AAMC exam 8 #49

[kabtq9s]

I understand why q>a

I just dont understand why q=qc, their explanation is not helping

http://mysowar.files.wordpress.com/2011/06/untaamc8-number-49.png

Any help is appreciated and thank in advance

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[Catburr]

I understand why q>a

I just dont understand why q=qc, their explanation is not helping

http://mysowar.files.wordpress.com/2011/06/untaamc8-number-49.png

Any help is appreciated and thank in advance

I never took test 8, maybe I'm missing something from the passage...is q the same as theta, and is qc the same as theta prime? (Boo at the lack of greek letters in this forum system ) Weird to suddenly switch it up like that. Or maybe I've forgotten that q ALWAYS refers to your incoming light beam, or something?

Anyway, if that's the case, then if you draw another dashed line right at the reflection where arrow 2 hits and arrow 3 starts heading up, you see that the beam represented by arrow 3 is also at angle alpha, with respect to that normal dashed line.

So, since it's the same 1.0 outside, 1.5 inside interface: Beam at angle alpha inside the material --> beam at angle theta outside the material. So the first two arrows are just vice versa of the second two arrows.

First two arrows, you're saying sin(theta1)/sin(theta2) = n2/n1.
Second two, you're just flipping it: sin(theta2)/sin(theta1) = n1/n2

 

[kabtq9s]

Hello again Catburr,

I'm sorry I lost you after the second paragraph

Also the "---> " is confusing me. Does it mean "to" or equal or "greater than" sign

But I did what you told me to do so far http://mysowar.files.wordpress.com/2011/06/untaamc8-number-491.png

Also: this question was a discrete so no passage. And I'm sure many people are also complaining about the symbol used in the choices.

 

[Catburr]

Also the "---> " is confusing me. Does it mean "to" or equal or "greater than" sign

I was going for like, a reaction arrow, so "to."

But I did what you told me to do so far http://mysowar.files.wordpress.com/2011/06/untaamc8-number-491.png

Also: this question was a discrete so no passage. And I'm sure many people are also complaining about the symbol used in the choices.

Yeah the switching between theta and q is a little weird. But anyway, your picture is good, shows you everything you need to know.

You have Beam 1, entering at angle theta, to the normal. Because it's going from n=1 to n=1.5, refraction occurs, and thus Beam 2 is at angle alpha. This is what you're given. You're also informed that Beam 2 reflects when it hits the bottom of the material.

Reflection takes Beam 2 and creates Beam 3, therefore we know Beam 3 is going to be at the same angle to the normal as Beam 2 was, because that's just how reflection works.

So when Beam 3 arives at the air/material interface, it is a beam at angle alpha, in a material of n=1.5, right? How do we describe Beam 4 then? Well it's just like Beam 1.

Why's that? Remember, when we went from Beam 1 to Beam 2, it was a situation where theta beam in n=1.0 environment refracts and becomes alpha bean in 1.5 environment. When we go from Beam 3 to Beam 4, we're just taking the opposite path: alpha beam in 1.5, refracts to create theta in 1.0.

Math behind this is just: sin(angle 1)/sin(angle 2) = index of refraction 2 / index of refraction 1

If that doesn't make sense in words, I'll do an ugly MS Paint pic or something.

 

[kabtq9s]

yes I think I get it now. I think this all boils down to the law of reflection and how it applies to parallel planes.

Thanks again Catburr, I really appreciate it

If you don't mind can you please take a shot at this question too
http://forums.studentdoctor.net/showthread.php?t=826201