Calculate Antireflective Coating Thickness for Blue Light on Glass Lens

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SUMMARY

The discussion focuses on calculating the minimum thickness required for an antireflective coating with a refractive index of 1.35 to eliminate reflections of blue light at a wavelength of 458 nm. The method involves applying the principle of destructive interference. The formula used for this calculation is straightforward, emphasizing the relationship between the wavelength of light and the coating thickness.

PREREQUISITES
  • Understanding of destructive interference principles
  • Knowledge of refractive indices, specifically for coatings
  • Familiarity with wavelength measurements in nanometers
  • Basic optics concepts related to light reflection and transmission
NEXT STEPS
  • Research the formula for calculating antireflective coating thickness
  • Explore the effects of different refractive indices on light reflection
  • Learn about the application of coatings in optical devices
  • Investigate the principles of interference in thin films
USEFUL FOR

Optical engineers, physicists, and anyone involved in the design and application of antireflective coatings on lenses and other optical components.

Jodi
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HI; Could someone please help me with this question: Calculate the minimum thickness needed for an antireflective coating (n=1.35) applied to a glass lens in order to eliminate blue (458 nm) reflections for light at normal incidence? How would I go about this question? Thanks.
 
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Use destructive interference.
 
You should have a very very simple formula to figure this out.
 

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