Question

White light is incident normally on a thin soap film having an index of refraction of 1.34. It reflects with an interference maximum at 684 nm and an interference minimum at 570 nm with no minima between those two values. The film has air on both sides of it. What is the thickness of the soap film?

Answer 1

Answer:

638 nm

Explanation:

n = index of refraction of soap

m = order

$\lambda _{max}$ = wavelength at maximum interference = 684 nm

$\lambda _{min}$ = wavelength at minimum interference = 570 nm

At constructive interference is given as

2nt = m $\lambda _{max}$                              eq-1

At destructive interference is given as

2nt = (m + 0.5) $\lambda _{min}$                   eq-2

using eq-1 and eq-2

m $\lambda _{max}$ = (m + 0.5) $\lambda _{min}$

m (684) = (m + 0.5) (570)

m = 2.5

using eq-2

2nt = (m + 0.5) $\lambda _{min}$    

2(1.34) t = (2.5 + 0.5) (570)

t = 638 nm