Discussion Forum

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

Answer:

Explanation:

 For constructive interference  through a thin film  in reflected ligh, thickness  of film  must be

 $t=\left(m+\frac{1}{2}\right)\frac{\lambda}{2n_{2}}$

 where, m=0,1,2,.....

 and $n_{2}$ = refractive index  of medium (soap film) with given values, we get

 $t= \frac{\lambda}{4n_{2}}$                   ( $t_{min}$ occurs with m=0)

 = $\frac{532 \times 10^{-9}}{4 \times 1.33}=100 n m$