Interference by amplitude division (double reflection) for wedges

In summary, the transparency of wax allows light to pass through it and create fringes on a glass plate. The phase difference between light that passes through the wax and light that passes through the glass determines where the fringes will be.
  • #1
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Homework Statement



Transparent wax of refractive index n=1.3 is deposited on top of a glass plate of width 1cm and refractive index n=1.5. The thickness of the wax is 0.01mm at one end of the plate and tapers uniformly to zero at the other end of the plate, which is defined to be at x=0. At this end the surface of the wax and of the glass form a small angle α. Light is incident on the plate from above, i.e. it goes through the wax, and is normal to the surface of the glass plate. Fringes are formed due to interference of light reflected from the top surface of the wax and from the glass.

i) Find the value of α.
ii) At what values of x do bright fringes occur if λ=520nm?

Homework Equations



Thickness of wedge, d = x tan(α) ≈ αx

The Attempt at a Solution



i) tan(α)=0.01/10 = 1×10-3 radians

ii) I'm not sure whether bright fringes occur at xn = ((ρ+0.5)λ)/2nα) or xn = (ρλ/2nα)
 
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  • #2
You see the wedge from above. The bright fringes occur at those places where the phase difference between the ray reflected from the wax and that reflected from the glass makes integer times 2pi. The phase can change upon reflection and during traveling through a medium. The phase change upon reflection is pi when the wave arrives from a lower index medium to the surface of a higher index material, and it is zero in the opposite case. What is the phase change when the wave reflects from the wax and what is it when it reflects from the glass?

ehild
 
  • #3
ehild said:
You see the wedge from above. The bright fringes occur at those places where the phase difference between the ray reflected from the wax and that reflected from the glass makes integer times 2pi. The phase can change upon reflection and during traveling through a medium. The phase change upon reflection is pi when the wave arrives from a lower index medium to the surface of a higher index material, and it is zero in the opposite case. What is the phase change when the wave reflects from the wax and what is it when it reflects from the glass?

ehild

Does that mean the phase change when light reflects from wax = pi and the phase change when light reflects from glass = pi?
How would I use this to find the values of x for which bright fringes occur?
 
Last edited:
  • #4
As the phase changes at the interfaces cancel, the total phase difference is the phase shift inside the layer. It is
(4pi/λ) nd =2kπ,
where k is integer and d is the thickness of the layer. That means that bright fringes occur where the thickness is d=kλ /2n.

ehild
 
  • #5
Thank you so much!
 
  • #6
You are welcome.ehild
 

1. What is the principle behind interference by amplitude division for wedges?

The principle behind interference by amplitude division for wedges is that when a wave is reflected off two surfaces, the reflected waves interfere with each other, resulting in a pattern of constructive and destructive interference.

2. How does the shape of the wedge affect interference by amplitude division?

The shape of the wedge affects interference by amplitude division because it determines the angle at which the waves are reflected and the path length difference between the two reflected waves. This can result in different patterns of interference.

3. What is the difference between single and double reflection interference?

Single reflection interference occurs when a wave is reflected off a single surface, while double reflection interference occurs when a wave is reflected off two surfaces. In double reflection interference, the reflected waves interfere with each other, resulting in a more complex pattern.

4. Can interference by amplitude division for wedges be observed in real-life situations?

Yes, interference by amplitude division for wedges can be observed in real-life situations. Examples include the interference patterns seen in oil films on water, the colors seen in soap bubbles, and the colors seen in thin layers of oil on the surface of water.

5. How is interference by amplitude division used in practical applications?

Interference by amplitude division is used in various practical applications, such as in optical coatings to create anti-reflective surfaces or in electronic devices such as optical filters and polarizers. It is also used in techniques such as interferometry for measuring small changes in distance or wavelength.

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