Agent_A
  • Agent_A
Refractive index question.........
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Agent_A
  • Agent_A
1 Attachment
anonymous
  • anonymous
I can walk you through it if you are on... send me a message
anonymous
  • anonymous
I would strongly recommend that you look at a single step at a time and actually try and reason this problem through

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Then once you have gotten the answer yourself you can check my to see if they agree.... This is by far the best practice since I can break down the steps for you
anonymous
  • anonymous
Ok so step one is to note that in a medium the velocity (v) of light in the material is related to is frequency times the wavelength of light in the material wavelength:
anonymous
  • anonymous
\[v_{m}=\nu \lambda_{m} \] Where \[ \nu \] (pronounced nu, different from v) is the frequency of the light.
anonymous
  • anonymous
Now the frequency of the light is special because it remains the same when light passes from one medium to another. This is because light can be treated as electromagnetic wave. Meaning it consists of oscillating electric and magnetic fields. At an interface, the oscillations must be coupled to one another such that at no time during the crossing do the two become unsynced. If they were to become unsynced then at some time during the passage from one medium to the other, there would be a discontinuity in the wave at the interface. Meaning the electric and magnetic fields jump discontinuously from one point in the first medium to another point in the second medium and continuing on at its new different frequency. Now this does not happen in mechanical waves, nor does it take place in EM waves. What does happen though is the speed of the transmitted wave and its associated wavelength change in such a way as to preserve the frequency.
anonymous
  • anonymous
Also, we can treat the air next to the glass as essentially a vacuum because the index of refraction of air is =1 up to the first 3 decimal places which is sufficient for this problem.
anonymous
  • anonymous
Therefore: \[v_a \approx v_{vac} = c = \lambda \nu \ \ \ \ \&\& \ \ \ \ v_{glass}= \lambda_{glass} \nu\] \[\rightarrow \ \ \ \ \frac{\lambda}{c} = \frac{1}{\nu} = \frac{\lambda_{glass}}{v_{glass}} \\ \rightarrow \ \ \ \ \lambda_{glass}=\frac{\lambda v_{glass}}{c}\] Given the definition for index of refraction: \[ v=\frac{c}{n} \ \ \ \ or \ \ \ \ n=\frac{c}{v} \] Therefore: \[ \lambda_{glass}=\frac{\lambda \frac{c}{n_{glass}}}{c} = \frac{\lambda}{n_{glass}} \] Which is also a valid definition of the index of refraction (ratio of the wavelengths in the two mediums).
anonymous
  • anonymous
Given that the wavelength of your light source in a vacuum is given as 638nm and the index of refraction of glass is 1.52; this gives: \[ \lambda_{glass}= \frac{\lambda}{n_{glass}} = \frac{638nm}{1.52} =419.73…nm \approx 420nm \]
anonymous
  • anonymous
Now applying Snell’s Law to the interface between the incident medium (glass) and the transmitted medium (liquid) yields: \[ \frac{ sin \theta_{i} }{ sin \theta_{t} } = \frac{ sin \theta_{glass} }{ sin \theta_{liquid} } = \frac{ \lambda_{glass} }{ \lambda_{liquid} } = \frac{ n_{liquid} }{ n_{glass} } \]
anonymous
  • anonymous
The critical angle for total internal reflectance is the angle of incidence that results in an angle of transmission of 90 degrees. If you draw the interface, you will see that the transmitted angle being 90 degree means the transmitted ray lies along the plane of the interface. Therefore: \[ n_i sin \theta_{i_{c}} = n_t sin 90^{o} = n_t \] Substituting in the known values: \[ n_t =n_{liquid_{A}} = n_i sin \theta_{i_{c}} = n_{glass} sin \theta_{c_{A}} = 1.52 * sin(52^{o}) = 1.1977…. \approx 1.20 \] Repeating this procedure for the other two liquids gives: \[ 1.06 \ for \ B \] and \[0.8998 \approx 0.9 \ for \ C \]
anonymous
  • anonymous
The critical angle for total internal reflectance is the angle of incidence that results in an angle of transmission of 90 degrees. If you draw the interface, you will see that the transmitted angle being 90 degree means the transmitted ray lies along the plane of the interface. Therefore: \[ n_i sin \theta_{i_{c}} = n_t sin \theta_t = n_t sin 90^{o} = n_t \\ \rightarrow n_t =n_{liquid_{A}} = n_i sin \theta_{i_{c}} = n_{glass} sin \theta_{c_{A}} \] Now I made a mistake here earlier and foolishly took the given values to be the desired critical angle inside the glass. However if you do this you will find the index of refraction of liquid C to be less than 1 which is (given the problem at hand) unphysical. As a result, I reread the question which then caused me to realize two things. First, we didn’t need to explicitly calculate the wavelength. It wasn't, as I thought, required for the question, so I guess you can leave that out. Second and right to the point, the given angles seem to correspond to the angles of incidence WITH RESPECT TO THE INITIAL AIR-GLASS INTERFACE. This then requires one more step to solve. First note that the critical angle given is with respect to the surface normal of the air-glass interface. Thus the transmitted critical angle (now in glass) will also be given with respect to the normal. Since parallel lines cut by a transversal have equal angles, the complimentary angle to the transmitted critical angle is equal to the critical angle (with respect to the surface normal) of the glass-liquid interface. The following diagram should help:
anonymous
  • anonymous
|dw:1442177367824:dw|
Agent_A
  • Agent_A
I also made the mistake of calculating the wavelength, but I changed it to something like your current solution so far (which is to calculate the angle, by subtracting it from 90 degrees). I didn't get the right answer. I'll post my solution, hold on.
