If I saw that problem on the test I would just take my graphing calculator and plug in y=1.66sin(x/2) and y=sin(x) and see where they intersect. This method gives the answer of ~67.78 degrees. However I pose my own question: would the proctor have accepted an answer of 0 degrees? Technically, it is correct because 0/2 is 0, but my only fear is that the proctor wouldn't be the best versed in grading and may just be a volunteer and be simply going off of the answer key.I'm very rusty on my trigonometry so I can't provide an elegant solution - https://www.microscopyu.com/tutorials/refraction but this gizmo leads me to believe the answers will be ~68 and ~34 degrees. Depending on the precision required in the answer, you could just make a quick table and plug values; you can narrow it down pretty quickly to +/- 1 degree that way.Thanks for posting this question. It's really quite interesting.Hello,
This question popped up in one of the tests I took recently.
"Light is incident on a piece of flint glass (n = 1.66) from the air in such a away that the angle of refraction is exactly half the angle of incidence. What are the values of the angles of incidence and refraction?"
I went ahead and tried to apply it to Snell's Law, to get this --->
Then, tried to isolate as much as I could, until I got to this point --->
However, more attempts to isolate were unsuccessful.
If anyone could help with this, I would greatly appreciate it.
From where you left off, my first idea is to graph it. Drop into a graphing calculator and find where the function intersects the x-axis over the interval (0,90).
When I was playing around with the graphing calculator I also noticed that there isn't always a solution for all indexes of refraction (for angles between 0 and 90). The only indexes that this property worked for were (sqrt(2)<n<2). I wonder why specifically sqrt 2 was the limit..
Just some food for thought
Nevertheless, I am not sure how to answer this question without the graphing calculator.