Sarahisme

ok now what i've got is that: rearraging the TDSE: since V(x) = 0 [math] \frac{-\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} = i \hbar \frac{\partial \Psi}{\partial t}[/math] (1) then for [math] \Psi_I [/math] : [math] \frac{\partial^2 \Psi}{\partial x^2} = -k^2Acos(kx - \omega t) [/math] and [math] \frac{\partial \Psi}{\partial t} = \omega ASin(kx - \omega t) [/math] subbing these into 1 gives: LHS = [math] \frac{\hbar^2 k^2 A}{2m}cos(kx -\omega t) [/math] and RHS = [math] i \hbar \omega A sin(kx - \omega t) [/math] but then i am a little bit stuck... :S

hey all, just having a bit of trouble with this question: what i think i have figured out so far is that, V(x) = 0 (because its a free particle) and i am also thinking that i should be using the time dependent SE? would anyone be able to me a pointer in the right direction? Cheers Sarah

ah ok, i think i see now, thanks for that!

Hi all, i am having a lot of trouble trying to do this problem, as my textbook does not discuss circuits and it is only covered in 3 very brief lecture slides. if anyone would be able to point me to some good websites on how to do these type of problems or be able to help me in any other way that would be great! Thanks guys & gals! -Sarah

ok here is a diagram i drew: the first is without the resulting vector (i.e. adding them all up) in it, and the second has it in there 2nd drawing: what do you think of those?

hmm just wondering if anyone else will back me up here, i don't believe that the solution given for this quesiton is drawn correctly: their answer: or is it me who is wrong?

thanks for all you help again swansont!

wait no i think i've got it, i was looking at the wrong interface, i had it going from air to the block when i should have been looking at the converse! so the answer for part (d) should be that there is refraction in (b) because the angle of refraction is 81°, however in © there is no refraction because you can't take the sin^{-1} of a number greater than 1 (and we have 1.21 in this case)... yay!

oh, and for part (d) is all you have to do is apply snells law for n2 = 1.4? but isnt it not possible to get refraction because if n2 = 1.4 and n1 = air = 1 then the critical angle: [math] \theta_c = sin^{-1}(\frac{n_2}{n_1}) [/math] which doesn't exist.....?????!

yeah i suppose, thats pretty much what i did, but its only an intuitive way of doing it isnt it? i mean its not really a rigorous proof or anything...

hmmm, from my diagram i can see that it looks like it should be 45°, but i just can't prove it to myself..... as for part © would the answer be 60° ? by which i mean for part © [math] a = r(1- \frac{\sqrt3}{2}) [/math] ??

part (a) i am pretty confident about, but part (b) i didn't actually do it properly, so if that answer is correct, it was just a fluke i think could you give me a hint or something as how to do part (b)?

hmmmm.... i am really stuck now , anyone?

hmmm, now i don't know how i got the value for theta!?!? aghhh going crazy

ok, now for part (b) i get: manipulating the triangles wthin the semi-circle i get: [math] sin(\theta) = \frac{r - a}{r} [/math] [math] \theta = \frac{\pi}{4} [/math] and so [math] a = r(1- \frac{1}{\sqrt 2}) [/math] hows that for part (b)??

hey all, i am having a bit of trouble with this optics problem involving whispering gallery stuff. here is the question: and my answer so far: for part (a) the critical angle of a ray going from air to some medium with refraction index of n is given by: [math] sin \theta_c = n [/math] so you contrcut a diagram such as this: from this diagram you can see that: [math] sin \theta = \frac{r-a}{r} [/math] setting this equal to the critical angle , we get: a = r(1-n) or a = |r(1-n)| = r(n-1) but then for part (b) i can draw a digram, but i don't really know what to do from there.... so can anyone help me out? -Sarah

i tryed doing the sign test, but it didnt work, so yeah. .... i dunno what to do

hey all, my textbook has this sample question... "Find the shape of a reectangular box with no top having given volume V and the least possible total surface area of its five faces." now the book has the soluition, i.e. you just do S = xy + 2yz + 2xz and V = xyz and you end up finding just one critical point. my question is why should this point be a minimum? -sarah

got it all now!

hmm yeah ok, i'll probably just give the first-order maxima then, because as you say, experimentally thats usually the most useful (easy to find) one. Thanks for all you help swansont!

ok yeah, so in terms of this question do you think it wants an algebraic or a numerical answer? or both? because i think its a bit unclear

ok, wait i think i've got the first bit, but not the second bit of the question (adding the two)

so i assume the correct answer to this question would not really involve plugging in numbers then? or do you just do it for m=1 or something? well the thing is in my textbook, there is a similar question, which just does it for m = 1, but isnt that not enitirely correcct, because it could be done for m =2, m=3,...etc. ??? the reason i ask is because well, since you are given numerical values then shouldn't i be producing a numerical answer?

so yep i think it all looks correct i gather you do too Thanks swansont!

hey i have a question about this question .... ok so for interference maxima the equation is [math] dsin \theta_m = m \lambda [/math] , where m = 0,1,2,... now for the first grating (4000 lines/cm) and the red wavelength: [math] \theta = sin^{-1}(\frac{m \lambda}{d}) [/math] now my question is can i let m be anything i like or whats the deal? because wouldn't there be multiple (theoretically infinite amount) angles that construct interference could happen at? -Sarah