How Is Film Thickness Calculated from Interference Patterns?

[timhunderwood]

Homework Statement

Light of wavelength $$\lambda$$ = 500nm, produced by an extended source, is incident at an angle of $$\phi$$= 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 $$\delta$$ = 2ndcos$$\theta$$*(2*pi/$$\lambda$$) where sin$$\phi$$=nsin$$\theta$$

Intensity proportional to sin²($$\delta$$/2)

The Attempt at a Solution

0th and first minima occur when the $$\delta$$=0 and $$\delta$$=2 pi respectively.

Hence:
2ndcos$$\theta$$*(2*pi/$$\lambda$$) = 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

 

[Jhero]

.

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.