Question

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 535 nm ? The index of refraction of the film is 1.32, and there is air on both sides of the film.

Answer 1

Answer:

0.203 micro meter

Explanation:

for destructive interference that appearsblack, use the formula

2 t = m λ / u (where m = 0 1 2 3 ... is order of minima)

where t = tickness,

u is the ref index = 1.32

Wavelenth λ = 535×10^-9 meter

for t (minimum) m = 1 (as m=0 is ruled out as t>0)

t = 1× 535×10^-9/2×1.32

t (min) = 202.65×10^-9 meter

OR

t (min) = 0.203×10^-6 meter = 0.203 micro meter

Answer 2

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

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.

                 $\lambda = \frac{\lambda_o}{n}$  

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 =$\frac{m \lambda }{2n}$  

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

        t = $\frac{1 \ 535 }{2 \ 1.32}$    

        t = 202.65 nm

Let's reduce this amount to millimeters.

         t = 202.65 nm  $(\frac{1m}{10^9 nm} ) ( \frac{10^3 mm}{1 m} )$

         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

Learn more here:  brainly.com/question/10179927