Where should a second hole be placed in order for the light [ ]?

  • Thread starter s3a
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  • #1
s3a
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"Where should a second hole be placed in order for the light [...]?"

Homework Statement


The problem and the solution are attached as jpg files. (Given that the problem depends on the drawing, I think it's more convenient for the reader to view the text in the image as well.)

Homework Equations


Snell's Law: ##n_1 sin θ_1 = n_2 sin θ_2##

The Attempt at a Solution


In the TheSolution.jpg, I don't, mathematically, get how it can it say ##2.54 sin 48.6° = 1.0 sin θ_2## since that would mean that ##asin(2.54 sin(48.6°))## would not exist (but, I do get total internal reflection when thinking about it theoretically rather than mathematically).

Also, how is the leftmost 11.4° angle (from the normal) found to be 11.4°?

Any input would be greatly appreciated!
 

Attachments

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  • TheSolution.jpg
    TheSolution.jpg
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Answers and Replies

  • #2
Doc Al
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In the TheSolution.jpg, I don't, mathematically, get how it can it say ##2.54 sin 48.6° = 1.0 sin θ_2## since that would mean that ##asin(2.54 sin(48.6°))## would not exist (but, I do get total internal reflection when thinking about it theoretically rather than mathematically).
The point is that the refraction ray cannot exist--thus you have total internal reflection. (Review how you'd determine the critical angle for total internal reflection.)

Also, how is the leftmost 11.4° angle (from the normal) found to be 11.4°?
Consider the triangle formed at the lower left of the prism, with one angle given as 60°.
 
  • #3
s3a
807
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The point is that the refraction ray cannot exist--thus you have total internal reflection. (Review how you'd determine the critical angle for total internal reflection.)


Consider the triangle formed at the lower left of the prism, with one angle given as 60°.
Thanks, I now get it! :)
 

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