A soap bubble is essentially a very thin film of water ( n = 1.33) surrounded by

Question

A soap bubble is essentially a very thin film of water (n = 1.33) surrounded by air. The colors that you see in soap bubbles are produced by interference, much like the colors of dichroic glass.

A- Derive an expression for the wavelengths ?_C for which constructive interference causes a strong reflection from a soap bubble of thickness d.
Hint: Think about the reflection phase shifts at both boundaries.

B- What visible wavelengths of light are strongly reflected from a 380-nm-thick soap bubble?

C- What color would such a soap bubble appear to be? (green, red, violet, blue?)

D- The shortest wave is? (yellow, green, violet, orange?)

Explanation / Answer

The light ray falling on the soap bubble reflects from two places. It reflects from
the surface of the bubble, and then it also reflects back from the back of the
bubble, after traveling an extra distance x = 2d (down and back), where d is
the thickness of the bubble. The wave reflecting from the top surface reflects from
a material having a higher index of refraction. This induces an extra phase shift
of , that the reflection from the back surface doesn't have.
The net phase of the rst wave, at a xed time which we take to be t = 0, is
1 = kx1+1+, where k is the wave number, and 1 is the intrinsic phase shift
due to the source. Similarly, the net phase of the second wave is2 = kx2 + 2.
The second wave doesn't have an extra phase shift because it's reflecting from the
air behind the bubble, which has a lower index of refraction. The phase shifts of
these two waves are equal since they both come from the same source, so 1 = 2.
The difference in waves is
= 2 ?? 1 = k (x2 ?? x1) + (x2 ?? x1) ?? = kx ?? :

2pi* m = 2pi*2d/(lambda/n) + intial phi = phase shift

In this case, when the light reflects off the bubble, the wave is shifted by pi, but when the waves go through the bubble reflect at the second water-air interface, the waves aren't shifted by pi, so the initial phase difference is equal to pi

2pi*m = 2pi*2dn/lambda + pi

pi(2m-1) = 4pi*dn/lambda

.5(m-.5) = dn/lambda

lambda= 2.66d/(m-.5)

Only m = 1 and 2 gives a re
ected wavelength in the visible spectrum. For
m = 1, = 692 nm, which is reddish, while for m = 2, = 415 nm, which is
violet. Together, they look purplish.

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