Light of wavelength 550 E-9 m is incident (in air) on a structure consisting of

Question

Light of wavelength 550 E-9 m is incident (in air) on a structure consisting of air / thin film / glass. The indices of refraction are, respectively, n1= 1.0 (air), n2 = 2.0, and n3 = 1.5

a) Consider two rays reflected from the structure. Ray #1 is reflected from the air/thin film interface, and ray #2 goes through the film and is reflected from the thin film/glass interface. Sketch changes in phase (if any) for each ray as they emerge in air after being reflected at the boundaries. Assume the incident phase of incoming wave to be 0 degrees or a sine wave or a positive pulse.

b) Determine the smallest thickness of the film that will give minimum reflection of this light. Asssume that the light is normally incident in air.

c) What is the smallest film thickness that will give maximum reflection from the structure?

Explanation / Answer

Part A)

The reflected ray from the air to thin film will have a 180 degree phase change and then when it hits the glass on the other side it will not change phases. The phase changes occur only when the ray goes from a lower index value to a higher one. At the air to film, its n = 1 to n = 2, thus a phase change. From film to glass, its n = 2 to n = 1.5 (smaller value) so no phase change.

Thus at the air/thin film, the pulse will swap to a negative pulse and remain negative when it hits the glass

Part B)

For reflection, apply 2nt = (m + .5)(wavelength)

2(2)(t) = (.5)(550 X 10-9)

t = 6.875 X 10-8 m which is 68.75 nm

Part C)

Apply 2nt = m(wavelength)

2(2)(t) = 1(550 X 10-9)

t = 1.375 X 10-7 m which is 137.5 nm

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