# Diffraction problem

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#### DeathandTaxes

##### Full Member
10+ Year Member

For the first one, don't understand TBR's explanation for why II doesn't work and why III does. How exactly does a smaller n difference affect Y, given Y = lambda * L / d? I also don't see how the third affects this equation either.

TBR says this for why III is valid:

"For Statement III, it is perhaps best to
visualize what happens when the hair is moved to either the left or right. If the
hair moves to the left, then the glass slides grow farther apart. If the hair moves
to the right, then the glass slides grow closer together.Taken to an extreme, if the
two slides touch one another, then all of the bright spots are gone (pushed off of
the plate if you will). Opening the gap creates the spots and closing the gap
eliminates the spots, so widening the gap must bring adjacent maxima closer
together. This makes Statement III a valid statement."

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For the first one, don't understand TBR's explanation for why II doesn't work and why III does. How exactly does a smaller n difference affect Y, given Y = lambda * L / d? I also don't see how the third affects this equation either.
No posts in 5 days. I'll take a stab at it. Both the small index of refraction and the 3rd option of increasing the angle could best be explained visually.

In the picture below you can see that given the index of refraction difference between air and glass the light will bend a distance "d" away from it's original path.
If the incident angle is zero "normal" then d=0. As the incident angle approaches 90 degrees, D is going to approach 2 centimeters. This is because if light entered nearly parallel to the glass surface and exited nearly parallel on the opposite side, the path was shifted by the width of the glass.

Related to the problem, If you lowered the index of refraction of the air on the bottom side of the glass the "exit" angle labeled G would be closer to normal and because in the example "d" is very small, increasing it by increasing the angle of the glass would cause the destructive bands to be closer together

For the second one, would my thinking be right to consider that if there are more slits, more light would appear- more light, more intensity, is that right? And they are spaced farther because there would be more likelihood for destructive interference.
For the second question, using a diffraction grating will decrease how much light is being blocked so intensity will increase. Also if you add a 3rd and 4th slit it becomes less likely that all the waves will be constructive. (The huge peaks are areas of total constructive interference)
The image provided is slightly deceiving because with a grating everything would be increased intensity but the "extra intensity" areas would be spaced farther apart. The picture makes it look like it's completely dark in between the peaks. In reality there will be slight banding in between the more significantly sharp peaks.

Hope that helps a little.

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Thanks for the help, though I'm still struggling with with the first part conceptually. Destructive bands only occur when there are phase shifts of 180 degrees, right? So how does increasing the angle of the glass affect construction/destruction?

Ya, probably easier to just rule out the other two options. lol.
I did find this online: http://en2k6.blogspot.com/2008/02/diffraction-of-light_18.html
It's a couple pages explaining this exact scenario with math and everything.

TLDR version is that the air gap between the two slides is a gradient of thickness. (smaller gap on left and bigger on right). Becauselight that enters in some parts is going to go through the top slide, shift a little and bounce off the bottom glass back up. When the wave comes back at itself at some points it will be in phase (constructive), and others it's will be out of phase (destructive), and that is going to be determined based on the length of the air gap which varies along a gradient.
Increasing the angle increases the "air wedge" gradient. The height varies faster so the banding is closer together.

Lol, Honestly a video would be a better way to explain it. Or graphical animation or something.

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Wow, thanks a ton for resources! I'll take a look through them

i dont know if this is oversimplifying it, or if this is even right but
dsin(theta)=nLambda

using this equation i think the question is referring to d.
1 says lower lambda, wich will lower d
2 says increase n which will increase d
3 says lower the angle theta (i think) which would increae d

so only 1 would work?

so i have found 2 formulas on this

sin theta = n lambda/ D

and lambda = delta x d/L

can anyone explain the difference?

Those equations are rearranged from what I used but the top one is to find the distance between maximum in a double slit interference experiment.
sin (theta) is the opposite side (distance to maxima) from the midline for the n'th maxima given a wavelength lambda and a screen distance of D.
Useful if you need to find the distance from midline for the 3rd maxima etc.
Same as equation here: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html#c1

2nd equation seems like the single slit equation to find the closest minima (interference). Not sure though.
A rearrangement of the equation found here: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinvar.html#c2

I think you only need to know how to use these Qualitatively. Just know all the parameters what will make the Maxima closer together and the opposite of that will make them farther apart.
Without a calculator the numbers would be too awful to solve during an mcat. Unless your taking a GS practice test. lol!

Thanks! can you confirm whether this is all correct?

single slit
theta used in equation is half the angle formed between middle of slit to both ends of the central fringe?
central fringe is twice as long as the others
how do you find width of the central fringe?
d is the width of the single slit?
light fringes are n= half integers and 0 and dark fringes are full integers

double slit
d is the width of both slits? or just one?
how do you find width of the central fringe?
light fringes are n= integers and dark fringes are half integers

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nevermind the 3rd one but if anyone can help explain the first 2?

Thanks! can you confirm whether this is all correct?

single slit
theta used in equation is half the angle formed between middle of slit to both ends of the central fringe?
central fringe is twice as long as the others
how do you find width of the central fringe?
d is the width of the single slit?
light fringes are n= half integers and 0 and dark fringes are full integers

double slit
d is the width of both slits? or just one?
how do you find width of the central fringe?
light fringes are n= integers and dark fringes are half integers
Single slit you find the distance to the minima intensity.
double slit you use the distance between slits.
I think you just can find the distance between the minima and maxima.

if anyone can help explain the first 2?

