Asked by Katie
I don't understand snell's law ( nsin(0)=nsin(0) ) and how to use it with refraction.
also how do i apply that to this problem:
'a light ray enters a square piece of plastci (n=1.39) near a corner. as it encounters the adjacent side it gets refracted along the surface (it hits at the critical angle). what was the original angle with which the ray hit the plastic from air?'
i know the answer (75 degrees) but i don't know how to get there.
please help!
also how do i apply that to this problem:
'a light ray enters a square piece of plastci (n=1.39) near a corner. as it encounters the adjacent side it gets refracted along the surface (it hits at the critical angle). what was the original angle with which the ray hit the plastic from air?'
i know the answer (75 degrees) but i don't know how to get there.
please help!
Answers
Answered by
bobpursley
What does "near a corner" mean? Does that mean the ray is internally reflected once?
Answered by
Katie
i have no idea, i just wrote what the question said. but yes that's what i interpreted it to mean
Answered by
bobpursley
Then start at the end of the ray.
1.39*sinThetainside=1*sin90
theta inside =46 deg.
Now take that angle back to where it was reflected, so it incident angle at the reflection was 46, then trace back to the corner side, so it leaves the surface refracted at 54 deg, no use Snell's law to find the ray angle outside ..
1.39sin54=1*sinTheta
and I think you will get your 75deg. Draw the figure to make certain you understand it.
1.39*sinThetainside=1*sin90
theta inside =46 deg.
Now take that angle back to where it was reflected, so it incident angle at the reflection was 46, then trace back to the corner side, so it leaves the surface refracted at 54 deg, no use Snell's law to find the ray angle outside ..
1.39sin54=1*sinTheta
and I think you will get your 75deg. Draw the figure to make certain you understand it.
Answered by
bobpursley
oops, typo it is 44 deg, not 54 deg. The sum has to be 90 between 46 and 44.
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