Rebecca SlaterLO #9

A substance that is normally clear and colorless can appear to give off an array of colors when it is found in a very thin layer

When light hits the surface (1), some is reflected (2), and some is transmitted through the new medium (3)

At the next barrier, some light is reflected (4) and then refracted through the first layer (5), and some is transmitted (6)

Constructive Destructive

Since the speed of light decreases in any medium other than a vacuum, the wavelength also decreases

Wavelength in a new medium is equal to the original wavelength divided by the index of refraction of the medium

OR

The type of reflection experienced by the light waves at each boundary depends on the refraction indexes of the two mediums:

Hard reflection – reflects off a medium with a higher refraction index resulting in a phase shift of λ/2

Soft reflection – reflects off a medium with a lower refraction index, resulting in no phase shift

If both waves undergo the same type of reflection, either both hard or both soft, use:

- because there is no phase shift between waves

If one wave undergoes a hard reflection, and the other a soft reflection, we use:

- because there is a phase shift of λ/2 for the wave that undergoes the hard reflection

**Where m is a whole number ≥ 0, t is the thickness of the thin film, and λn is the wavelength in medium

Again, there are two cases:

- use this if there are an even number of hard reflections, so that the end result is a λ/2 phase shift

- use this if there are an odd number of hard reflections, to keep the waves at their λ/2 phase shift

**Where m is a whole number ≥ 0, t is the thickness of the thin film, and λn is the wavelength in medium

Since each colour of light has a different wavelength, different thicknesses of the film layer correspond to different colours interfering constructively/destructively

This creates a rainbow-like effect as the thickness of the film changes at different points

A beam of light is aimed almost perpendicular to

a soap bubble floating through the air. The air and

soapy water have refraction indexes of 1.00 and

1.33, respectively. At the point on the bubble

where the light beam is aimed, the bubble appears

to be green. What is the thinnest layer of soapy

water that would provide this result?

(λgreen = 510 nm)

Since the light beam is being reflected off of a medium with a higher index of refraction (1.33>1.00), a hard reflection will occur

Because there is one hard and one soft reflection in a soap bubble, and the green light waves must interfere constructively, to counteract the λ/2 phase shift that occurs due to the hard reflection, we use the formula:

Since the question asks for the thinnest possible film, we use m = 0

Subbing in 0 for m and rearranging the formula to find t, we get:

Now, to find λn:

Lastly, just sub in all our values!

Therefore, the thinnest soapy water layer that would produce a green-looking surface, is

95.8 nm!

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/oilfilm.html

http://en.wikipedia.org/wiki/Iridescence http://www.quora.com/Are-there-any-real-life-

examples-or-cases-of-light-waves-undergoing-constructive-or-destructive-interference

http://imgarcade.com/1/images-of-peacock-feathers/ http://dev.physicslab.org/Document.aspx?doctype=5

&filename=PhysicalOptics_ThinFilmInterference.xml http://www.webexhibits.org/causesofcolor/15.html http://www.hk-

phy.org/iq/oil_rainbow/oil_rainbow_e.html https://alicethroughthemacrolens.wordpress.com/tag

/oil-and-water/