Final answer:
Wavelengths of visible light most strongly destroyed from a soap bubble with a wall thickness of 600 nm are those satisfying the condition of destructive interference, calculated using the bubble's thickness and index of refraction.
Step-by-step explanation:
Conditions for Destructive Interference in Soap Bubbles
The question involves thin film interference, a phenomenon where light waves reflected from the top and bottom surfaces of a thin film, like a soap bubble, interfere with each other. The condition for destructive interference is that the reflected waves are out of phase by λ/2 (where λ is the wavelength of the light in the medium). In the case of a soap bubble with a wall thickness of 600 nm and an index of refraction of n=1.33, the same as plain water, we look for wavelengths that will result in destructive interference in air.
The path difference for the two reflected rays must be an integer multiple of the wavelength (mλ) plus half a wavelength (λ/2) for destructive interference to occur, essentially being mλ + λ/2. Considering the film is within the bubble, we must account for the fact that the light travels a distance of 2t within the bubble wall, where t is the thickness of the wall. Therefore, to find the specific wavelengths in air, we use the adjusted thickness (2t/n) and set it to mλ + λ/2 for the first order of destructive interference (m=0), which yields 2t/n + λ/2. Using 600 nm for t, we find the wavelengths in air that satisfy this condition fall within the visible spectrum of 400nm to 700nm.
The most strongly destroyed wavelength in air for this scenario would be the one closest to a half-integer multiple (including zero) of the wavelength in the bubble medium. By plugging in the values, we can solve for the specific wavelengths that result in destructive interference, and thus, are 'destroyed' or minimized in the reflected light.