How much "Light and Optics" was on the April MCAT?

[mac_kin]

Hey guys,

So the MCAT is fast approaching and I have pretty much gotten through all of the physics except Optics. However, I also have a lot of O-Chem to review.

Last summer I remember there were a few questions about this and I had to guess on a couple. How was it in april?


PS. I know this belongs in the MCAT forum but I've had a lot of trouble accessing this site all day and since it finally let me in, I'm just going to post it here. Feel free to move it.

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[MollyMalone]

No prob. Moving!

(P.S., I had plenty of light and optics, but that was last year.)

 

[beaverfetus]

hmmm, I think there was some, but i never really bothered learning those passages, and I did pretty good. I guess I knew light pretty well, just not optics, drawing rays are fore sissies. I'll concave you!

 

[DrBowtie]

It wouldn't take long to learn the basics. Probably an hour or so just to CYA.

 

[oxeye]

Hmmm, mine had a total internal reflection question. That's all I remember specifically, though.

I was glad I had studied that formula! It was one of the topics I was having trouble with and I grilled it into my brain a few days before the test.

I don't remember overall how much stuff there was on light/optics, though. Sorry. Plus, there are a lot of different forms.

 

[Kikaku21]

Hmmm... so what do we need to know for light/optics? Just a few things!!

1/f = 1/o + 1/i
f = (n-1)*(1/r1 - 1/r2)

All of your signs... thats the confusing part, really. Just make a table for convex and concave mirrors and then one for convex and concave lenses.

n1*sin(a) = n2*sin(b) The critical angle bit and total internal reflection.

Also 400 - 700 nm is visible light, that is worth knowing. ROYGBV is the color order. Ultraviolet is < 400 and infrared is > 700.

Finally interference. A*sin(theta) = n*lambda
This is for a single slit. It gives the locatoin of the dark fringes.

The double slit formula is d*sin(theta) = m*lambda
It gives the location of the bright fringes.

Am I forgeting anything?

Oh yeah....

 

[TMP-SMX]

Yeah you definitely have to know wavelengths and their relationship to frequency.

Same with v=lambda frequency

And the forms of light ranging from infrared and microwaves through visible light to UV and above.

 

[Omyss]

A*sin(theta) = n*lambda


The double slit formula is d*sin(theta) = m*lambda

f = (n-1)*(1/r1 - 1/r2)

i have never come across these equations in either the kaplan or TPR books so i don';t think we need to know these.

 

[Kikaku21]

A*sin(theta) = n*lambda


The double slit formula is d*sin(theta) = m*lambda

f = (n-1)*(1/r1 - 1/r2)

i have never come across these equations in either the kaplan or TPR books so i don';t think we need to know these.

Hmmm.. They are in my Kaplan books, which is why I mentioned them..

Also-
There was a question on an AAMC test that talks about what happens to the distance between the fringes as you increase wavelength... which applies to the 2nd equation above....

 

[mac_kin]

great!

much thanks for the help!

 

[deleted106503]

Hmmm.. They are in my Kaplan books, which is why I mentioned them..

Also-
There was a question on an AAMC test that talks about what happens to the distance between the fringes as you increase wavelength... which applies to the 2nd equation above....

Hmm, EK doesn't have this equation. The only thing they mentioned was that longer wavelengths (i.e. Red) would bend more through the slit than shorter ones (i.e. violet). So red would show up on the opposing wall further from an imaginary straight line drawn through the opening. Is this what you mean or do we actually have to calculate distances between the bands of light?

 

[TMP-SMX]

The slits are all about constructive/destructive interference. Not colors...

 

[medicin]

is it worth memorizing the lens maker equation?

 

[gridiron]

Hmm, EK doesn't have this equation. The only thing they mentioned was that longer wavelengths (i.e. Red) would bend more through the slit than shorter ones (i.e. violet). So red would show up on the opposing wall further from an imaginary straight line drawn through the opening. Is this what you mean or do we actually have to calculate distances between the bands of light?

Correct me if I am wrong, but I believe the magnitude to which light will bend is dependent on the spacing between the slits---longer slits, wavelengths much bigger than the light, means less spreading and shorter slits, about the wavelength of light, the more spreading.

 

[TMP-SMX]

is it worth memorizing the lens maker equation?

No. Just know what it means at a basic level.

