Could you explain how we know theta1=pi/6 in the first part of the question? And also, theta1 is representing the incident angle of the beam, right? But it told us in the problem that the beam is parallel to the wire going through the center, as it’s drawn in the diagram..?
Drop a perpendicular down from A to a point on the horizontal line passing through A, and call that point B. Then AB is 1m, AC is 2m, and angle ABC is a right angle. Therefore this is a 30, 60, 90 triangle, so the angle ACB is pi/6. Therefore (by corresponding angles) theta1 is pi/6.
The angle of incidence of the beam is the angle between the beam and the normal to the surface at the point of entry, which is theta1.
Got it, thank you so much!
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