Question

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 550 nm

Answer 1

Answer:

The expression for destructive interference in thin films allows to find the result for the smallest thickness of the films is:

         t = 2.03 10⁻⁸ mm

Explanation:

Given parameters

Incident wavelength lamo = 535 nm

Refractive index of the film n = 1.32

To find

The minimum thickness for destructive interference

The interference phenomenon occurs when the path of two rays scattered by an obstacle have different optical paths. In the case of thin films we must take into account:

The reflected wave has a phase change of 180º when it goes from a medium with a lower refractive index to a medium with a higher index.

Inside the film medium the wavelength is modulated by the refractive index.

                 

In the attachment we see an outline of these events and the expression for destructive interference remains.

               2 n t = m λ₀

Where n is the refractive index, t the thickness of the film, λ₀ the wavelength in the vacuum and m an integer indicating the order of interference.  

             t =  

The first destructive interference occurs for m = 1, let's calculate.

       t =    

       t = 202.65 nm

Let's reduce this amount to millimeters.

        t = 202.65 nm  

        t = 2,027 10⁻⁸ mm

In conclusion, using the expression for destructive interference in thin films we can find the result for the smallest thickness of the films is:

         t = 2.03 10⁻⁸ mm