Question

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 535 nm ? The index of refraction of the film is 1.32, and there is air on both sides of the film.

Answer 1

Answer:

0.203 micro meter

Explanation:

for destructive interference that appearsblack, use the formula

2 t = m λ / u (where m = 0 1 2 3 ... is order of minima)

where t = tickness,

u is the ref index = 1.32

Wavelenth λ = 535×10^-9 meter

for t (minimum) m = 1 (as m=0 is ruled out as t>0)

t = 1× 535×10^-9/2×1.32

t (min) = 202.65×10^-9 meter

OR

t (min) = 0.203×10^-6 meter = 0.203 micro meter