Question

A good quality camera "lens" is actually a system of lenses, rather than a single lens, but a side effect is that a reflection from the surface of one lens can bounce around many times within the system, creating artifacts in the photograph. To counteract this problem, one of the lenses in such a system is coated with a thin layer of material (n=1.28) on one side. The index of refraction of the lens glass is 1.68. What is the smallest thickness of the coating that reduces the reflection at 640 nm by destructive interference? (In other words, the coating’s effect is to be optimized for λ=640nm.)

Answer 1

The smallest thickness of a coating for destructive interference at a wavelength of 640 nm is found by dividing the wavelength by four times the refractive index of the coating. The resulting thickness causes a path difference of half a wavelength within the coating, leading to destructive interference and reduced reflection.

Step-by-step explanation:

To determine the smallest thickness of a coating that reduces reflection at a wavelength of 640 nm by destructive interference, we follow the principle of thin-film interference. The condition for destructive interference in thin films occurs when the path difference between the two reflected rays equals half an odd number of wavelengths inside the film (mλ/2, where m is an odd integer). Here's how we calculate it:

1. 
   
3. Determine the wavelength of light within the film by dividing the vacuum wavelength by the index of refraction of the coating: λfilm = λ / n = 640 nm / 1.28.
4. 
   
6. For minimum thickness for the first order (m=1) of destructive interference, the thickness (t) should be such that the path difference is half the wavelength inside the film: 2t = λfilm / 2.
7. 
   
9. Calculate the thickness: t = λfilm / 4.

By following these steps, we can calculate the smallest thickness of the coating to achieve the desired effect for a given wavelength.