Minimum thickness n/wavelength equation?

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SUMMARY

The minimum thickness of a soap film (n = 1.33) that constructively reflects light of wavelength 477 nm is calculated using the formula \(\frac{1}{4}\frac{\lambda}{n}\), resulting in a thickness of 89.66 nm. The next minimum thickness for constructive interference is found by multiplying the first thickness by 3, yielding 269 nm. This approach is based on understanding the phase changes of light waves upon reflection and transmission at boundaries of different refractive indices.

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Homework Statement



White light is incident normally on a thin soap film (n = 1.33) suspended in air.
1. What is the minimum thickness of the soap film that will constructively reflect light of wavelength 477 nm?
2. What is the next minimum thickness of the soap film that will constructively reflect light of wavelength 477 nm?

Homework Equations


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The Attempt at a Solution


for #1 i found the answer is \frac{\frac{\lambda}{n}}{4} so 89.66nm was the answer to this, but i don't know why i divided by 4. can someone give an equation/explanation of why i had to do this?
for #2 i have no idea but i foudn to get the right answer i just multiplied my answer for #1 by 3. so 86.99*3=269nm.
 
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It may help you to draw a picture. Draw the film and an incoming incident wave. Suppose the wave is at a crest when it hits the surface of the film. A few questions to guide your sketch:
- What happens to the phase of a wave when it reflects from a material of higher index? Use this to draw the wave reflected from the front surface of the film. (Call this wave 1)
- What happens to the phase of a wave when it is transmitted from a material of lower index to one of higher index? Use this to draw the part of the incident wave that passes through the front surface of the film. (Call this wave 2)
- What happens to the phase of a wave when it reflects from a material of lower index? Use this to draw the part of wave 2 that is reflected from the back surface of the film. (Call this wave 3)
- What happens to the phase of a wave when it is transmitted from a lower index to a higher index? Use this to draw the part of wave 3 that passes back through the front surface of the film. (Call this wave 4)

If you get your phases right and draw all of this, you will see that for wave 1 and wave 4 to constructively interfere, the smallest possible film thickness is ##\frac{1}{4}\frac{\lambda}{n}## (where ##\frac{\lambda}{n}## is the wavelength of the light inside the film). You will also be able to see what the second, and so on, smallest thickness is, and so figure out the general expression.
 
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