DIRECTIONS: Work the problem You must to receive credit, and you must return thi
ID: 1557041 • Letter: D
Question
DIRECTIONS: Work the problem You must to receive credit, and you must return this sheet. The problem is worth 20 points. If you have to construct a graph, use the graph paper provided in the laboratory. Student was asked to repeat experiment on Snell's Law using the laser beam, instead of the white light beam, and the Optics Ray Table whose scale is devided into 360 equal parts. A Optics Ray Table like the one she used to study reflection and refraction is shown in Figure (a). The laser beam passes over the zero mark on the Ray Table. When a semicircular transparent plate has its straight edge aligned with AA', the refracted beam is observed to disappear. Point R is at 75.0 degrees. (i) What is the critical angle for total internal reflection for this substance? Please explain your answer. The semicircular transparent plate is then reoriented, as in Figure (b), to study the Snell's law of refraction. (ii) If point A in Figure (b) is at 0 degrees, and point B is at 50.0 degrees, determine the position of point C, the position of the refracted beam on the Ray Table. Please explain your answer. (iii) what is the velocity of the light in the acrylic semicircle? (iv) What are the positions of N and N' on the Ray Table? Please explain your answer. Recall that c = 3.00 times 10^8 m/sExplanation / Answer
i) When total internal reflection happens i = r
Here ic + r = 75
Hence ic = 75/2 = 37.5o
ii) Here using same relation i = 25o
Now using critical angle we can fing refractive index of transparent material
u = 1/sin ic = 1.6427
Now using Snell's law
angle of refraction r can be found as
sin r = sin i/u = sin 25/1.6427
r = 14.91o
iii) Let the velocity be v = c/u = 3*10^8/1.6427 = 182626164.2 m/s
iv) N - N' = 25o - 205o
As N lies 25 degrees above 0 degree line and N' lies 180 degrees opposite to N so 180+25 = 205degrees
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.