Physical Optics - Thin Film Interference

Forum Expert

Posted on
31 Dec 2010 20:26:41 IST

Posts: 7537

31 Dec 2010 20:26:41 IST

Like 0 people liked this

342

Physical Optics - Thin Film Interference

academic , physical optics thin film interference

Just as pulses along a rope when reflected off a fixed-end are inverted,

or when reflected off a free-end, remain in phase,

light waves also exhibit these behaviors when they encounter an interface between two mediums. When a light ray is reflected off an interface which has a less optically-dense medium above a more optically-dense medium the reflected ray is inverted. For example, air over water, air over glass, water over glass.

This reflection produces a phase difference of ½λ from the original wave and is called a fixed-end reflection. Shown below is a waveform diagram of this type of reflection. The incident crest (blue C) is reflected as a trough (orange T).

When a light ray is reflected off an interface which has a more optically-dense medium above a less optically-dense medium the reflected ray is not inverted. For example, water over air [soap bubbles], glass over air [glass wedge], glass over water.

This is called a free-end reflection. This reflection produces no phase difference with the original wave. Shown below is a waveform diagram of this type of reflection. The incident crest (blue C) is reflected as a crest (orange C). REMEMBER: the refracted rays NEVER undergo a phase change!

Oil and Soap Films

These reflective properties are critical to our understanding of the colors in such thin films as soap bubbles, coatings on camera lenses, colors in a butterfly's wings or peacock's feathers, or oil spills. Below is a diagram of a thin film and the light rays associated with the reflections and refractions as light impinges on the film.

When ray 1 strikes the top interface, some of the light is partially reflected, ray 2, and the rest is refracted, ray 3.
When ray 3 strikes the bottom interface, some of it is reflected, ray 4, and the remainder is refracted, ray 6.
When ray 4 strikes the top interface from underneath, some is reflected (not shown) and some is refracted, ray 5.

It is the interference between rays 2 and 5 that produces a thin film's color when the film is viewed from above.

The formula used for thin film interference is

EPD = 2t + Φ

where

t represents the thickness of the thin film

Φ is the net phase inversion between rays 2 and 5

0 λ

½ λ

EPD represents the extreme path difference

constructive interference: mλ where m {0, 1, 2, 3…}

destructive interference: ½(2m -1)λ where m {1, 2, 3...}

The net phase inversion Φ is the difference in the phases between the two reflected rays - one off the top interface (#2) and the second off the top of the bottom interface (#5). These are the rays that span the thickness of the film. Remember that since we are dealing with the behavior of the light in the thin film, you must ALWAYS use the light's wavelength IN THE FILM. Also remember that refracted rays remain in-phase with their initial rays.

If both reflections occur at interfaces of less/more, then both are inverted and their net phase difference is zero.
If both reflections occur at interfaces of more/less, then neither is inverted and the net phase difference between them is zero.
However, if one reflection occurs at less/more and the other at more/less, then there is a net phase difference between the reflections of ½ λ.

Solve the above equation for t by calculating its value in terms of the appropriate wavelength in the thin film; that is, if the film is made of air, use the wavelength in air; if the film is of water, use the wavelength in water. Remember that the wavelength in a medium whose refractive index is n can be calculated with the formula

λ_n = λ_vacuum/n

and that the refractive indices for air, water, and glass are: 1, 1.33, and 1.5 respectively. Since the wavelength DECREASES when the light travels through an optically-denser medium, its average speed decreases.

Color Addition and Subtraction

Consequently, a given thickness of film can cancel or reinforce specific wavelengths of light resulting in the thin film displaying unique colors.

If the film thickness cancels red, then you see its complementary color cyan.

If the film thickness cancels green, then you see its complementary color magenta.

If the film thickness cancels blue, then you see its complementary color yellow.

Refer to the following information for the next eight questions.

A thin soap film is formed by dipping a plastic rectangular wand into a solution of soapy water which has a refractive index of 1.4. When viewed in daylight, one portion of the film reflects blue light of wavelength 475 nm. Estimate the minimum thickness of that section of the film.

