Hello. This is lauressa.

Well today in class we went over page 12 in our green booklet; waves in two dimensions, and learned about Snell's Law. Snell's Law shows that the ratio of the sine of the angle of incidence and the sine of the angle of refraction is a constant. The constant we use in Snell's Law is referred to as the index of refraction. With Snell's Law it can be used to determine a number of things, for instance the speeds and the wavelengths of incident or refracted waves by this equation

For those who aren't quite sure what refraction is it occurs when waves approach a boundary between two water depths at an angle causing the direction of travel of the refracted waves change. Since the value of n are ratio it doesn't have any specific unit of measurement, but the value of n, is specific for certain materials like air, glass, and water. For Snell's Law we also used old equations but with a little twist. As we all know velocity V, can be found by multiplying frequency F x wavelength λ. Well we can use that equation for help with Snell's Law by putting deep .vs. shallow in front the the characters to solve for the velocity of the deep and shallow water.

After we learned about Snell's law Ms.K handed out an assignment to test how much we could put our knowledge to use. Here was the example questions and the answers if you didn't manage to get them in class.

A water wave of frequency 10.0Hz and speed 40cm/s is traveling in deep water. It then moves into shallow water where its speed is 30cm/s. The angle of refraction is 30°

Find
A) The index of refraction
n=Vdeep /V shallow

n=(40cm/s)/(30cm/s)

n=1.33

B) The wavelengths in the two media

Deep
λdeep=Vdeep/F

=(40cm/s)/10.0Hz

λdeep=4cm

Shallow
λshallow=Vshallow/F

=(30cm/s)/10.0Hz

λshallow=3cm

C)the angle of refraction in the shallow water.
For this question there is more than one way to solve it .

~solution 1~
n=sinθi/sinθr

sinθr=sin30°/1.33

sinθr=0.375 or 22°


~solution 2~
sinθi/sinθr=Vi/Vr

sinθr=(sinθi x Vr)/Vi

=(sin30° x 30cm/s)/(40cm/s)

sinθr=0.375 or 22°


~solution 3~
sinθi/sinθr=λdeep/λshallow

sinθr=(sinθi x λshallow)/λdeep

=(sin30° x 3 cm)/4cm

sinθr=0.375 or 22°

Well that was pretty much it for our class on Friday, but for those who haven't already, the assignment for the week were questions 5-8 in our green booklet, and our wave research project is due on Tuesday. See you all on Monday!!
O yea the next scribe is Melissa

If you're thinking you should realize that question #6 is a repeat of the question we did on the whiteboard.