Maths/Physics Question

[Jeros]

Ok a little help, i dont want the answer, just nudge me in the right direction

A Wave moves through a later at a Velocity of = 400 km per sec, at 15 metres it breaks into layer 2 where the wave moves at a of Velocity = 3500 km/s for a further 19 metres

the critical angle is 4 degrees

What are are the thickness's of the layers (i think you can only calculate one)

 

[Zenith.UK]

You only want a nudge in the direction?

It's plain trigonometry and the velocities aren't needed.

Layer 1:
Critical angle is 4deg and the hypotenuse is 15m.
Sine is a ratio of the opposite side from the angle divided by the hypotenuse.
Sine of 4deg = 0.06976
So the length of the vertical side of the triangle is 0.06976 of the hypotenuse length. Simple algebra can derive the length from the information you have.

Layer 2:
Critical angle is still 4deg and the hypotenuse is 19m.
Same calculation as for layer 1.

The results for each layer are the vertical thickness of each layer

My answer under the spoiler. Bear in mind that I might be completely missing the point. I've not taken into account change in angle of incidence due to refraction.

Spoiler

Layer 1
0.0698 = opp/15
opp = 0.06976*15
opp = 1.046m

Layer 2
0.0698 = opp/19
opp = 0.06976*19
opp = 1.325m

Total = 2.371m

 

[Jugvayne]

^^ I'm not entirely sure that's right. If it had an angle of incidence of 4 degrees surely none of the wave would transmit to the second layer?

I'm not sure, the question seems a little unclear, but I think this might be a Snells law question.

 

[Jeros]

I think this might be a Snells law question.

It is :/

 

[Zenith.UK]

In that case, sine 4deg still stands for the critical angle from normal in the first material.

sine(crit.angle) / sine(refracted angle) = velocity1/velocity2

You know the critical angle and the 2 velocities, so a bit of algebra will give you the sine of the refracted angle.
Inverse the sine operation (Sin-1) to give you the angle of the wave direction in the 2nd layer.
Now you have two angles, take the complement of each (90 - angle), sine the complement and multiply it by the hypotenuse (the distance the wave travels in the layer).

My answer (which is probably wrong...)

Spoiler

0.06976 / sine(refr.angle) = 400/3500 = 0.1143
0.06976 / sine(refr.angle) = 0.1143
sine(refr.angle) = 0.06976 / 0.1143
sine(refr.angle) = 0.6103
refr.angle = Sin-1(0.6103)
refr.angle = 37.61deg

Considering the wave speeds up in the 2nd layer, the refraction angle will be larger than the incidence angle. So this fits with expectations.

Now you take the complement of each angle, sine it, multiply by the hypotenuse length and that gives you the vertical thickness of each layer.

Layer 1
90-4 = 86deg
sine 86deg = 0.9976 = opp/hyp
opp = 0.9976*15
opp = 14.964m

Layer 2
90-37.61 = 52.39deg
sine 52.39deg = 0.7922 = opp/hyp
opp = 0.7922*19
opp = 15.052m

Thickness of layer 1 is 14.964m.
Thickness of layer 2 is 15.052m.

Just for good measure, the time it takes for that wave to travel the 34m is...
((1/400000)*15)+((1/3500000)*19) = 0.0000429s = 4.29*10^-5s

 

[Jugvayne]

It is :/

Ok, well I think you might be confusing incident angle with critical angle, so I'll take 4deg to be the incident angle.

If this is the case, you can find the refracted angle using snells law:

sin(incident angle)/sin(refracted angle) = v1/v2, so rearrange and use the speeds to get the refracted angle.

Then as Zenith said just use trigonometry to get the thicknesses (except he has it wrong - you need to use cos and not sin, make sure to draw a diagram and this should become clearer)

Hopefully that's right

Edit: Sorry just seen his second post - the answer for the first thickness is now right, though the second thickness I think is wrong, you need to take sin of the refracted angle not take it away from 90. Again just draw a diagram and should be more clear