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Interference Applications

Interference Applications. Physics 202 Professor Lee Carkner Lecture 25. PAL #23 Interference. Light with l = 400 nm passing through n=1.6 and n=1.5 material D N = (L/ l )( D n) L = D N l / D n = (5.75)(400)/(0.1) = 23000 nm Compare to L = 2.6X10 -5 m

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Interference Applications

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  1. Interference Applications Physics 202 Professor Lee Carkner Lecture 25

  2. PAL #23 Interference • Light with l = 400 nm passing through n=1.6 and n=1.5 material • DN = (L/l)(Dn) • L = DNl/Dn = (5.75)(400)/(0.1) = 23000 nm • Compare to L = 2.6X10-5 m • DN = (2.6X10-5)(0.1)/(400X10-9) = 6.5 • 6.5 l is total destructive interference and so the above situation is brighter (5.75 l)

  3. What directions will the beam be bent towards as it enters A, B and C? • Up, up, up • Down, down, down • Up, down, up • Up, up, down • Down, up, down A B C n=1.5 n=1.3 n=1.4 n=1

  4. Rank the 3 materials by the speed of light in them, greatest first. • A, B, C • B, C, A • C, A, B • A, C, B • Speed is the same in all A B C n=1.5 n=1.3 n=1.4 n=1

  5. What happens to the distance between the fringes if the distance between the slits increases? • Increases • Decreases • Stays the same

  6. What happens to the distance between the fringes if the light is switched from red to green? • Increases • Decreases • Stays the same

  7. What happens to the distance between the fringes if the entire apparatus in submerged in a clear liquid? • Increases • Decreases • Stays the same

  8. Orders • At the center is the 0th order maxima, flanked by the 0th order minima, next is the 1st order maxima etc. • The orders are symmetric e.g. the 5th order maxima is located both to the left and the right of the center at the same distance • The intensity varies sinusoidally between minima and maxima

  9. Intensity of Interference Patterns • How bright are the fringes? • The phase difference is related to the path length difference and the wavelength and is given by:  = (2pd sin q) / l • Where d is the distance between the slits, and q is the angle to the point in question • f is in radians

  10. Intensity • The intensity can be found from the electric field vector E: I  E2 I = 4 I0 cos2 (½ f) • For any given point on the screen we can find the intensity if we know q,d,l and I0 • The average intensity is 2I0 with a maximum and minimum of 4I0 and 0

  11. Intensity Variation

  12. Thin Film Interference • Camera lenses often look bluish • Light that is reflected from both the front and the back of the film has a path length difference and thus may also have a phase difference and show interference

  13. Reflection Phase Shifts • The phase shift depends on the relative indices of refraction • If light is incident on a material with lower n, the phase shift is 0 wavelength • Example: • If light is incident on a material with higher n, the phase shift is 0.5 wavelength • Example: • The total phase shift is the sum of reflection and path length shifts

  14. Reflection and Thin Films • Since nfilm > nair and nglass > nfilm • Example: optical antireflection coatings • Since nfilm > nair and nair < nfilm • Have to add 0.5 wavelength shift to effects of path length difference • Example: soap bubble

  15. Path Length and Thin Films • For light incident on a thin film, the light is reflected once off of the top and once off of the bottom • If the light is incident nearly straight on (perpendicular to the surface) the path length difference is 2 times the thickness or 2L • Don’t forget to include reflection shifts

  16. Reflection and Interference • What kind of interference will we get for a particular thickness? • The wavelength of light in the film is equal to: l2 = l/n2 • For an anti-reflective coating (no net reflection shift), the two reflected rays are in phase and they will produce destructive interference if 2L is equal to 1/2 a wavelength 2L = (m + ½) (l/n2) -- dark film 2L = m (l/n2) -- bright film

  17. Interference Dependencies • For a film in air (soap bubble) the equations are reversed • Soap film can appear bright or dark depending on the thickness • Since the interference depends also on l, soap films of a particular thickness can produce strong constructive interference at a particular l • This is why films show colors

  18. Color of Film • What color does a soap film (n=1.33) appear to be if it is 500 nm thick? • We need to find the wavelength of the maxima: 2L = (m + ½) (l/n) l = [(2) (500nm) (1.33)] / (m + ½) l = 2660 nm, 887 nm, 532 nm, 380 nm … • Only 532 nm is in the visible region and is green

  19. Next Time • Read: 36.1-36.6 • Homework: Ch 35, P: 40, 53, Ch 36, P: 2, 17

  20. Interference: Summary • Interference occurs when light beams that are out of phase combine • The interference can be constructive or destructive, producing bright or dark regions • The type of interference can depend on the wavelength, the path length difference, or the index of refraction • What types of interference are there?

  21. Reflection • Depends on: n • Example: thin films • Equations: • n1 > n2 -- phase shift = 0 • antireflective coating • n1 < n2 -- phase shift = 0.5 • soap bubble

  22. Path Length Difference • Depends on: L and l • Example: double slit interference • Equations: • d sin q = ml -- maxima • d sin q = (m + ½)l -- minima

  23. Different Index of Refraction • Depends on: L, l, n • Example: combine beams from two media • Equations: • N2 - N1 = (L/l)(n2 -n1)

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