First question is just interference, The waves are 180 degrees out of phase like the picture shows so they cancel out.
The second one both waves at in phase at point p, and they are apparently saying intensity = Power = v * mu * (A^2) * (w^2)
http://www.schoolphysics.co.uk/age1...rties/text/Intensity_and_amplitude/index.html

Single slit you find the distance to the minima intensity.
not sure what this means, but n=half integers is where the bright fringes are right? but how do you find the width of the fringe if all they give you is distance between the light and the wall. will this distance become D? i have read conflicting things on this. i have also seen some people use Tan instead of Sin
double slit you use the distance between slits.
not sure what this means. the distance occupied by the slit is D isnt it? this is tru for both single and double slits?

First question is just interference, The waves are 180 degrees out of phase like the picture shows so they cancel out.
The second one both waves at in phase at point p, and they are apparently saying intensity = Power = v * mu * (A^2) * (w^2)
so yes the waves are 180 degrees out of phase interfering destructively, but how are we supposed to know this?
one wave is in a straight line and the other is meeting it at an angle.
what if the angle was big enough to allow for the meeting point of the waves to be constructive?
are we supposed to assume this just based on half a lambda difference?

thanks!

Best I can do is make a recommendation. Browse this page ==> http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

It has almost everything you will need to know to thoroughly grasp this concept, and most of MCAT related physics. (equations included)

so yes the waves are 180 degrees out of phase interfering destructively, but how are we supposed to know this?
one wave is in a straight line and the other is meeting it at an angle.
what if the angle was big enough to allow for the meeting point of the waves to be constructive?
are we supposed to assume this just based on half a lambda difference?
thanks!

You don't need angle to solve. The image is a generalization of the concept that does not relate to this problem (at least for calculations).

You know that it will be constructive interference because of the distance. For example, if you take a Cosine wave and you go 100 wavelengths down the wave. You know the value is going to be 1 because it started at 1.
Here you are going 22.5 and 45 wavelengths respectively to reach point p.

45 wavelengths lands you right where you started.
22.5 (or) anything and 1/2 gets you the opposite of where you started (180 out of phase)

The critical statement is line 1. "Two light sources produce light in phase". So anywhere you are, you know that if point "p" is an integer wavelength from both sources it will be constructive. (In the example they are not both integer wavelengths from point p)

Regarding slits, I had a question on this T/F from Pre-med Wiki:

I'm a bit confused by the explanation. My thought process was that if we have dsin(theta) = n(Lambda), then if Lambda goes up (because red light is higher wavelength), then d goes up to keep equivalency. Thus it would make the fringes farther apart.
Would it be accurate to think of it in this way? (And thanks for great support btw!)

My thought process was that if we have dsin(theta) = n(Lambda), then if Lambda goes up (because red light is higher wavelength), then d goes up to keep equivalency.
You can't increase d because d is the distance between the two slits. That would be cheating. =p
so if d is constant then you have to increase theta giving a greater fringe separation.

If you rearrange to the equation y = mL(lambda)/d then you see that as you increase wavelength (red) you will get a greater fringe separation.
btw, y is fringe separation, m is fringe number, L is length to screen, lambda = wavelength, d is slit separation distance.

I suggest just hitting a video on it. (watch at double speed if you only have 2 minutes instead of 4)
He uses the equation y = L delta /d
delta is just m*lambda in one term.

Best I can do is make a recommendation. Browse this page ==> http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

It has almost everything you will need to know to thoroughly grasp this concept, and most of MCAT related physics. (equations included)

You don't need angle to solve. The image is a generalization of the concept that does not relate to this problem (at least for calculations).

You know that it will be constructive interference because of the distance. For example, if you take a Cosine wave and you go 100 wavelengths down the wave. You know the value is going to be 1 because it started at 1.
Here you are going 22.5 and 45 wavelengths respectively to reach point p.

45 wavelengths lands you right where you started.
22.5 (or) anything and 1/2 gets you the opposite of where you started (180 out of phase)

The critical statement is line 1. "Two light sources produce light in phase". So anywhere you are, you know that if point "p" is an integer wavelength from both sources it will be constructive. (In the example they are not both integer wavelengths from point p)

what i am trying to say is that if these waves were both traveling in straiht lines, then yes they would be half a wavelength apart as you explained.
BUT, since one wave is travelin at an angle, (and is therefore traveling a greater distance then the first wave) how can you tell that it will still be exatly HALF a wavelength apart from the first wave once it reaches the point of intersection?

what i am trying to say is that if these waves were both traveling in straiht lines, then yes they would be half a wavelength apart as you explained.
BUT, since one wave is travelin at an angle, (and is therefore traveling a greater distance then the first wave) how can you tell that it will still be exatly HALF a wavelength apart from the first wave once it reaches the point of intersection?
So the question says Two light sources produce light in phase with wavelength lambda. The sources are placed 22.5 and 45 wavelengths from point p.
Right there it tells you how far the waves are traveling to reach point P. If you assume the velocity of the waves in the medium is the same, and it says they both start in phase. You don't need to know the angle between them.
Wave A will have traveled 22.5 wavelengths to get to point P.
Wave B will have traveled 45 wavelengths to get to point P.

From that you can extrapolate that wave B will be at the same amplitude it was when it started.
Wave A will be the opposite amplitude from when it started causing interference.

1 user
So the question says Two light sources produce light in phase with wavelength lambda. The sources are placed 22.5 and 45 wavelengths from point p.
Right there it tells you how far the waves are traveling to reach point P. If you assume the velocity of the waves in the medium is the same, and it says they both start in phase. You don't need to know the angle between them.
Wave A will have traveled 22.5 wavelengths to get to point P.
Wave B will have traveled 45 wavelengths to get to point P.

From that you can extrapolate that wave B will be at the same amplitude it was when it started.
Wave A will be the opposite amplitude from when it started causing interference.
that makes sense, thank you!