## Optics B/C

jonboyage
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Posts: 102
Joined: December 13th, 2016, 8:32 am
State: PA

### 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?
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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Rustin '19
UPenn '23

gryphaea1635
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State: MA
Location: touching my spaghet

### 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?
ESKKEEEETTTIIITT

jkang
Member
Posts: 107
Joined: October 17th, 2014, 8:49 pm
State: TX

### 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?
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 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.
UT Austin '19
Liberal Arts and Science Academy '15

ericlepanda
Member
Posts: 58
Joined: February 9th, 2016, 2:53 pm
Division: C
State: IL

### 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?
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?
ntso
quack quack

0ddrenaline
Member
Posts: 117
Joined: May 21st, 2015, 6:36 pm
State: MI

### 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?
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
Member
Posts: 107
Joined: October 17th, 2014, 8:49 pm
State: TX

### 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
UT Austin '19
Liberal Arts and Science Academy '15

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Posts: 66
Joined: January 6th, 2017, 4:51 pm
State: PA
Location: Swarthmore College

### 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
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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Posts: 3
Joined: July 11th, 2016, 1:48 pm
Division: B
State: OH

### 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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Joined: February 26th, 2014, 10:54 am
State: MI
Location: Robinson Twp.

### 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
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).
Si Quaeris Peninsulam Amoenam Circumspice

SPP SciO
Member
Posts: 261
Joined: March 24th, 2015, 8:21 am
Division: B
State: NY
Location: Brooklyn

### 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
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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