Q)	Using the same numbering scheme as in the lesson, will ray 2 be in-phase or out-of-phase with ray 1?
		The outer interface is between air and the soap film. Since the refractive index of soap is greater than that of air, this interface will act like a fixed-end reflector and ray 2 will be reflected out-of-phase. This makes rays 1 and 2 out-of-phase with a phase inversion of ½ λ.

Q)	Using the same numbering scheme as in the lesson, will ray 5 be in-phase or out-of-phase with ray 1?
		The inner interface is also between air and the soap film. But, the light is incident from soap to air, so this interface will act like a free-end reflector and ray 2 will be reflected in phase. This makes rays 3 and 4, in-phase, which translates into rays 1 and 5 being in-phase since refracted light is never inverted. Thus rays 1 and 5 are in-phase with no phase inversion.

Q	What is Φ, the net phase inversion, for this series of reflections?
		Since one reflection is ½ λ and the other is 0 λ, the net phase inversion equals ½ λ.

Q)	To see blue, are you going to use the formula for constructive or destructive interference?
		constructive interference - a bright

Q)	What expression will represent the EPD in this problem?
		EPD = mλ where m {0, 1, 2, 3...}

Q)	An average blue wavelength is 475 nm in air, what is its wavelength in the soap film?
		339 nm
		In the soap film the effective wavelength is λ_n = 475/1.4 = 339 nm

Q)	Write the equation for thin film interference as it applies to this problem.
		EPD = 2t + Φ_{rays 2 & 5} mλ = 2t + ½ λ

Q)	What is the minimum, non-zero thickness of the film?
		84.8 nm
		Since we are determining the thickness of the film, we must use the light's wavelength in the film. 2t = mλ_n - ½λ_n t = ½(m - ½)λ_n t = ½(1 - ½)(339) t = ¼(339) t = 84.8 nm

Wedges

A thin film interference pattern will be a series of alternating bright and dark fringes of the same color when it is created by a wedge illuminated by monochromatic light. For example: (1) an air wedge formed between two pieces of glass, or (2) a sliver of glass surrounded by air or water.


thick wedge	thin wedge

As the wedge becomes thinner (θ → 0), the fringes are less numerous as they spread further apart and get wider. Note that the point of contact is a dark fringe. This results not from the thickness of the wedge, but from the net phase inversion between the two reflected rays equals ½ λ. If the plates are placed flat, directly on top of one another, the plates would be entirely dark. If the plates remained parallel, but were slowly raised apart, the pattern would alternate between completely dark and completely bright as the distance between the plates changed by ¼ wavelengths. [2t = ½ λ].

Refer to the following information for the next six questions.

A small sliver of glass is perfectly wedge-shaped. When viewed with monochromatic 475 nm blue light, 17 tightly-spaced bright blue bands can be viewed along its 1-centimeter length.

Q)	Why are the fringes uniformly spaced along the length of the sliver?
		Because the angle is so small, the thickness of the sliver is increasing uniformly. For very small angles, tan θ behaves linearly.

Q)	Are rays 2 and 5 in-phase or out-of-phase with respect to each other?
		Ray 2 is out of phase with ray 1. Ray 5 is in-phase with ray 1. Therefore there is a net phase inversion of ½ λ.

Q)	Since the edge of the wedge ends on a bright fringe, would you use the relationship for constructive or destructive interference?
		constructive interference

Q)	State the equation for thin films that you will use to determine the thickness of the wedge at its widest edge.
		17 λ = 2t + ½ λ
		Since the final fringe is bright, EDP = mλ where m = 17

Q)	What is the wavelength of the light in the glass sliver?
		317 nm
		λ_n = 475/1.5 = 317 nm

Q)	How thick is the glass sliver?
		2.62 x 10^-6 m
		Since we are determining the thickness of the glass sliver, we will use the wavelength of light in glass. 17 λ = 2t + ½ λ 2t = 16.5 λ_n t = 8.25(317) t = 2620 nm

Share this article on:

Comments (1)

sauhard koolwal

Cool goIITian

Joined: 19 Jul 2009 22:15:09 IST

Posts: 92

21 Jan 2012 11:03:18 IST

Like 0 people liked this

unable to see the full article clearly.. Please repost.

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

Free Sign Up!

Comments (1)

Quick Reply

Reply

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info