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  #1  
Old 12-14-2004, 04:31 PM
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Any physics gurus out there?

( I am posting this here because i know there are alot of smart folks here)
i have looked all over the web and in my book and i feel like a real dummy
i can not figure out where to start on this problem.
can anyone help me please

"The drawing shows two plane mirrors that intersect at an angle of 50 degrees. An incident ray reflects from one mirror and then the other. what is the angle between the incident and outgoing ray?

http://www.villagephotos.com/pubbrowse.asp?folder_id=1141305

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File Type: bmp physics reflection.bmp (24.1 KB, 333 views)
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  #2  
Old 12-14-2004, 05:02 PM
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say: Jook-Ur-Pah
 
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do you know the angle of attack of the incident ray?

get a protractor.. ha ha!
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  #3  
Old 12-14-2004, 05:47 PM
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The angle of the outgoing ray is 100 degrees based upon the diagram.

The logic is as follows:

Say the incoming ray is incident on the first reflector at a 60 degree angle.

The outgoing ray from the first reflector is also at a 60 degree angle. This adds to the 50 degree angle provided to result in 110 degrees. Therefore the incoming ray on the second reflector is 70 degrees (180-110).

The outgoing ray off the second reflector is also at 70 degrees leaving the angle between the rays of 40 degrees (180-70-70).

You know the angle between the rays on the first reflector: 60 degrees (180-60-60).

So, the remaining angle is: 80 degrees. (180-40-60).

The angle that you are looking for is 100 degrees. (180-80).

The beautiful thing is that the angle remains at 100 degrees no matter what the angle is of the incoming ray (within limits; the incoming angle cannot equal or be less than 40 degrees).

Good luck.
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Old 12-14-2004, 07:14 PM
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thats kinda where i was heaed, but since the ray has no numbered angles, i am starting think she, ( the instructor) is tricking us with the mirrors. this may be a very simple geometry problem. since a triangle equals 180 deg you can label all three corners A,B, and C. lets call the corner near the unknwn angle C. the angle opposite C is also C. so the unkown angle is equal to A+B.
what do you think? does that make since?
thanks for this help, I really do appreciate it.
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  #5  
Old 12-14-2004, 07:55 PM
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OK, you forced me to derive it. We will initially solve for C.

Follow carefully.

C= angle adjacent the angle we are looking for.
A= included angle at first incidence (per your definition)
B= included angle at second incidence (per your definition)
I= angle of first incidence (relative to mirror)
J= angle of second incidence (relative to mirror)

C=180-(A+B)

A=180-2I
B=180-2J
J=180-I-50

Substitute in formula for C:

180-((180-2I)+(180-2J))

180-((180-2I)+(180-2(180-I-50)))

Solve for C:

180-180+2I-180+360-2I-100=80

The angle we want is 180-C.

180-80=100.
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  #6  
Old 12-16-2004, 03:54 PM
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=180 - 2(50) = 80 degrees

hope that helps

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