Question

5. Wave Optics – Thin Films

A thin film 4.0X10^-5cm thick interfaces with air (n=1) on each its front and back surfaces and is illuminated by white light normal to its front surface. Its index of refraction is 1.5. What wavelengths within the visible spectrum will be intensified in the reflected beam? (visible light is taken to be wavelengths between 390 to 700 nm) Follow the solution outline below:

1. Sketch the situation and label the index of refraction on each side of each interface (you will be graded on this)
2. Given the phase of the incoming wave and knowing that the transmitted wave does not experience a phase change, what is the phase change of the reflected wave at each interface of the thin film?
   • First interface: (identify) change, same
   • Second interface: (identify) change, same
   • How many phase changes total: (identify) 0 or 2, 1

1. Greater intensity of the reflected beam corresponds to which type of interference?
2. Choose the thin film equation for the type of interference and number of phase changes of this situation and solve for the wavelength (s) in the visible range. (may need to evaluate multiple m’s)

Answer 1

At top interface , reflectd light gets a phase difference equivalent to a path difference .

Where is wavelength of the incident light radiation . This is because reflection takes place when light ray incident on a medium of higher refractive index compare to medium of incoming light ray

At bottom interface , light is reflected by a medium of lower index of refraction, there is no phase difference and hence no path difference.

Reflected light is intensified , due to constructive interference between reflected light at top inteface and reflected light at bottom interface. Constructive interference is possible if path difference between the reflected light rays equal m , where m = 0,1,2... etc

For constructive interference relected light should get path difference /2 , 3/2 , 5/2 etc. since there is no phase difference for reflected light at bottom interface, optical path length travelled by light ray in the medium of thin film should get this path difference . Since light ray travels the depth of film twice, for constructive interference we should have , ( n t ) = (2m+1) /4 , where n is refractive index of thin film material and m = 0, 1, 2 etc

Let m = 0 ,   n t = /4   , hence = 4 n t = 4 1.5 0.4 10-6 = 2.4 10-6 m = 2400 nm

Above wavelength is not in visible range.

Let m = 1, n t = 3/4 , hence = (4 /3 ) n t = ( 4 / 3 ) 1.5 0.4 10-6 = 0.8 10-6 m = 800 nm

Above wavelength is not in visible range.

Let m = 2, n t = 5/4 , hence = (4 /5 ) n t = ( 4 / 5 ) 1.5 0.4 10-6 = 0.48 10-6 m = 480 nm

Hence light of wavelength 480 nm is intensified due to constructive interference