Internal Reflection and Snell's Law TPR FL 1 Spoiler

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betterfuture

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Can someone explain this

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I totally guessed on this but I still want to know how do I figure this out. I know that internal reflection occurs when the medium has an index of refraction greater than the other medium. But what I don't understand it what results when that same medium is put into another fluid and how that affects the critical angle. Thanks!

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Well, you know that the crit angle is arcsin n(small)/n(large). n(small) used to be 1, then increased to some value between 1 and n(large). Since the fraction increases, so would arcsin.
 
Imagine some object with index of refraction n. Light going from the object to the air (measured from the inside) can undergo total internal reflection since n > n(air).

Using snell's law: nsin(theta1) = n(air)sin(theta2). In TIR, sin(theta2) = 1 and theta1 = sin_inverse(n(air) / n).

Putting the object in fluid still requires measuring from the inside since n > n(fluid), but now the light goes from the object to fluid instead of air so the equation above gets n(air) replaced by n(fluid). The result is a larger theta1, which is the critical angle.
 
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Imagine some object with index of refraction n. Light going from the object to the air (measured from the inside) can undergo total internal reflection since n > n(air).

Using snell's law: nsin(theta1) = n(air)sin(theta2). In TIR, sin(theta2) = 1 and theta1 = sin_inverse(n(air) / n).

Putting the object in fluid still requires measuring from the inside since n > n(fluid), but now the light goes from the object to fluid instead of air so the equation above gets n(air) replaced by n(fluid). The result is a smaller theta1, which is the critical angle.

Don't you mean that the result is a larger theta1 and thus a larger critical angle? Everything you said sounded solid except that last sentence, I believe. Since n(air) is replaced by a larger value, the critical angle will be larger too.
 
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