Question

You are standing in air and are looking at a flat piece of glass on which there is a layer of transparent plastic . Light whose wavelength is 589 nm is vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

Answer 1

t_min,1 = 183 nm

t_min,2 = 366 nm

Step-by-step explanation:

Given:-

- The wavelength of light , λ = 589 nm

- The refractive index of glass, ng = 1.52

- The refractive index of transparent plastic, np = 1.61

Find:-

Find the two smallest possible nonzero values for the thickness of the layer.

Solution:-

- The concept behind this problem is of destructive interference of light ( "looks dark" ) that occurs when the path difference between lights is odd multiples of 1/2 wavelengths.

- When light travels from air to plastic the phase difference of incident light is:

λ / 2

- When the light falls on plastic-glass boundary, there is no phase change, hence the net change in phase is λ / 2.

- So for destructive difference we have the following relation for thickness (t):

2t + λ / 2 = ( m + 0.5)*λ

- We will solve for first order of destructive interference, m = 1 as first possible smallest value:

2t = 1.5*λ - λ / 2 = λ

t = λ / 2*np

t_min,1 = 589 / 2*1.61

t_min,1= 183 nm

- We will solve for first order of destructive interference, m = 2 as second possible smallest value:

2t = 2.5* λ - 0.5*λ = 2 λ

t = λ /np

t_min,2 = 589 / 1.61

t_min,2 = 366 nm