Irodov Problem 5.81
Physics · optics
Problem
To decrease light losses due to reflection from the glass surface the latter is coated with a thin layer of substance whose refractive index n' = -17n where n is the refractive index of the glass. In this case the amplitudes of electromagnetic oscillations reflected from both coated surfaces are equal. At what thickness of that coating is the glass reflectivity in the direction of the normal equal to zero for light with wavelength X?Intuitive explanation
The key insight is that destructive interference between reflections from the two surfaces (air–coating and coating–glass boundaries) can cancel reflected light entirely. Since the coating has a negative refractive index, the reflection phase at each surface is already controlled; you need to choose the coating thickness so that the two reflected waves arrive back at the air interface exactly out of phase. With equal reflection amplitudes guaranteed by the given refractive index relationship, destructive interference occurs when the extra path length traveled in the coating creates a specific phase difference—typically an odd multiple of half-wavelengths in that medium.
Intuitive explanation
The key insight is that destructive interference between reflections from the two surfaces (air–coating and coating–glass boundaries) can cancel reflected light entirely. Since the coating has a negative refractive index, the reflection phase at each surface is already controlled; you need to choose the coating thickness so that the two reflected waves arrive back at the air interface exactly out of phase. With equal reflection amplitudes guaranteed by the given refractive index relationship, destructive interference occurs when the extra path length traveled in the coating creates a specific phase difference—typically an odd multiple of half-wavelengths in that medium.