Question

If a soap film of refractive index 1.33, surrounded by air, is illuminated by a red (633 nm) and a green (515 nm) light, find the minimum film thickness at which the reflected light will appear red. Repeat to find the thickness at which is will appear green.​

Answer 1

The minimum thickness of the soap film for red light is approximately 238 nm, and for green light, it is approximately 194 nm, based on wave optics principles of thin film interference and constructive interference.

Step-by-step explanation:

To find the minimum thickness of a soap film at which the reflected light will appear red, we need to consider the wave optics principle of thin film interference. For constructive interference of reflected waves from the top and bottom surfaces of the film, the path difference must be a multiple of the wavelength inside the film.

For the red light (633 nm), the formula for constructive interference is given by:

1. mλ = 2nt, where m is the order of the interference (beginning with m=1 for the first minimum thickness), λ is the wavelength of light in vacuum, n is the refractive index, and t is the thickness of the film.

The wavelength inside the film is shorter by a factor of the refractive index, so λ' = λ/n. The minimum thickness (m=1) for red light is thus:

t = (mλ)/(2n) = (1 × 633 nm)/(2 × 1.33) ≈ 238 nm

Repeating the calculation for the green light (515 nm) gives:

1. t = (mλ)/(2n) = (1 × 515 nm)/(2 × 1.33) ≈ 194 nm