Agent_A
  • Agent_A
\[n_(glassblock) = 1.52 --> Let: n_g\] \[n_asin(\theta_a) = n_gsin(\theta_b)\] \[n_asin(52) = (1.52)\sin(90-52)\] \[n_a = 1.19\] Which is not correct.
anonymous
  • anonymous
So again, by Snell’s Law (again): \[ n_{air} sin \theta_{air_{c}} = n_{glass} \ sin \theta_{glass_c} \\ \rightarrow \theta_{air_{c}} = sin^{-1} ( \frac{ sin \theta_{glass_{c}} }{ n_{glass}} ) \] And thus from the final relation given in the post above: \[ n_{liquid_{A}} = n_{glass} \ sin \theta_{c_{A}} = n_{glass} \ sin ( 90^{o} - sin^{-1}( \frac{ sin \theta_{glass_{c}} }{ n_{glass}} )) \] So from here just plug in your knowns.
Agent_A
  • Agent_A
Hmm let me try your solution.
anonymous
  • anonymous
Oh hello yes sorry about that I mistyped its really hard to edit equations in this interface by browser is lagging.
Agent_A
  • Agent_A
No problem!
Agent_A
  • Agent_A
Okay, I'm confused now....
Agent_A
  • Agent_A
WAIT, NO IT'S CORRECT!!!!!
anonymous
  • anonymous
Liquid A: \[ n_{liquid_{A}} = 1.25 \ sin ( 90^{o} - sin^{-1}( \frac{ sin(52^{o})}{1.25}) \approx 1.23 \]
Agent_A
  • Agent_A
Thank You, @PlasmaFuzer!!!!
Agent_A
  • Agent_A
I'm just confused about the arcsin part.
anonymous
  • anonymous
Wait a minute try did liquids B and C come out to be 1.35 and 1.40 respectively (rounded)?
anonymous
  • anonymous
Yea sorry about that I am getting frustrated I made this mistake in the first place so I condensed all the algebra so I could just knock it out in one post.
Agent_A
  • Agent_A
That's fine! I should be okay with the rest, and it's just that one part that I don't get....
anonymous
  • anonymous
I have an issue with those indicies of refraction though... usually total internal reflection happens when you are travelling from a more dense medium to a less dense medium so the fact their indicies are higher than that of glass (1.25) bothers me.
Agent_A
  • Agent_A
Yeah, I was wondering that, too.
anonymous
  • anonymous
Is it an computer input system? If so plug in the answer for B to see if it accepts it. I wouldn't be shocked I think this question is worded kind of confusingly. This should be an easy problem.
Agent_A
  • Agent_A
I thought it was easy, but I didn't get it on my own. The correct answers are "1.23, 1.35, and 1.40", a, b, and c respectively.
Agent_A
  • Agent_A
One last thing, how did you get arcsin? :)
anonymous
  • anonymous
Seriously what is going on (scratches head)? I have either completely forgotten something or my reasoning is flawed....
