WEBVTT
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in this problem, we are asked to find the
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following angle. So we have a interface with air
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in a glass. So this is a light that
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strikes this place and then bounces back and some off
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that is refracted here. So we call this angle
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. This is angle of incident. Um and this
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is Anglo for a fraction. So that our and
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here we have there, Toby, they don't be
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reflection. So we are asked to find this angle
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on that. Then we are asked to far in
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this angles of the angle between the glass and surface
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off air. So those angles so it. But
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maybe it has to find this so we can find
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this angle by finding the angle of refraction which is
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actually equal to angle off incident. So the angle
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off incident we have already that is given. Then
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we can just subtract a 90 a minus Anglo for
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incident which is equal to the angle of refraction.
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So here we have angle off, eh? Is
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equal to the angle of refraction. An angle is
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given so we can just plug this value here in
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subject. Then we can get the angle between glass
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and reflect beam, which is 47.5. What,
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seven or five degrees? Then in part A.
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We can find that to be here. And then
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from there we can just subtract from 90. Tend
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to be and then we can find this angle.
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So, using a Snell's law and a done sign
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there angle that a is equal to physical tow N
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B. So I'm signed Refractive Index and be time
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sign. There'd be, um then we console for
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that. I'd be here from the relation by parking
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go values in N and B than the value here
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we get here is B is 24 degrees very 40
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degrees. Then, as we did previously here by
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subtracting 24 from 90 Um, so then we get
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this angle. So 90 minus 24 that gives a
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66 degrees. So this angle here is 66 degrees
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, and this angle here is 47.5 degrees. End
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off the problem. Thank you for watching