Question

Light of frequency 6.00 × 1014 hz illuminates a soap film (n = 1.33) having air on both sides of it. when viewing the film by reflected light, what is the minimum thickness of the film that will give an interference maximum when the light is incident normally on it? (c = 3.00 × 108 m/s)

Answer 1

Ans: Minimum thickness of the film = t = 94nm

Explanation:

The condition for the (constructive) interference is:
$2nt = ( m + \frac{1}{2})\lambda$ --- (1)

n = refractive index = 1.33
t = thickness of the film = ?
m = integer = for minimum thickness, m = 0
λ = wavelength = $\frac{c}{f}$ = $\frac{3*10^8}{6*10^{14}}$ = $5 * 10^{-7}m$

Plug in the values in equation (1):
$2nt = (0 + \frac{1}{2} )\lambda$

$t = \frac{\lambda}{4n}$

$t = \frac{5*10^{-7}}{4*1.33}$

$t = 94*10^{-9}m$
t = 94nm