Question

What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is $\text{750 nm}$? Assume that the refractive index for the film is $\mu =\mathrm{1.33.}$

• A. $\text{282 nm}$
• B. $\text{70.5 nm}$
• C. $\text{141 nm}$
• D. $\text{387 nm}$

Answer 1

The correct option is C $\text{141 nm}$

Given: The wavelength of the light incident on the soap film is $750nm$.

The refractive index of the film is $1.33$.

To find: The minimum thickness of the soap film required for constructive interference.

The thickness of the film for the constructive interference is given by:

$2t=(m+\frac{1}{2}){\lambda }^{\prime }$

The new wavelength in the medium is given as:

${\lambda }^{\prime }=\frac{\lambda }{\mu }$

So, the minimum thickness of the film ${}^{\prime }{t}^{\prime }$ will be at $m=0$ :

$2t_{min}=(0+\frac{1}{2})\frac{750\times {10}^{-9}}{1.33}$

$t_{min}=\frac{750\times {10}^{-9}}{4\times 1.33}$

$=1.41\times {10}^{-7}or141nm$