Question

A light wave with a wavelength of - 1.46) on water (1.33). Assign a value to A (475 nm) 3. is normally incident (i.e., hits at 90°) on a thin layer of oil (n a) What is the wavelength of the light wave as it travels through the oil (3 points)? Some of the light reflects at the air-oil boundary, and some of the light reflects at the oil- water boundary. At which boundaries will the reflected light wave experience a phase shift? Explain your reasoning (3 points). b) c) What minimum film thickness is necessary in order for the interaction between the reflected light waves to result in total deconstructive interference (3 points)?

Answer 1

a) The wavelength of the light as it travels through the oil is

$\lambda_{oil}=\lambda/n_{oil}=475/1.46=325.34\,nm$

b) The phase shift due to reflection occurs when the light is incident on a medium that has a larger refractive index than the medium from which the light is traveling. Hence there will be a phase shift upon reflection from the air -oil boundary. Because the refractive index of oil is larger than the refractive index of air.

c) The condition for the total deconstructive interference of reflected light is

$\2n_{oil}d=m\lambda$

where m is an integer. For m = 1, one obtains the minimum thickness for the oil film

$d=\lambda/(2n_{oil})=475/(2\times 1.46)=162.67\,nm=162.67\times 10^{-9}\,m$