## Optics B/C

MrGood
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### Re: Optics B/C

I haven't been able to find any good sources for the films chemicals and dies section of the absorbtion spectra topic, does anyone have good sources they know of or general inforation?

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### Re: Optics B/C

I'm interested in checking this again a little later in the season- what type of scores are people averaging on the LS? Additionally, how difficult has the barrier mirror typically been for you? i.e. at a convenient 45° angle vs. 180° away from the laser.
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

breakingankles
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### Re: Optics B/C

The people doing it on my team are able to get close to a perfect score almost every time (all 5 mirrors + barrier mirror + very accurate). It's just the test that kills us....

kenniky
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### Re: Optics B/C

Does anyone know what a minimum dispersion angle is? (not minimum deviation). it came up on a test we took

10. What is the index of refraction for an equilateral prism with a minimum dispersion angle of 30 degrees?
a. 1.50 b. 1.52 c. 1.58 d. 1.62

If you use the formula for minimum deviation you get 1.41. The answer is [hide]1.62[/hide]
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ericlepanda
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### Re: Optics B/C

How are some of you able to get the barrier mirror? Since it can be placed at any angle, isn't it impossible to use?
Marie Murphy

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jonboyage
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### Re: Optics B/C

ericlepanda wrote:How are some of you able to get the barrier mirror? Since it can be placed at any angle, isn't it impossible to use?

It is certainly not impossible to use. There are many different ways the barrier mirror can be placed, however the judge should not make it an impossible mirror, for example too close to any wall. There are a lot of interesting solutions that my partners and I have found for various barrier locations simply by trying out different ways the mirror can be placed. I would recommend playing around with an LSS that you can make for yourself if you don't have one already. There are some really good suggestions earlier in this thread. Good luck!
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gryphaea1635
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### Re: Optics B/C

Hi guys,

Can someone explain how to solve this problem? It was on the question marathon (page 3 or 4 I think, kenniky's problem on angle of deviation in prisms)
Thanks!

Question:
In the prism below, α is 35° and θ is 19°. The prism has index of refraction 1.45 and is in a vacuum

What is the deviation?

(It was 19 degrees between the light ray and prism face, not the normal)

I tried finding the initial angle of refraction, and then using that to find the second angle of incidence out of the prism by making a triangle with the apex angle, the complementary of the second angle of incidence, and the complementary of the first angle of refraction. But then I got around 95 degrees as the angle of incidence within the prism, which doesn't make sense, and that's as far as I got haha.
Am I taking the right approach? What do?
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jkang
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### Re: Optics B/C

gryphaea1635 wrote:Hi guys,
Can someone explain how to solve this problem? It was on the question marathon (page 3 or 4 I think, kenniky's problem on angle of deviation in prisms)
Thanks!
Question:
In the prism below, α is 35° and θ is 19°. The prism has index of refraction 1.45 and is in a vacuum
What is the deviation?

(It was 19 degrees between the light ray and prism face, not the normal)
I tried finding the initial angle of refraction, and then using that to find the second angle of incidence out of the prism by making a triangle with the apex angle, the complementary of the second angle of incidence, and the complementary of the first angle of refraction. But then I got around 95 degrees as the angle of incidence within the prism, which doesn't make sense, and that's as far as I got haha.
Am I taking the right approach? What do?

For future reference, inserting the image or linking to the original problem (here) can be useful for the visuals. In terms of solving the problem, refer to the last equation found on the Wikipedia for minimum deviation. Here iwould be 90-19=71 degrees, since we are calculating for the angle of incidence. We know that A[/] is 35 degrees, and [i]n is 1.45. Plugging in these numbers, we find our angle of deviation to be 27.72 degrees (28 technically with the given significant figures), which seemed to have been the correct solution. You can probably derive this equation for yourself without too much difficulty through the use of Snell's law and arbitrary angles.
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ericlepanda
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### Re: Optics B/C

jonboyage wrote:
ericlepanda wrote:How are some of you able to get the barrier mirror? Since it can be placed at any angle, isn't it impossible to use?

It is certainly not impossible to use. There are many different ways the barrier mirror can be placed, however the judge should not make it an impossible mirror, for example too close to any wall. There are a lot of interesting solutions that my partners and I have found for various barrier locations simply by trying out different ways the mirror can be placed. I would recommend playing around with an LSS that you can make for yourself if you don't have one already. There are some really good suggestions earlier in this thread. Good luck!

