Interference by amplitude division (double reflection) for wedges

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SUMMARY

The discussion focuses on the interference of light due to amplitude division in a wedge formed by transparent wax (n=1.3) on a glass plate (n=1.5). The thickness of the wax varies from 0.01mm to 0mm, creating a wedge angle α calculated as tan(α) = 0.01/10 = 1×10-3 radians. Bright fringes occur where the phase difference between light reflected from the wax and glass is an integer multiple of 2π, leading to the formula for fringe positions: d = kλ / (2n), where k is an integer, λ is the wavelength (520nm), and n is the refractive index.

PREREQUISITES
  • Understanding of wave optics and interference patterns
  • Knowledge of refractive indices and their implications in phase changes
  • Familiarity with the concept of wedge-shaped films in optics
  • Basic algebra for manipulating equations related to light interference
NEXT STEPS
  • Study the derivation of the interference condition for thin films
  • Learn about the effects of varying refractive indices on light reflection
  • Explore the application of the formula d = kλ / (2n) in different optical setups
  • Investigate the phenomenon of phase shifts upon reflection in various media
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Students and educators in physics, particularly those focusing on optics, as well as researchers exploring interference patterns in thin films and wedges.

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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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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
 
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:
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
 
Thank you so much!
 
You are welcome.ehild
 

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