Download
1 / 7
70 likes | 148 Views
L17/P221 Follow Up. P221: Theory. Apply Snell ’ s Law successively to the two boundaries with q i being the angle of incidence on the first boundary and 90 o being the angle of refraction from the second boundary, one can show: sin q i = n sin(90 o – sin -1 (1/ n ))
E N D
P221: Theory Apply Snell’s Law successively to the two boundaries with qi being the angle of incidence on the first boundary and 90o being the angle of refraction from the second boundary, one can show: sinqi = nsin(90o – sin-1(1/n)) = ncos(sin-1(1/n)) Then qi = sin-1(ncos(sin-1(1/n))). This assumes that the index of refraction of the surrounding medium is 1.
Getting rid of the nested trig functions By the Pythagorean theorem, the side adjacent to a is (n2 – 1)1/2. The cosine of a is then cosa = (n2 – 1)1/2/n. Then ncosa = (n2 – 1)1/2 and qi = sin-1(n2 – 1)1/2. We want to know: What is the cosine of the angle whose sine is 1/n? Set up a right triangle with an angle α that has a sine of 1/n as shown to the right.
Application For water, n = 1.33, and qi = 61.3o. Since (n2 – 1)1/2 must be <= 1, then n must be <= 21/2 or about 1.414. If the index of refraction of the plastic block is greater than the square root of 2, there is no minimum angle for which ray emerges at 90o.
L17 results for the index of refraction of the plastic block Class of 2013 1.75 1.69 1.60 1.58 1.58 1.57 1.55 1.54 1.51 1.49 1.45 1.44 1.41 1.41 1.38 1.02 Mean: 1.50 Class of 2014 1.82 1.55 1.49 1.48 1.44 1.41 1.39 1.32 1.30 1.29 1.25 1.42
