whispering gallery rays (light)

[Sarahisme]

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

[Sarahisme]

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)??

[Sarahisme]

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

[Sarahisme]

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

[swansont]

the approach for a and b look OK to me.

 

For part c, look at the symmetry of the problem. Where does the second internal reflection hit? And Snell's law is also a symmetry to use in constructing the angles, as you saw in part b.

[Sarahisme]

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)?

[swansont]

part (a) i am pretty confident about' date=' 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)?[/quote']

 

From the symmetry the internal reflection angle has to be 45 degrees, which is what it looked like you did.

[Sarahisme]

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] ??

[swansont]

hmmm' date=' 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] ??

 

For both, the system has to look symmetric - if you flip it it has to look the same. So for two reflections the second ray has to be vertical, so it changes by 90 degrees (The path looks like half of a square) - 45 on either side of the normal.

 

For the three reflection case, the path looks like half of a hexagon. Since the interior angle is 120 degrees, the reflection agle has to be 60, as you concluded.

[Sarahisme]

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...

[Sarahisme]

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.....?????!

[Sarahisme]

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!

[Sarahisme]

thanks for all you help again swansont!