The critical angle for light passing from glycerine to air is 42.9 degrees.
What is the index of refraction of glycerine?
If you could show your working out id really appreciate it
What is the index of refraction of glycerine?
If you could show your working out id really appreciate it
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Well Snell's Law is N1*sinθ1 = N2*sinθ2 (θ means the angle)
We know:
N2 (air) = 1
θ1 = 42.9°
We know that if the critical angle is 42.9°, so the refracted ray must be traveling along the interface between the two media, which is a refracted angle of 90°
If we put these numbers in the equation:
N1 * sin42.9° = 1 * sin90°
sin90° = 1, so we can simplify the equation to
N1 * sin42.9° = 1
so N1 = 1/sin42.9
N1=1/sin42.9
N1=1.47
Hope this helped.
We know:
N2 (air) = 1
θ1 = 42.9°
We know that if the critical angle is 42.9°, so the refracted ray must be traveling along the interface between the two media, which is a refracted angle of 90°
If we put these numbers in the equation:
N1 * sin42.9° = 1 * sin90°
sin90° = 1, so we can simplify the equation to
N1 * sin42.9° = 1
so N1 = 1/sin42.9
N1=1/sin42.9
N1=1.47
Hope this helped.
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sin theta1/sin theta2 = n2/n1
Since we know the critical angle is 42.9, that means if theta1 is 42.9, then theta2 is 90 degrees. Air we take to have an index of 1. Then just solve for n2.
sin 42.9degrees / sin 90 degrees = 1 / n2
n2 = 1 / sin 42.9 degrees
n2 = 1 / 0.6807
n2 = 1.47
Since we know the critical angle is 42.9, that means if theta1 is 42.9, then theta2 is 90 degrees. Air we take to have an index of 1. Then just solve for n2.
sin 42.9degrees / sin 90 degrees = 1 / n2
n2 = 1 / sin 42.9 degrees
n2 = 1 / 0.6807
n2 = 1.47
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