 

[deleted106503]

Correct me if I am wrong, but I believe the magnitude to which light will bend is dependent on the spacing between the slits---longer slits, wavelengths much bigger than the light, means less spreading and shorter slits, about the wavelength of light, the more spreading.

I'm confused, did you say in the first half your'e using a double slit wall, but for the second half you're talking about specific slits?

I was using the EK example in the book and they used a diagram with one slit (not the constructive/destructive picture they have). They said that a smaller slit will make the light bend more holding wavelenght constant. They also said that increasing the wavelength while holding the slit opening constant will cause the light to bend more. So this means that red will appear on the outer most fringes compared to violet, right? I'm not sure how this would work for a double slit opening.

 

[gridiron]

I'm confused, did you say in the first half your'e using a double slit wall, but for the second half you're talking about specific slits?

I was using the EK example in the book and they used a diagram with one slit (not the constructive/destructive picture they have). They said that a smaller slit will make the light bend more holding wavelenght constant. They also said that increasing the wavelength while holding the slit opening constant will cause the light to bend more. So this means that red will appear on the outer most fringes compared to violet, right? I'm not sure how this would work for a double slit opening.

Yes, I do believe your generalization is correct. What I meant to characterize was a double slit---more specifically young's double slit experiment. Your explanation is right, and I hope I didn't confuse you. With double slits, it is a little harder to explain in terms of color---it is used to more to prove that light is a wave.

 

[Omyss]

darn.. could someone provide me with a link or thread on this double/single slit stuff. that we need to know, .b/c my books don't have any information on them.

 

[deleted106503]

Yes, I do believe your generalization is correct. What I meant to characterize was a double slit---more specifically young's double slit experiment. Your explanation is right, and I hope I didn't confuse you. With double slits, it is a little harder to explain in terms of color---it is used to more to prove that light is a wave.

This might help the others here about wave/particle nature of light. An EK question said something like: if light was made of only particles, what would the result of a slit experiment show against the wall? The answer was that it would show a single spot the same size as the hole.

Since light is also wave nature, it bends and creates fringes rather than a direct spot on the wall. Hopefully this will help someone for the MCAT.

By the way, do you have any good links talking about a double slit experiment? EK only mentions that it will create dark and light spots. Maybe we don't have to know the equation?

 

[gridiron]

This might help the others here about wave/particle nature of light. An EK question said something like: if light was made of only particles, what would the result of a slit experiment show against the wall? The answer was that it would show a single spot the same size as the hole.

Since light is also wave nature, it bends and creates fringes rather than a direct spot on the wall. Hopefully this will help someone for the MCAT.

By the way, do you have any good links talking about a double slit experiment? EK only mentions that it will create dark and light spots. Maybe we don't have to know the equation?

Actually, I have two very reliable sources. One is a link:
http://theory.uwinnipeg.ca/physics/light/node9.html

And the other is a physics book:
Halliday, Resnick and Walker, Fundamentals of physics, 6th edition.

The website is a good source to understand the equation but the book is awesome to understand the theory. I also recommend the book to anyone having trouble with physics concepts. It has conceptual questions, like you would see on a typical MCAT, at the end of each section of each chapter that tests the material from the specific question. I hope this helps!!!

 

[TMP-SMX]

The point of the slit experiments have nothing to do with color. As has been said, it is to show that light is a wave and they can either superimpose positively/negatively (constructively light spot/destructively darkspot).

The whole point of that is to show that electrons, if run through the same experiment, also have wave properties. (A particle can't be in two places at once and interfere with itself unless it is a wave).

 

[Omyss]

So do we only need to know these:

If the rays were in phase when they passed through the slits, then the condition for constructive interference at the screen is:

dsintheta = mlamba ,m = 1, 2,...

whereas the condition for destructive interference at the screen is:
dsintheta = (m + (1/2)lamda) ,m = 1, 2,...

or do we need to know thesee also:

In the case that y , the distance from the interference fringe to the point of the screen opposite the center of the slits (see Fig.22.10) is much less than L ( y < < L ), one can use the approximate formula:
sintheta= y/L

so that the formulas specifying the y - coordinates of the bright and dark spots, respectively are:
y Bm = m x lamba x L/d brightspots

y Dm = (m +0.5) x lamba x L darkspots

The spacing between the dark spots is
y = lamdaxL/d