MgF2 Thin Film Reflection: Visible Intensification?

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SUMMARY

The discussion centers on the optical properties of a thin film of magnesium fluoride (MgF2) with a thickness of 1.00x10^-5 cm and a refractive index of 1.38, specifically regarding its effect on visible light reflection. The key equation used is thickness = wavelength/4n, which relates the film thickness to the wavelengths that will experience constructive interference. It is established that certain wavelengths in the visible spectrum can be intensified due to constructive interference, while others may be diminished due to destructive interference. The phase shifts occurring at the air/MgF2 and MgF2/glass interfaces are crucial for determining the conditions for constructive interference.

PREREQUISITES
  • Understanding of thin film interference principles
  • Familiarity with the refractive index and its implications
  • Knowledge of wavelength and frequency relationships in optics
  • Basic grasp of phase shifts in wave reflections
NEXT STEPS
  • Explore the concept of thin film interference in detail
  • Learn about the calculation of wavelengths that lead to constructive interference in thin films
  • Investigate the effects of varying refractive indices on light reflection
  • Study the principles of phase changes during reflection at boundaries
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Optics students, physicists, and engineers involved in lens design and coatings, as well as anyone interested in the practical applications of thin film interference in enhancing optical performance.

FizX
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Homework Statement


A thin 1.00x10^-5 cm-thick film of MgF2 (n=1.38) is used to coat a camera lens. Are there any wavelengths in visible spectrum intensified in reflected light?


Homework Equations


thickness = wavelength/4n (n is refraction index)
?


The Attempt at a Solution


I am guessing that the incoming light in the material reflects inside the material to cause constructive interference at certain wavelengths (which will be multiples of the thickness of the MgF2).
My main question is, how do I relate possible wavelengths to thickness of the material?
 
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FizX said:

Homework Statement


A thin 1.00x10^-5 cm-thick film of MgF2 (n=1.38) is used to coat a camera lens. Are there any wavelengths in visible spectrum intensified in reflected light?


Homework Equations


thickness = wavelength/4n (n is refraction index)
?


The Attempt at a Solution


I am guessing that the incoming light in the material reflects inside the material to cause constructive interference at certain wavelengths (which will be multiples of the thickness of the MgF2).
My main question is, how do I relate possible wavelengths to thickness of the material?
Reflection cannot add to the incident light intensity. What happens is that some frequencies are not reflected as intensely due to destructive interference within the wave coating so the frequencies which experience constructive interference appear to be more intense.

So the question is: at what frequency does the light reflected from the first surface (Air/MgF2) constructively interfere with light reflected from the second surface (MgF2/Glass).

The reflections at both surfaces undergo a phase change of 180 so these phase shifts cancel out. In order to have constructive interference, what must the path from surface 1 to 2 and back to 1 represent in terms of wavelength of the light in the MgF2 medium?

Work out the frequency of the light from the wavelength.

AM
 

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