Interference Applications Physics 202 Professor Lee Carkner Lecture 25.

Interference Applications

Physics 202Professor Lee

CarknerLecture 25

PAL #23 Interference

Light with = 400 nm passing through n=1.6 and n=1.5 material N = (L/)(n) L = N/n = (5.75)(400)/(0.1) =

Compare to L = 2.6X10-5 m N = (2.6X10-5)(0.1)/(400X10-9) = 6.5

Orders

At the center is the 0th order maxima, flanked by the 0th order minima, next is the 1st order maxima etc.

The intensity varies sinusoidally between minima and maxima

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:

= (2d sin ) /

is in radians

Intensity The intensity can be found from the electric

field vector E:

I = 4 I0 cos2 (½ )

For any given point on the screen we can find the intensity if we know ,d, and I0 The average intensity is 2I0 with a maximum and

minimum of 4I0 and 0

Intensity Variation

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

Thin Film

Reflection Phase Shifts In addition to the path length shift there can also be

a phase shift due to reflection

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

Reflection Phase Change

Reflection and Thin Films If the thin film covers glass, both reflection

phase shifts will be 0.5

Interference is due only to path length difference Example:

If the thin film is in air, the first shift is 0.5 and the second is zero

Have to add 0.5 wavelength shift to effects of path length difference Example:

Path Length and Thin Films

What is the path length difference between the two reflected rays?

Don’t forget to include reflection shifts

Anti-reflective Coating

Reflection and Interference What kind of interference will we get for a particular

thickness?

The wavelength of light in the film is equal to:

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) (/n2) -- The two rays will produce constructive interference if 2L is

equal to a wavelength2L = m (/n2) --

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 soap films of a particular thickness can produce strong constructive interference at a particular

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:

= (2Ln) / (m + ½)

= 1330 nm / (m + ½)

= Only 532 nm is in the visible region and is green

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?

Reflection

Depends on: n Example: thin films Equations:

• n1 > n2 -- phase shift = 0• antireflective coating

• n1 < n2 -- phase shift = 0.5• soap bubble

Path Length Difference

Depends on: L and Example: double slit interference Equations:

d sin = m -- maxima d sin = (m + ½) -- minima

Different Index of Refraction

Depends on: L, , n Example: combine beams from two

media Equations:

N2 - N1 = (L/)(n2 -n1)