130_Lec21_Nov21 2022

Phase Difference Due to Refraction

• The wavelength of light shortens when passing through a refractive medium.

• When light enters a refractive medium:

  • Frequency (f) remains constant.

  • Velocity (v) decreases.

  • Consequently, the wavelength ( ( \lambda )) also decreases.

• In air: ( f, \lambda, c )

• In glass: ( f, \lambda', v_1, n_1 )

• A phase difference exists between rays after one has passed through glass.

Phase Difference Representation

• Phase difference (( \Delta f )) can be expressed in two ways:

  • Number of cycles.

  • Radians.

Example 1: Monochromatic Light Interference

• Given:

  • Wavelength ( \lambda = 500 \text{ nm} )

  • Thickness of glass flake ( L = 0.010 \text{ mm} = 10 \text{ \mu m} )

  • Index of refraction ( n = 1.5 )

• Light inside the flake:

  • Undergoes exactly 10 more cycles than light outside.

  • Since this is an integer, the rays are in phase.

  • Observed interference: constructive interference.

Thin Film Interference

• Incoming light may be:

  • Partially reflected

  • Partially transmitted at each interface.

• If reflected rays B and C are in phase:

  • Constructive interference occurs.

• For simplification, assume:

  • Incident light is nearly perpendicular (( q_i \approx 0 )).

  • The path length of ray C inside the film is ( 2L ).

Notes on Reflection and Interference

• For transparent media:

  • If reflected rays interfere destructively, light is transmitted into the underlying layer.

Practical Example: Rainbow Effect in Thin Films

• Interference of white light in thin diesel fuel film on wet pavement.

  • Appearance of a "rainbow" effect is due to variable thickness of the film (h).

  • Type of interference depends on the relationship between wavelength ( \lambda ) and thickness h.

Phase Difference Determination

• Complication: Phase change upon reflection.

  • Light reflecting off an interface undergoes a phase change (( \Delta f = \pi )) if ( n_2 > n_1 ).

Reflection Phase Changes

• Recall:

  • Reflection from a discontinuity may lead to a phase change.

• Example insights can be found in demonstrations provided online.

Thin Film Interference Problems

• Important considerations:

  1. Calculate phase difference due to path-length difference (( = 2h )) for both reflected rays.

  2. Account for any phase difference due to reflection between the two rays.

  3. Remember to adjust the wavelength inside refractive media: ( \lambda' = \frac{\lambda}{n} ).

Example 2: Analyzing Phase Differences in Films

• Two scenarios (A, B): Light reflects from the top and bottom interfaces of a thin film viewed from above.

• Questions to address:

  • (a) Which cases have zero phase difference due to reflection?

  • (b) If thickness ( h = \frac{\lambda'}{4} ), what type of interference will be observed in each case?

Example 2 Answers

• (a) Cases A and B have zero phase difference due to reflection.

• (b) Observed interference:

  • Cases A & B: Destructive (dark) interference.

  • Cases C & D: Constructive (bright) interference.

Criteria for Interference

• General relationship:

  • Wavelength in the film ( \lambda' = \frac{\lambda}{n} ).

  • Equivalent to constructive and destructive interference criteria from the textbook.

Example 3: Eliminating Reflection

• Problem statement:

  • To eliminate light reflection with ( \lambda = 550 \text{ nm} ) in air:

    • A flat glass plate is coated with a thin layer of plastic (( n = 1.20 )).

    • Index of refraction of glass is 1.50.

    • Find minimum thickness of plastic required (assume nearly zero angle of incidence).

• Answer: Minimum thickness required is 115 nm.