At a science fair, one exhibition is a very large soap film...

At a science fair, one exhibition is a very large soap film that has a fairly consistent width. It is illuminated by a light with a wavelength of $432 \mathrm{nm},$ and nearly the entire surface appears to be a lovely shade of purple. What would you see in the following situations? a. the film thickness was doubled b. the film thickness was increased by half a wavelength of the illuminating light c. the film thickness was decreased by one quarter of a wavelength of the illuminating light

At a science fair, one exhibition is
a very large soap film that has a fairly consistent width. It is illuminated by a light with a wavelength of $432 \mathrm{nm},$ and nearly the entire surface appears to be a lovely shade of purple. What would you see in the following situations?
a. the film thickness was doubled
b. the film thickness was increased by half a wavelength of the illuminating light
c. the film thickness was decreased by one quarter of a wavelength of the illuminating light

Key Concepts

Thin Film Interference

This phenomenon occurs when light waves reflect off the different boundaries of a thin film, such as a soap film, and interfere with each other. The interference pattern—whether constructive (bright) or destructive (dark)—is determined by the film’s thickness, the wavelength of the light, and the relative phase shifts upon reflection. It is a classic example used to illustrate wave interference and is responsible for the colorful patterns seen in thin films.

Optical Path Difference

The optical path difference is a key factor in determining the interference outcome and is calculated by considering the extra distance traveled by the light within the film, multiplied by the refractive index. In a thin film, this path difference is essentially 2t (where t is the thickness of the film), and it dictates whether the waves will interfere constructively or destructively depending on how it compares with the wavelength of the light.

Phase Change Upon Reflection

When light reflects off a medium of higher optical density, such as the transition from air to the soap film, it undergoes a half-wavelength (? radians) phase shift. This additional phase shift must be taken into account along with the optical path difference to determine the conditions for constructive or destructive interference in thin films.

Constructive and Destructive Interference

The conditions for interference are determined by both the optical path difference and the phase changes at the film's surfaces. Constructive interference occurs when the total phase difference is an integer multiple of 2?, leading to bright (or enhanced color) regions, while destructive interference occurs when it is an odd multiple of ?, producing dark regions. Small changes in the film's thickness can shift these conditions, altering the interference pattern and the resulting color spectrum observed.

Transcript

00:01 In this question, the equations that we would use are 2d equals m plus 1 half lambda over n film, and 2d equals n lambda over n film, where m and n are whole numbers.

00:53 Now, i'm going to replace lambda over n film with just lambda film.

01:04 So that's going to be the wavelength of light when it's in the film.

01:09 So the first equation becomes 2d equals.

01:15 I'm also going to add those together.

01:21 So that would be 2m plus 1 over 2.

01:40 And then i'm gonna write lambda film.

01:47 And the second equation is going to be 2d equals n lambda.

01:55 A film.

01:59 Now, second thing that we know from part a is that we have constructive interference because the bubble is a bright purple color.

02:17 And so, for constructive interference with one inversion, which is what we have here in the bubble, we are using the first equation.

02:28 So, we know that the first equation is applying.

02:36 But then in part a, we're going to double the thickness of the bubble.

02:48 And so, not the thickness of the bubble, but the thickness of the soap film, the thickness of the soap film.

02:55 And so the new diameter is going to be two times the old diameter.

03:05 Times d.

03:09 Okay, what i want to do is i want to rearrange that first equation.

03:14 So i'm going to divide by two on both sides.

03:17 So i'm going to get d equals 2m plus 1 over 4 lambda fim.

03:34 But now d new is 2d.

03:41 So it's going to be 2.

03:43 And then, then i'm going to substitute this d in to m plus 1 over 4 lambda film.

04:09 If we rearrange that, then we will see that 2d new is 2m plus 1 times lambda film.

04:30 And 2m plus 1, if m is an integer, is just an integer.

04:34 So we could write that's another integer, and that's 2d new.

04:46 And then we compare.

04:49 This equation is the same as this equation, and this is for destructive, and this is for constructive.

05:16 So there will be complete destructive interference, and so the bubble will appear dark.

05:31 B.

05:35 This is a long question for this textbook...

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