Destructive Interference of a sheet of mica

In summary, the problem involves finding the thickness of a sheet of mica (n = 1.5) on top of glass (n = 1.8) that will produce destructive interference for light with a wavelength of 420 nm. The formula used is 2t/(wavelength/n) = 1/2, where t represents the thickness of the mica. The final answer is 58.3 nm.
  • #1
conniechiwa
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Homework Statement


A sheet of mica sits on glass (n = 1.8). How thick should the sheet of mica (n = 1.5) be if it is to be as thin as possible and still give rise to destructive interference for reflection of light in the blue part of the spectrum (use l = 420 nm)? (All parts of the mica except the bottom surface are still surrounded by air.)


Homework Equations


1/2 + 2t/(wavelength/n) = m
2t/(wavelength/n) = 1/2
t = (wavelength/n)/4


The Attempt at a Solution


t = (wavelength/n)/4
t = (420nm/1.8/)4
t = 58.3 nm

I'm not sure what I'm doing wrong...
 
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  • #2
Hi conniechiwa,

conniechiwa said:

Homework Statement


A sheet of mica sits on glass (n = 1.8). How thick should the sheet of mica (n = 1.5) be if it is to be as thin as possible and still give rise to destructive interference for reflection of light in the blue part of the spectrum (use l = 420 nm)? (All parts of the mica except the bottom surface are still surrounded by air.)


Homework Equations


1/2 + 2t/(wavelength/n) = m
2t/(wavelength/n) = 1/2
t = (wavelength/n)/4


The Attempt at a Solution


t = (wavelength/n)/4
t = (420nm/1.8/)4
t = 58.3 nm

I'm not sure what I'm doing wrong...

The interference comes about from considering two rays of light: one reflects off the air/mica interface (not entering the mica), and the other enters the mica and reflects off the mica/glass interface.

The n in the formula is due to the fact that the wavelength in the mica is smaller than the wavelength in air. Do you see what you did wrong in the above work?
 
  • #3
Yeah I get it now. Thanks!
 

1. What is destructive interference?

Destructive interference occurs when two waves of equal frequency and amplitude overlap and cancel each other out, resulting in a decrease or total elimination of the wave's amplitude.

2. How does a sheet of mica cause destructive interference?

A sheet of mica is a thin, transparent mineral that has parallel layers. When waves of light pass through the mica, they are reflected and refracted by each layer, resulting in destructive interference as the waves overlap and cancel each other out.

3. What factors affect the degree of destructive interference in a sheet of mica?

The thickness of the mica sheet, the angle at which the waves pass through the sheet, and the wavelength of the waves all play a role in the degree of destructive interference that occurs.

4. How is destructive interference of a sheet of mica useful in scientific research?

Destructive interference of mica can be used to study the properties of light and the behavior of waves. It is also used in various optical devices, such as polarizers and filters, to manipulate and control the intensity and direction of light waves.

5. Can destructive interference occur in other materials besides mica?

Yes, destructive interference can occur in any material that has layers or barriers that reflect and refract waves, such as thin films, gratings, and diffraction gratings. It is a common phenomenon in optics and is also observed in other fields, such as acoustics and quantum mechanics.

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