A soap bubble, diameter 3.0 cm, containing helium floats in the air. Determine the thickness of the bubble, as a multiple of the wavelength, 650 nm, of red light, assuming soap solution has a density of 1.00 x 10^3kg m-3. The density of the air and helium are, respectively, 1.28 kg m-3and 0.18 kgm-3.

### 1 Answer | Add Yours

The soap bubble of diameter 3.0 cm, containing helium floats in the air. The soap solution has a density of 1.00 x 10^3 kg m^-3. The density of the air and helium are, respectively, 1.28 kg m^-3 and 0.18 kg m^-3.

The problem has been solved with the assumption that the bubble is a sphere. If the bubble floats, the combined density of the helium and the soap is equal to that of air.

A sphere with radius 1.5 cm has volume `(4/3)*pi*(1.5/100)^3` m^3. The surface area of the sphere is `4*pi*(1.5/100)^2` .

If the thickness of the bubble is t, the mass of soap solution that forms the bubble is `4*pi*(1.5/100)^2*t*1.00*10^3` . The mass of the bubble is `4*pi*(1.5/100)^2*t*1.00*10^3` + `(4/3)*pi*(1.5/100)^3*0.18`

Equating this to the mass of a sphere of radius 1.5 cm gives:

`4*pi*(1.5/100)^2*t*1.00*10^3 + (4/3)*pi*(1.5/100)^3*0.18` = `(4/3)*pi*(1.5/100)^3*1.28`

=> `t*10^3 + (1/3)*(1.5/100)*0.18 = (1/3)*1.28`

=> `t = 12773/30000000` m

As a multiple of the wavelength of red light at 650 nm, `12773/30000000 ~~ 655`

The thickness of the soap bubble is approximately 655 times the wavelength of red light.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes