Thin film interference

Your Name].In summary, to find the thickness of the dielectric film in this problem, we can use the equation for phase difference and set it equal to the angular separation multiplied by 2*pi/\lambda. By solving for the angular separation and using it to find the thickness for the second minimum, we can determine that the thickness of the film is approximately 1453nm.
  • #1
timhunderwood
12
0

Homework Statement



Light of wavelength [tex]\lambda[/tex] = 500nm, produced by an extended source, is incident at an angle of [tex]\phi[/tex]= 30 degrees from the normal upon a dielectric film of refractive index, n=2, supported on a solid planar substrate. Reflectivity minima are observed to have an angular separation of 0.05 radians. What is the thickness, d, of the film?

Homework Equations



Phase difference [tex]\delta[/tex] = 2ndcos[tex]\theta[/tex]*(2*pi/[tex]\lambda[/tex]) where sin[tex]\phi[/tex]=nsin[tex]\theta[/tex]

Intensity proportional to sin2([tex]\delta[/tex]/2)



The Attempt at a Solution



0th and first minima occur when the [tex]\delta[/tex]=0 and [tex]\delta[/tex]=2 pi respectively.

Hence:
2ndcos[tex]\theta[/tex]*(2*pi/[tex]\lambda[/tex]) = 2pi

which gives d=258nm

which seems very small and I didn't have to use the angular separation to get my answer which leads me to believe I'm doing something wrong.

I think my mistake is to do with the fact I don't now how to involve the angular separation of 0.05 radians into the solution?

Help appreciated
 
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  • #2
.

Thank you for your post. Your approach to finding the thickness of the film is correct. However, you are right in thinking that the angular separation of 0.05 radians needs to be incorporated into the solution.

To do this, we can use the equation for the phase difference, \delta, and set it equal to the angular separation, \Delta\theta, multiplied by 2*pi/\lambda:

\delta = \Delta\theta*(2*pi/\lambda)

We know that for the first minimum, \delta=2pi, so we can substitute that into the equation:

2pi = \Delta\theta*(2*pi/\lambda)

Rearranging for \Delta\theta, we get:

\Delta\theta = \lambda/(2nd)

Substituting in the values given in the problem, we get:

\Delta\theta = (500nm)/(2*2*258nm) = 0.48 radians

This is the angular separation for the first minimum. We can then use this value to find the thickness of the film for the second minimum:

\Delta\theta = 0.48 radians = 0.05 radians + \Delta\theta_{2nd}

where \Delta\theta_{2nd} is the angular separation for the second minimum.

Solving for \Delta\theta_{2nd}, we get:

\Delta\theta_{2nd} = 0.48 radians - 0.05 radians = 0.43 radians

Finally, we can use this value to find the thickness of the film for the second minimum:

\Delta\theta_{2nd} = \lambda/(2nd)

0.43 radians = (500nm)/(2*2*d)

Rearranging for d, we get:

d = (500nm)/(2*2*0.43 radians) = 1453nm

So the thickness of the film is approximately 1453nm. I hope this helps clarify the solution. Let me know if you have any further questions.
 

1. What is thin film interference?

Thin film interference is a phenomenon that occurs when light waves reflect off of the top and bottom surfaces of a thin film. Depending on the thickness of the film, the light waves may interfere constructively or destructively, resulting in different colors or patterns.

2. How does thin film interference happen?

Thin film interference occurs due to the difference in refractive index between the film and the surrounding medium. When light waves travel through the film, they are partially reflected at each surface. If the reflected waves are in phase, they will interfere constructively and create a bright spot. If they are out of phase, they will interfere destructively and create a dark spot.

3. What is the difference between constructive and destructive interference in thin films?

Constructive interference in thin films occurs when the reflected waves are in phase and add together, resulting in a bright spot or color. Destructive interference occurs when the reflected waves are out of phase and cancel each other out, resulting in a dark spot or color.

4. What factors affect thin film interference?

The thickness of the film, the angle of the incident light, and the refractive index of the film and the surrounding medium all affect thin film interference. Thicker films and larger refractive index differences can result in more pronounced interference effects.

5. What are some real-world applications of thin film interference?

Thin film interference is commonly used in anti-reflective coatings on glasses and camera lenses to reduce glare. It is also used in the production of thin-film solar panels, as well as in the creation of colorful patterns on soap bubbles and oil slicks.

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