Agent_A
  • Agent_A
I think I'm going to ask my professor about the indices.... He might agree with us.
anonymous
  • anonymous
Ummm that part came from applying snell's law to the air-glass boundary... and solving for the transmitted critical angle (in glass).... so basically I have sin (theta glass_c) = stuff and I take the arcsin of that to get just the angle.
Agent_A
  • Agent_A
Like this? \[1.52\sin(\theta_a) = \sin(52)\]?
Agent_A
  • Agent_A
because I know we're solving for n, so we have two unknowns, yes?
anonymous
  • anonymous
Yea I am remember this criteria correclty... If we have total internal reflction then: \[ n_i sin \theta_{i_{c}} = n_t sin \theta_t = n_t sin 90^{o} = n_t \\ \rightarrow n_t =n_{liquid_{A}} = n_i sin \theta_{i_{c}} = n_{glass} sin \theta_{c_{A}} \\ \rightarrow sin \theta_{c_{A}} = \frac{n_{liquid_{A}}}{n_{glass}} > 1 \] Which cant be... Unless we are allowing for complex indicies of refraction.
anonymous
  • anonymous
Yes that is correct... and take the arcsin of both sides to get the angle itself
anonymous
  • anonymous
After dividing by 1.52 that is
Agent_A
  • Agent_A
but Snell's Law is \[n_asin(\theta_a) = n_bsin(\theta_b) \] if we are solving for theta, we would need to know another n too, right?
Agent_A
  • Agent_A
Just wondering where that went, in \[\theta = \sin^{-1} (\frac{ \sin(52) }{ 1.52 })\]
anonymous
  • anonymous
Oh I mentioned that above... It comes from the fact that the index of refraction of air is almost the index of refraction of the vacuum (at least to 3 decimal places)... So n_a is approx 1
Agent_A
  • Agent_A
Oh yes, okay. Wow.
Agent_A
  • Agent_A
Yes, that's right. It was there. It was just "one" so you didn't put it anymore.
Agent_A
  • Agent_A
sorry
anonymous
  • anonymous
Its ok.... Please ask your professor what is going on with this question and why 2 of the indicies of refraction are greater than that of glass. I would be curious to know if I am overlooking something fundamental or if it is just a bad problem.
anonymous
  • anonymous
When you do please message me with the answer I am really curious to know :D
Agent_A
  • Agent_A
Yes, of course! :)
anonymous
  • anonymous
Btw I know its wordy, but if you havent already I would go over all my posts. I break it down fairly well and if anything is unclear it may be helpful.
anonymous
  • anonymous
Thank you :D :D
Agent_A
  • Agent_A
Ah well, for the most part, looks like things are sorted out. Yes, I'll go over them later :). Thank YOU!
anonymous
  • anonymous
WAIT WAIT!!!!
Agent_A
  • Agent_A
???
anonymous
  • anonymous
I just realized (facepalms) I am kinda tired I think I need some coffee.... there isnt an issue here because the angles are different.... look:
anonymous
  • anonymous
Hold on my chat box is screwing up I have to refresh :/
Agent_A
  • Agent_A
No problem!
anonymous
  • anonymous
Oh nm :( it turned out to be a dead end... please do ask your professor... Hopefully it isnt me and I am just really tired. Anyways good luck I will see you around :D
Agent_A
  • Agent_A
Get some rest! Hope my problem didn't wear you out that much. LOL. Thanks! See you around! :)
Agent_A
  • Agent_A
I'll send you a message when I get to ask my prof.!
anonymous
  • anonymous
Hahaha nah youre fine Im doing my own homework right now which is why Im tired... thought it has been a while since I have done optics.... Still wondering about that problem.
anonymous
  • anonymous
It probably comes down to the fact that I didn't really use the wavelength of the light anywhere in the final calculation... It didn't show up at all.
anonymous
  • anonymous
Oh wow I really am tired (double facepalm).... I used 1.52 everywhere in my calculation as the index of refraction of glass... but then freaked out because I transposed the digits on my board here... I wrote 1.25 on all the subsequent comparisons not 1.52... The resulting indicies of the liquids are all < 1.52 and the answers were all correct because I used 1.52... *sigh
Agent_A
  • Agent_A
I was just about to get back to you. Glad you figured it out! On my end, I was wondering why it was greater than 1.00! But yes, the answers are all less than 1.52. It's funny how just as I was about to ask my teacher the question, I figured it out on my own too. Oh well. :P That's good, I guess?

Looking for something else?

Not the answer you are looking for? Search for more explanations.