Question

A sheet of glass is coated with a 500-nm-thick layer of oil 1n = 1.422.

a. For what visible wavelengths of light do the reflected waves interfere constructively?

b. For what visible wavelengths of light do the reflected waves interfere destructively?

c. What is the color of reflected light? What is the color of transmitted light?

Short answer

a) The visible wavelength is (473, 710) nm

b) The visible wavelength is (406, 568) nm

c) Same as before hitting the film.

Step-by-step solution

The concept of reflected waves 

The changes in the direction of propagation of a wave when it meets the boundaries which follow the law of reflection.

Understanding the wavelengths and their constructive and destructive interference. 

The constructive interference in the traveling beam in the film will be multiple of $2\mathrm{\pi }$.

The thickness is d and the wavelength is role="math" localid="1649150515694" $\lambda f=\frac{\lambda }{n_f}$

It can be written as $∆\varnothing =2\mathrm{\pi }\frac{2\mathrm{d}}{{\mathrm{\lambda }}_{\mathrm{f}}}=\mathrm{m}·2\mathrm{\pi }$

Thus, $\frac{2\mathrm{dnf}}{\mathrm{\lambda }}=\mathrm{m}=\mathrm{\lambda }=\frac{2{\mathrm{dn}}_{\mathrm{f}}}{\mathrm{m}}$

The m integer is missing then it is ${\lambda }_{min}=\frac{2·5·{10}^{-7}·1.42}{m_{min}}=4·{10}^{-7}\Rightarrow m_{min}=3.55$

Similarly ${\lambda }_{max}=\frac{2·5·{10}^{-7}·1.42}{m_{max}}=8·{10}^{-7}\Rightarrow m_{max}=1.775$

Thus, when the values are m=2 and m=3 the wavelengths will be role="math" localid="1649150962999" $\boxed{{\lambda }_1=710nm,{\lambda }_2=473nm}$

The destructive interference is m=2.5 and m=3.5 Thus, the interference will be $\boxed{{\lambda }_1=568nm,{\lambda }_2=406nm}$

Once the light goes back to the air the length becomes the similar.