Snell's Law Refraction

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arc5005

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Picture of the Question from TBR Physics II Diagnostic Passage #2.2

Additionally, I've uploaded the passage this comes from; however, it doesn't seem necessary to answering this question. But if you are interested click it.

ANSWER:

Here is the answer they have: click here please. :)

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Okay so I just need clarification for this question!

I understand how to use snell's law in this situation, but it is time consuming to write all this down.

so n1 = index of refraction of air correct, so n1 and n4 are the same, because n1 and n4 are both air?

n2 = crown glass, and n3 = flint glass?

Is there a simpler way to do this, because overall this seems like it could be time consuming... is there a generalized rule for this situation?

Thanks! :)

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I don't remember where I saw this, but a video said think of it a car driving on different surfaces. Roads have a low index, dirt and sand are high. From sand to pavement the car will speed up and deviate away from the direction the first tire hits (it speeds up first). The reverse is also true, car leaves pavement for sand, that first tire digs in and turns the car in that direction. This should allow you to ditch some answer choices without any calculations. I think that's what you were asking anyway?
 
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Assuming each medium is parallel as they are in this case, there is no need to go through the entire process. Even if the fourth medium was not air, you could still equate n1sin1=n4sin4. This is because n1sin1=n2sin2=n3sin3=n4sin4. An exception to this might be if TIF occurs, such that the incident light is reflected at the interface of a higher n to a lower n. In this case no light would exit.
 
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Assuming each medium is parallel as they are in this case, there is no need to go through the entire process. Even if the fourth medium was not air, you could still equate n1sin1=n4sin4. This is because n1sin1=n2sin2=n3sin3=n4sin4. An exception to this might be if TIF occurs, such that the incident light is reflected at the interface of a higher n to a lower n. In this case no light would exit.

oh awesome. thanks
 
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Assuming each medium is parallel as they are in this case, there is no need to go through the entire process. Even if the fourth medium was not air, you could still equate n1sin1=n4sin4. This is because n1sin1=n2sin2=n3sin3=n4sin4. An exception to this might be if TIF occurs, such that the incident light is reflected at the interface of a higher n to a lower n. In this case no light would exit.


ahhh i remember this from KA video i watched a long time ago.
 
Questions like that are why I love TBR so much. They give two answers, a long one that follows traditional physics convention so I can see how it works and then a short and easy approach for future reference.
 
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