirrors can cause fringes of equal thickness.
Accurate length measurements are accomplished by fringe counting as one of the mirrors is moved.
Compensator
Plate
Detector

31. 42. Monochromatic light is beamed into a Michelson interferometer
42. Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced mm, causing the interferometer pattern to reproduce itself times. Determine the wavelength of the light. What color is it?

32. 5. Young’s double-slit experiment is performed with 589-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits.
7. Two narrow, parallel slits separated by mm are illuminated by green light (λ = nm). The interference pattern is observed on a screen 1.20 m away from the plane of the slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands.
17. In Figure 37.5, let L = 120 cm and d = cm. The slits are illuminated with coherent 600-nm light. Calculate the distance y above the central maximum for which the average intensity on the screen is 75.0% of the maximum.

33. 32. A thin film of oil (n = 1. 25) is located on a smooth wet pavement
32. A thin film of oil (n = 1.25) is located on a smooth wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640 nm and reflects no blue light at 512 nm. How thick is the oil film?
33. A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?
37. A beam of 580-nm light passes through two closely spaced glass plates, as shown in Figure P For what minimum nonzero value of the plate separation d is the transmitted light bright?

34. Mach Zehnder Interferometer
Detector
No compensating plate needed
Test cell easily inserted in one leg.
No factor of two as in the Michelson interferometer.
Difficult to align

35. Sagnac Interferometer
Detector
Ring laser gyro.
The rotation effectively shortens the path taken by one direction over the other.

36. Phase Angle