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## Homework Statement

The interference pattern formed is given by I(θ)=4I

_{0}cos

^{2}(∏dsin(θ)/λ). For d=1×10

^{-5}m and λ=500nm plot the intensity pattern as a function of θ for small θ. How would this change if a block of material of thickness 500nm and refractive index n=1.5 were placed over one slit (without altering the intensity from that slit)?

Please see the attachment

## Homework Equations

sin(θ)≈θ

I(θ)=4I

_{0}cos

^{2}(∏dsin(θ)/λ)

maybe these:

Maxima at d sin(θ)=mλ for m=0,1,2, ...

Minima at d sin(θ)=(m+0.5)λ for m=0,1,2, ...

## The Attempt at a Solution

1st maxima at θ=sin

^{-1}(0)=0 radians

1st minima at θ=sin

^{-1}(λ/2d)=0.025 radians

2nd maxima at θ=sin

^{-1}(λ/d)=0.05 radians

2nd minima at θ=sin

^{-1}(3λ/2d)=0.075 radians

3rd maxima at θ=sin

^{-1}(2λ/d)=0.1 radians

Then I subbed these into I(θ)=4I

_{0}cos

^{2}(∏dsin(θ)/λ) & plotted this with θ on the x-axis and I(θ) on the y-axis. I'm not sure is that's right but I can't seem to figure out the second part

#### Attachments

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