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## Homework Statement

Monochromatic laser light is shown through a double-slit apparatus with a slit separation of onto a screen, but no

interference pattern is seen because the thin transparent coating (index of refraction n=1.35) on the screen is creating a thin-film destructive interference effect. This is known because when the light is shown on an old uncoated screen, they saw an interference pattern with the first bright fringe deflected at an angle of 3º.Find the minimum thickness of coating that needs to be scrubbed off so that the interference pattern for that light will

appear as brightly on the screen as if it were uncoated (i.e. no destructive interference at all).

## Homework Equations

[tex]dsin(\theta) = m\lambda \\

t = \frac{\lambda}{2n} (constructive)\\

t = \frac{\lambda}{4n} (destructive)

[/tex]

## The Attempt at a Solution

We know that m = 1 at theta = 3 degrees since that's where the first interference pattern is, so the first equation becomes lambda = d*sin(theta). Plug in lambda = d*sin(theta) into t = lambda / 4*n to find the thickness at which destructive interference occurs, so t = d*sin(theta)/(4*n) (where n = 1.35) and get 106.6 nm. If 106.6nm of the coating is removed, then constructive interference on the film will occur, so 106.6nm has to be removed.