Question

A soap bubble is essentially a thin film of water surrounded by air. the colors you see in soap bubbles are produced by interference. part a what visible wavelengths of light are strongly reflected from a 390-nm-thick soap bubble?

Answer 1

Visible wavelengths of light that strongly reflected from a 390-nm-thick soap bubble are  the complementary color to blue-violet

Further explanation

Electromagnetic radiation in this range of wavelengths is called visible light. A typical human eye will respond to wavelengths about 380 to 740 nanometers. In terms of frequency, this corresponds to a band in the vicinity of 430–770 THz. This narrow band of visible light is affectionately known as ROYGBIV - red (R), orange (O), yellow (Y), green (G), blue (B), and violet (V).

According to the Visible wavelengths spectrum, The color you would see of 390-nm would be the complementary color to blue-violet. The red wavelengths of light are the longer wavelengths and the violet wavelengths of light are the shorter wavelengths. Between red and violet there is a continuous range or spectrum of wavelengths.

Hence 390 nm thick soap bubble is around the same thickness as the violet-indigo-blue end of the spectrum and will reflect these colors. The color reflected is not the color seen, as light waves reflect off both the inside and outside and then recombine out of phase and interfere with each other and cancelling out.

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1. Learn more about  Visible wavelengths brainly.com/question/5499524

Answer details

Grade:  9

Subject:  physics

Chapter:  wavelengths

Keywords:  Visible wavelengths

Answer 2

The colors we see in the bubble are produced by the interference between two waves: a) the wave reflected by the surface of the bubble b) the wave that travels inside the bubble and it is reflected by its back. 
The condition for the constructive interference is that the phase shift between the two waves is an integer multiple of $2\pi$. The phase difference between the two waves is
$\Delta \phi = k \Delta x - \pi$
where $\Delta x = 2t$ is twice the thickness of the bubble (since the second wave travels inside the bubble and it is reflected from the back), while $\pi$ is due to the fact that the first wave has an extra phase shift $\pi$ because it is reflected from a material (soap) with higher refraction index than the air.

So, in order to have constructive interference, we should require
$k \Delta x - \pi = 2m\pi$
where m is an integer. Substituting $\Delta x=2t$ and $k= \frac{2 \pi}{\lambda}$, we have
$\lambda = \frac{2t}{m+ \frac{1}{2} }$
But here $\lambda$ is the wavelength in the soap; we need instead the wavelength in the air, which is 
$\lambda' = n \lambda$
where n=1.33 is the refraction index.

Therefore, we have
$\lambda' = \frac{2nt}{m+ \frac{1}{2} }$

Using t=390 nm, and using different values of m, we find tha only m=1 and m=2 have wavelength in the visible spectrum: $\lambda'=415 nm$ and $\lambda'=692nm$. These wavelengths correspond to red and violet, so the bubble appears as red-violet.