Destructive Interference in Thin Film Interference

In summary, the conversation discussed the observation of bright and dark lines when a "sodium vapor lam" is placed vertically above two oblong pieces of plane glass separated by a thin sheet of cellophane. The relation between the distance between point C and point A (or B), the length of the wedge-shaped space, the thickness of the cellophane, and the wavelength was found to be x=\frac{l}{2d} \lambda(n + \frac{1}{2}), taking into account the additional path difference when the wave is reflected from a denser medium. The conversation referenced a figure shown on a website for further clarification.
  • #1
willydavidjr
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Two oblong pieces of plane glass are separated from each other by a thin sheet of cellophane and held together with rubber bands, as shown in figure(I provided a website for the figure below), where the amount of separation is much exaggerated. If you place a "sodium vapor lam" vertically above the glass plates, you will observe an alternate series of very close bright and dark lines. A light wave is reflected at point A and simply forms a returning wave. Some of the original light wave crosses the narrow air gap and is reflected at point B. If it advances across the air gap as a trough, then it is reflected at point B as a crest.

Question: The length of the wedge-shaped space between the two glasses is l, the thickness of the thin sheet of cellophane is d, and the distance between point C and point A (or B) is x as shown in the figure.Supposing the result is destructive interference or darkness. Find the relation between x, [tex]\lambda[/tex],l and d.

My idea:
I think their relation is x=[tex]\frac{l}{2d} \lambda(n + \frac{1}{2}) [/tex]


This is the website: www.geocities.com/willydavidjr/interference.html
 

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  • #2
At A the ray is reflected from rarer medium (air) and at B it is reflected from denser medium (glass). When a wave is reflected from a denser medium an additional path difference of (lembda/2) is to be considered. Did you considered this additional path difference?
 

1. What is destructive interference in thin film interference?

Destructive interference in thin film interference is a phenomenon that occurs when two waves of light with opposite phases meet and cancel each other out, resulting in a decrease in the overall intensity of the light. This can happen when light passes through a thin film, such as a soap bubble or an oil slick, and reflects off of both the top and bottom surfaces.

2. How does destructive interference affect the appearance of thin films?

Destructive interference can cause thin films to appear darker or even completely black, depending on the thickness of the film and the wavelengths of light involved. This is because the canceled out waves result in a decrease in the amount of light that is reflected back to our eyes.

3. What factors can affect destructive interference in thin film interference?

The thickness of the thin film, the refractive index of the material, and the wavelengths of light involved are all factors that can affect destructive interference in thin film interference. Additionally, the angle at which the light hits the film and the polarization of the light can also play a role.

4. How is destructive interference used in practical applications?

Destructive interference in thin film interference is used in many practical applications, such as anti-reflective coatings on glasses and camera lenses, thin film solar cells, and optical filters. In these cases, the goal is to control the destructive interference to either enhance or reduce certain wavelengths of light.

5. Can destructive interference occur in other types of waves besides light?

Yes, destructive interference can occur in other types of waves besides light, such as sound waves and water waves. The same principles of wave interference apply, where waves can cancel each other out if they have opposite phases, resulting in a decrease in intensity.

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