But how are you able to account for the mirror being at any angle? If they put a barrier mirror at a weird angle like (for example) 157 degrees, how would anyone be able to use it?
Marie Murphy

Wright State\Regional\State\Nationals
Optics: 4th\2nd\2nd\6th
Food Science: 8th\1st\12th\14th
Hovercraft: 12th\1st\1st\2nd
Ecology 10th\3rd\5th\11th

0ddrenaline
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### Re: Optics B/C

ericlepanda wrote:
jonboyage wrote:
ericlepanda wrote:How are some of you able to get the barrier mirror? Since it can be placed at any angle, isn't it impossible to use?

It is certainly not impossible to use. There are many different ways the barrier mirror can be placed, however the judge should not make it an impossible mirror, for example too close to any wall. There are a lot of interesting solutions that my partners and I have found for various barrier locations simply by trying out different ways the mirror can be placed. I would recommend playing around with an LSS that you can make for yourself if you don't have one already. There are some really good suggestions earlier in this thread. Good luck!

But how are you able to account for the mirror being at any angle? If they put a barrier mirror at a weird angle like (for example) 157 degrees, how would anyone be able to use it?

Work backwards. Find the angle from your target location to the barrier, then reflect off from the barrier to find the location where you should place the mirror.

jkang
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### Re: Optics B/C

With the UT Regional now over, here are my test and key for the tournament. I included explanations for each of my answers in the key this time, so hopefully that will help. Topics are a bit heavy on waves so unlikely to be seen at states/nats, but wave optics >> geometric optics
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### Re: Optics B/C

jkang wrote:With the UT Regional now over, here are my test and key for the tournament. I included explanations for each of my answers in the key this time, so hopefully that will help. Topics are a bit heavy on waves so unlikely to be seen at states/nats, but wave optics >> geometric optics

Thanks for sharing that, and I very much agree
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

whyiamafool
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### Re: Optics B/C

Hello,

This question popped up in one of the tests I took recently.

"Light is incident on a piece of flint glass (n = 1.66) from the air in such a away that the angle of refraction is exactly half the angle of incidence. What are the values of the angles of incidence and refraction?"

I went ahead and tried to apply it to Snell's Law, to get this ---> $sin(x)=1.66*sin(x/2)$

Then, tried to isolate $x$ as much as I could, until I got to this point ---> $(sin(x)/sin(x/2))=1.66$

However, more attempts to isolate $x$ were unsuccessful.

If anyone could help with this, I would greatly appreciate it.

Thanks,
Whyiamafool
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John Richardsim
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### Re: Optics B/C

whyiamafool wrote:Hello,

This question popped up in one of the tests I took recently.

"Light is incident on a piece of flint glass (n = 1.66) from the air in such a away that the angle of refraction is exactly half the angle of incidence. What are the values of the angles of incidence and refraction?"

I went ahead and tried to apply it to Snell's Law, to get this ---> $sin(x)=1.66*sin(x/2)$

Then, tried to isolate $x$ as much as I could, until I got to this point ---> $(sin(x)/sin(x/2))=1.66$

However, more attempts to isolate $x$ were unsuccessful.

If anyone could help with this, I would greatly appreciate it.

Thanks,
Whyiamafool

Thanks for posting this question. It's really quite interesting.

From where you left off, my first idea is to graph it. Drop $f(x) = (sin(x)/sin(x/2)) - 1.66$ into a graphing calculator and find where the function intersects the x-axis over the interval (0,90).
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### Re: Optics B/C

John Richardsim wrote:
whyiamafool wrote:Hello,

This question popped up in one of the tests I took recently.

"Light is incident on a piece of flint glass (n = 1.66) from the air in such a away that the angle of refraction is exactly half the angle of incidence. What are the values of the angles of incidence and refraction?"

I went ahead and tried to apply it to Snell's Law, to get this ---> $sin(x)=1.66*sin(x/2)$

Then, tried to isolate $x$ as much as I could, until I got to this point ---> $(sin(x)/sin(x/2))=1.66$

However, more attempts to isolate $x$ were unsuccessful.

If anyone could help with this, I would greatly appreciate it.

Thanks,
Whyiamafool

Thanks for posting this question. It's really quite interesting.

From where you left off, my first idea is to graph it. Drop $f(x) = (sin(x)/sin(x/2)) - 1.66$ into a graphing calculator and find where the function intersects the x-axis over the interval (0,90).

I'm very rusty on my trigonometry so I can't provide an elegant solution - https://www.microscopyu.com/tutorials/refraction but this gizmo leads me to believe the answers will be ~68 and ~34 degrees. Depending on the precision required in the answer, you could just make a quick table and plug values; you can narrow it down pretty quickly to +/- 1 degree that way.
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