How Thick is the Mica in the Double-Slit Experiment?

[kreil]

1. The problem statements, all variables and given/known data
1) A think flake of mica (n=1.58) is used to cover one slit of a double-slit arrangement. The central point on the screen is now occupied by what had been the seventh bright side fringe (m=7) before the mica was used. If the wavelength=550nm, what is the thickness of the mica?

2)The reflection of perpendicularly incident white light by a soap film in air has an interference maximum at 600nm and a minimum at 450nm, with no min inbetween. If n=1.33 for the film, what is the film thickness, assumed uniform?

Homework Equations

2nt=m \lambda (for destructive interference)

2nt=(m+ \frac{1}{2}) \lambda (for constructive interference)

d sin(\theta)=m \lambda (condition for bright fringes)

d sin(\theta)= (m+ \frac{1}{2}) \lambda (condition for dark fringes)

The Attempt at a Solution

1) What confuses me here is that the mica only covers one slit. How is this taken into account? Can I just use the equations above?

2) Very lost on this one. Do I use the fringe equations to solve for m, which can then be used in another equation to solve for thickness.

Any other info on these questions is more than welcome.

Thanks,
Josh

 

[OlderDan]

Let's get one of these before we try doing two of them.

An important concept in interference is that of the "optical path" of light. The optical path is the number of wavelengths light must travel between two points. So the optical path of light in a material with index of refraction > 1 is greater for a given physical length than it is in air or vacuum. In your first problem, before the mica is put in place the optical path difference between the rays from the two slits is zero at the central maximum, one wavelength at the adjacent maximum, then two at the next, etc. At the seventh bright side fringe (m = 7) the optical path difference is 7 wavelengths. If that fringe is moved to center by the addition of the mica to one slit, then the optical path difference between the ray through the mica and the ray that bypasses the mica must be 7 wavelengths.

 

[kreil]

I don't understand exactly how that relates to the question. Could you explain?

 

[OlderDan]

I don't understand exactly how that relates to the question. Could you explain?

Because the index of refraction of mica is > 1, the light that goes through the mica covered slit has to travel additional wavelengths compared to the light that goes the same distance through air to the uncovered slit. In your problem you are being told that the number of additional wavelengths in the mica path is seven.