Angle of incidence and total internal reflection

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The discussion centers on the concept of total internal reflection and the angles involved. It clarifies that while the angle of incidence at point A is 40°, the answer key refers to the maximum angle of incidence being just less than 90° to achieve total internal reflection at point B. This is necessary to ensure the maximum angle of refraction is also considered. Participants are encouraged to think about ways to increase the angle of incidence at point B. Understanding these angles is crucial for applying the principles of total internal reflection effectively.
ellieee
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Homework Statement
qn: explain why a ray of light that enters the glass through surface A cannot be totally internally reflected at surface B.
Relevant Equations
n = sin I / sin r
the angle of incidence at A is 40°, then why did the answer key say "angle of incidence at A is just less than 90°". it is obvious that 40° is quite far from 90° right?
 

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Hi, and
:welcome: !​

Breaks my neck to look at your picture :mad:
ellieee said:
it is obvious that 40° is quite far from 90° right?
Yes, that is obvious. But to get a possible total internal refection at B, you need a maximum angle of refraction, and therefore a maximum angle of incidence. So they observe ( correctly :smile: !) that for a maximum angle of incidence at A
ellieee said:
angle of incidence at A is just less than 90°
##\ ##
 
BvU said:
you need a maximum angle of refraction, and therefore a maximum angle of incidence.
just to clarify, is it maximum angle of refraction and incidence at B ?
 
Can you think of a way to get a bigger angle of incidence at B ?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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