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This is problem 62 of chapter 23.
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For this problem, we're asked to analyze a glass wedge of a certain refractive index where a 5 -nometer wavelength of light is shined onto it, and we're looking for the thickness of the wedge such that constructive interference occurs.
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So let's go ahead and write down what we know.
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We know the refractive index of air.
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We're going to write them down as 1 .00.
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We have the refractive index of the glass.
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This is given as 1 .64.
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The wavelength is lambda equal to 500 nanometers.
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So this is equal to 500 times 10 to the negative 9 meters.
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Let's go ahead and draw our wedge.
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So we have a horizontal surface on the wedge.
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Here is the height of the wedge and then it comes all the way down here now this wedge is coated here at the bottom so it's coated with silver at the bottom and what that's going to do that's going to cause a reflection once the light hits this glass silver interface it's going to reflect away okay we're also given the height of the wedge as one times 10 to the negative 5 meters and what we're looking for so this is here the silver coating at the bottom of the plate what we're looking for is the distance from some origin so if we take our origin here we're looking for the distance from this origin to some position here x such that when light comes here and it reflects there's going to be constructive interference so let's go ahead and draw a light ray diagram one more thing that we should write before we start is the that the refractive index for silver.
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Well, since it's going to be reflective here, here that refractive index here for silver, once it bounces off of silver, it's going to experience a pi -degree phase shift, okay? pi -radian phase shift, or 180 -degree phase -shift.
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So let's start with a ray diagram.
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So as light is coming in here, it reaches the interface between the glass and the air, since it's going from low to high, the reflected light from this interface is going to be shifted by 180 degrees.
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Or is going to experience a high shift.
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So 180 degree shift.
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Now that corresponds to a path length difference, delta ref.
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Okay, so here this is delta reflection.
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It's going to be equal to lambda over 2.
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So light comes here, it strikes the glass, since the glass has a higher refractive index than the air, the reflected light gets shifted by 180 degrees and that corresponds to a path length difference of lambda over 2.
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Okay, the light that's transmitted continues to go here, this distance d.
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So it's going to be a distance d.
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Now keep in mind that this wedge is very thin, so even though i'm drawing it at an angle here, the light is really going to come down here like so.
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Like so.
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So actually let me go ahead and rearrange this orientation of the light.
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So as the light comes in here, it's going to get reflected.
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And then part of it is going to get transmitted.
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Here it is transmitted.
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It's going to travel this distance d here.
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And then part of it is going to get here.
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It gets reflected because of the silver coating.
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And when it gets reflected, it's also shifted.
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180 degrees.
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So there's another shift of delta ref.
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Delta ref for reflection.
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That occurs here at the bottom interface of the glass.
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And then the light is going to continue to move through the glass and then eventually out through it.
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Like so.
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So we're looking for this thickness here, d, such that these two reflected rays constructively interfere.
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The net path length difference, we're going to symbolize that with delta.
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This is going to be equal to the path length difference associated with the first reflection, so delta ref, plus the path length difference associated with the light going through the glass wedge.
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So we're going to call that delta t.
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And then we have another pi shift with its associated path length difference here.
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So delta ref again.
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For constructive interference, the net path length difference must be equal to some integer multiple of wavelengths, lambda.
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So in this case, m is going to be equal to 0, 1, 2, 3, etc.
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So an integer multiple.
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But since these are non -zero quantities, we can go ahead and start at 1.
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1, 2, 3, 4.
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Or because this thickness here is going to be non -zero, so there's no way that this could be equal to zero.
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So here we are starting at 1.
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Now let's go ahead and make some substitutions here because each delta ref or the path length difference associated with 180 -degree phase shift is lambda over 2.
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We can make this substitution in this equation.
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So let's go ahead and do that.
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We have delta is equal to lambda over 2 plus delta.
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T plus another lambda over 2 is going to be equal to an integer multiple of lambda.
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Notice that this lambda over 2 and that lambda over 2 combine.
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So we have lambda plus delta t is equal to some integer multiple of lambda.
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So that tells us that delta t must be equal to some integer multiple of lambda.
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Now the important thing to note here is that the integer multiple.
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Multiple of lambda that delta t has to be equal to is in reference to the wavelength of the light in the material.
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So in this case, it's the wavelength of the light in the glass...