# Underwater ridge Physics question

I am stuck on part b of this question :https://isaacphysics.org/questions/underwater_ridge_sym?board=5e80a230-1ea2-4a5e-baf8-434804da2836&stage=all
Initially I got the length of the path to be L/(sin(cos^-1(sqrt(d_1/d_2))), and then divide by the speed sqrt(g*c_1), however, I believe I should work out the two distances in the different ridges, so the distance in the d_1 medium and the distance in the d_2 medium and then divide by their corresponding velocities. However, any help would be appreciated.
Original post by Javier García
I am stuck on part b of this question :https://isaacphysics.org/questions/underwater_ridge_sym?board=5e80a230-1ea2-4a5e-baf8-434804da2836&stage=all
Initially I got the length of the path to be L/(sin(cos^-1(sqrt(d_1/d_2))), and then divide by the speed sqrt(g*c_1), however, I believe I should work out the two distances in the different ridges, so the distance in the d_1 medium and the distance in the d_2 medium and then divide by their corresponding velocities. However, any help would be appreciated.

After a bit of trial and error and ... Hint 2 says "Find the time between when the parallel wave will first reach B and when the last part of the totally internally reflected wave arrives" so youre after the time difference. The time when the wave travelling along the ridge (parallel tothe boundary) reaches B is easy to find, though you have a typo in the OP for the speed. The time when the reflected wave reaches B is based on theta being the TIR angle, so the value worked out in part a). Like part a) there are no trig (or arctrig) functions necessary in the solution, just ratios.
Original post by mqb2766
After a bit of trial and error and ... Hint 2 says "Find the time between when the parallel wave will first reach B and when the last part of the totally internally reflected wave arrives" so youre after the time difference. The time when the wave travelling along the ridge (parallel tothe boundary) reaches B is easy to find, though you have a typo in the OP for the speed. The time when the reflected wave reaches B is based on theta being the TIR angle, so the value worked out in part a). Like part a) there are no trig (or arctrig) functions necessary in the solution, just ratios.

Ok, for the time the parallel wave takes to go to B is just L/(sqrt(g*d_1)), and then to find the length of the path that reflects off the boundary with the TIR angle cos^-1(sqrt(d_1/d_2)) I have to do trigonometry? Also, what do you mean by just ratios, the TIR angle is cos^-1(sqrt(d_1/d_2)), right?
Original post by Javier García
Ok, for the time the parallel wave takes to go to B is just L/(sqrt(g*d_1)), and then to find the length of the path that reflects off the boundary with the TIR angle cos^-1(sqrt(d_1/d_2)) I have to do trigonometry? Also, what do you mean by just ratios, the TIR angle is cos^-1(sqrt(d_1/d_2)), right?

You wnat the length of the hypotenuse and if you know theta=arccos(sqrt(d1/d2)) you should be able to get the approriate trig(theta) as a ratio.
Original post by mqb2766
You wnat the length of the hypotenuse and if you know theta=arccos(sqrt(d1/d2)) you should be able to get the approriate trig(theta) as a ratio.

I dont think I really understand what the question is asking, I believe I have to find the time taken for the parallel wave, and the time taken for the reflected wave, that is, the hypothenuse of the little triangle when the wave approaches the line between d_1 and d_2 and the other hypothenuse when the wave reflects. That whole length is L/(sin(cos^-1(sqrt(d_1/d_2)))) then the time taken to go through that path is L/(sin(cos^-1(sqrt(d-1/d_2)))*sqrt(g*d_1)) and then I substract this time with L/sqrt(g*d_1). If it is not like this, then do I have to calculate the hypothenuse of the traingle when the wave reflects back? if so, I get that length to be (L+x_2)/(2*sin(theta)), where x_2 is the length of the right part of the "L" length bisected by the line normal to the point where the wave reflects. I'm sorry for being this complicated
Original post by mqb2766
You wnat the length of the hypotenuse and if you know theta=arccos(sqrt(d1/d2)) you should be able to get the approriate trig(theta) as a ratio.

This is wha I get
Original post by Javier García
This is wha I get

The theta in part a) is the angle that the hypotenuse makes with AB, and its complementary to the theta_c in hint 5, so the trig(arctrig bit will simplify and you should be good.
Original post by mqb2766
The theta in part a) is the angle that the hypotenuse makes with AB, and its complementary to the theta_c in hint 5, so the trig(arctrig bit will simplify and you should be good.
Thanks so much, my mistake was that I was assuming that the Total Internal Reflection angle was the same as the critical angle, however in the drawing you can tell it is 90º-theta. Anyways, it is its complementary, hence when having cos(cos^-1(sqrt(d_1/d_2))) it simplified to sqrt(d_1/d_2),