Calculate the minimum thickness of a soap film (n= 1.33) that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is 532 nm.

Options

A. 113nm
B. 100nm
C. 200nm
D. 226nm

Answer

B

Explanation

For constructive interference through a thin film in reflected light, thickness of film must be
`t= (m + (1)/(2)) (lamda)/(2n_(2))`
where, m = 0, 1,2….
and `n_(2)` = refractive index of medium (soap film) With given values, we get
`t = (lamda)/(4n_(2))` (`i_("min")` occurs with m = 0)
`= (532 xx 10^(-9))/( 4 xx 1.33) = 100nm`