Forest ElemPhysics Optics ThinFilms

Prperties of Refection and Refraction

1.) Waves reflect from higher index of refraction materials will have a half wave shift.

2.) Frequency of light stays the same no matter what the medium change

3.) The wavelength of light changes with the index of refraction.

[math]n \equiv \frac{c}{v}[/math]

[math]\frac{\lambda_0} {\lambda} = \frac{v_o/f}{v/f} = \frac{c/n_0}{c/n} = \frac{n}{n_0} = n : n_0 =1 [/math]if air

Thin Film Applet

http://physics.bu.edu/~duffy/semester2/c26_thinfilm.html

Thin Film of MgF2 on glass

1.) [math]n_{air} = 1 \lt n_{MgF2} = 1.38 \lt n_{glass} = 1.5[/math]

This means that reflection on all the interfaces will have a half wavelength [math](\pi)[/math] phase shift because wave is reflected from a higher index of refraction both times.

Consider the interference pattern observed when the wave from the MgF2 recombines with the reflect wave in the air.

The path difference between the waves = (Path difference from traveling in MgF2) - (Path difference from reflection)

(Path difference from reflection) = [math]\lambda_0[/math]/2

(Path difference from traveling in MgF2) = 2 t = 2 [math]\times[/math] thickness of MgF2 film

BUT

The wavelength is shorter in the MgF2 than in the air.

Let's instead use the effective distance d_o to represent the distance traveled in the MgF2 film if the wavelength didn't shorten.

d_0 = n \time t

We can now treat the recombination of the light from the MgF2 as having the same wavelength as the light it is recombining with in air.

The path difference between the waves =[math] 2nt - \lambda_0/2[/math]

If we are interested in the

Conditions for DESTRUCTIVE Interference

Then [math]2nt - \lambda_0/2 = (m) \lambda_0[/math]

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