Solve Visible Wavelengths in Reflected Light from Soap Film w/ n=1.90

[rachiebaby17]

Homework Statement

A soap film has an index of refraction n = 1.90. The film is viewed in reflected light.
(a) At a spot where the film thickness is 910.0 nm, which visible wavelengths are missing in the reflected light? (Give the three wavelengths which correspond to the lowest such m values, in order of increasing m.)

Which visible wavelengths are strongest in reflected light? (Give the three wavelengths which correspond to the lowest such m values, in order of increasing m.)

Homework Equations

EPD = 2t + Φ
constructive interference: mλ where m {0, 1, 2, 3…}
destructive interference: ½(2m -1)λ where m {1, 2, 3...}

The Attempt at a Solution

no idea - help me please!

 

[anathema]

Hello there! It looks like you're trying to understand the phenomenon of interference in thin films. Let me break it down for you.

First, let's talk about the missing wavelengths in reflected light. When light is reflected off of a thin film, it undergoes interference. This means that the waves of light are either reinforcing each other (constructive interference) or canceling each other out (destructive interference). The equation for constructive interference is mλ, where m is an integer and λ is the wavelength of light. This means that at certain points on the film, only certain wavelengths of light will be reinforced and therefore visible. The wavelengths that are missing will correspond to the lowest values of m. In this case, since the film has an index of refraction of 1.90, the visible wavelengths that will be missing are those with m values of 0, 1, and 2. To find the corresponding wavelengths, we can use the equation EPD = 2t + Φ, where EPD is the path difference (the difference in distance traveled by the two waves of light), t is the thickness of the film, and Φ is the phase shift (which is usually 0 for reflected light). We can rearrange this equation to solve for λ, giving us λ = (2t + Φ)/m. Plugging in the given thickness of 910.0 nm and m values of 0, 1, and 2, we get the corresponding wavelengths of 910.0 nm, 1820.0 nm, and 2730.0 nm. These are the wavelengths that will be missing in the reflected light.

Now, let's talk about the strongest wavelengths in reflected light. These will correspond to the m values that are the lowest (since these are the wavelengths that are reinforced). In this case, since the film has an index of refraction of 1.90, the strongest wavelengths will correspond to m values of 0, 1, and 2. Using the same equation as before, we can find the corresponding wavelengths of 910.0 nm, 1820.0 nm, and 2730.0 nm.

I hope this helps you understand the phenomenon of interference in thin films. Keep in mind that this is a simplified explanation and there are other factors that can affect the interference pattern, such as the angle of incidence and the polarization of light. Let me know if you have

 

[nlightner]

I would approach this problem by first understanding the concept of interference in thin films. When light reflects off a thin film, it undergoes both constructive and destructive interference, resulting in certain wavelengths being enhanced or diminished in the reflected light. The equations for constructive and destructive interference, as shown above, can help us determine which wavelengths will be present or absent in the reflected light.

In this case, the soap film has an index of refraction of 1.90. This means that the speed of light is slower in the film than in air, causing a phase shift in the reflected light. To solve for the missing and strongest wavelengths, we can use the equation EPD = 2t + Φ, where EPD is the effective path difference, t is the thickness of the film, and Φ is the phase shift.

(a) At a spot where the film thickness is 910.0 nm, we can plug in the values and solve for EPD:
EPD = 2(910.0 nm) + 1.90(910.0 nm) = 3650 nm
This value represents the total path difference between the reflected light from the top and bottom of the film.

To find the visible wavelengths that are missing, we can use the equation for destructive interference, ½(2m -1)λ, where m is the order of the interference. We know that the lowest possible m value for destructive interference is 1, so we can plug in m = 1 and solve for λ:
½(2(1) -1)λ = 3650 nm
λ = 7300 nm
This means that the visible wavelength of 7300 nm is missing in the reflected light.

To find the other two missing wavelengths, we can plug in m = 2 and m = 3 and solve for λ:
½(2(2) -1)λ = 3650 nm
λ = 3650 nm
½(2(3) -1)λ = 3650 nm
λ = 2433.33 nm
Therefore, the three missing visible wavelengths in the reflected light are 7300 nm, 3650 nm, and 2433.33 nm.

To determine the strongest wavelengths in the reflected light, we can use the equation for constructive interference, mλ. Again, the lowest possible m value for constructive interference is 0, so we can plug in