Question

A soap bubble is essentially a very thin film of water 1n = 1.332 surrounded by air. The colors that you see in soap bubbles are produced by interference.

a. Derive an expression for the wavelengths lC for which constructive interference causes a strong reflection from a soap bubble of thickness d. Hint: Think about the reflection phase shifts at both boundaries.

b. What visible wavelengths of light are strongly reflected from a 390-nm-thick soap bubble? What color would such a soap bubble appear to be?

Short answer

a) The constructive interference is ${\lambda }_c=\frac{4_{n_{w}d}}{4_{n+1}}$

b) The visible wavelength is 691. 6 nm dark red and 414 .96 nm dark blue.

Step-by-step solution

The concept of the wavelength 

The distance line between the waves is known as the wavelength.

Explanation of the equation 

The wavelength of the light is traveling with speed thus $2d=(2n+1)\mathrm{\pi }$

Thus, the formula of the wavelength is $2\mathrm{\pi }\frac{2\mathrm{d}}{{\mathrm{\lambda }}_{\mathrm{w}}}=(2\mathrm{n}+1)\mathrm{\pi }\Rightarrow \frac{4{\mathrm{n}}_{\mathrm{w}}\mathrm{d}}{{\mathrm{\lambda }}_{\mathrm{c}}}=2\mathrm{n}+1\Rightarrow \boxed{{\mathrm{\lambda }}_{\mathrm{c}}\Rightarrow \frac{4{\mathrm{n}}_{\mathrm{w}}\mathrm{d}}{2\mathrm{n}+1}}$

Thus, the spectrum is (400-800)nm thus the calculation of the value n is

$2\mathrm{n}+1=\frac{4{\mathrm{n}}_{\mathrm{w}}\mathrm{d}}{\mathrm{\lambda }}\Rightarrow n=\frac{4{\mathrm{n}}_{\mathrm{w}}\mathrm{d}-\mathrm{\lambda }}{2\lambda }$

the extremum will be $n_1=\frac{4·1.33·3.9·{10}^{-7}-4·{10}^{-7}}{2*4·{10}^{-7}}$

The other one will be $n_2=\frac{4·1.33·3.9·{10}^{-7}-8·{10}^{-7}}{2*8·{10}^{-7}}$

The integer values of n are 1 and 2. The interferences of the integers are 691.6nm and 414. 96 nm.