## Optics B/C

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

### 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.
If I saw that problem on the test I would just take my graphing calculator and plug in y=1.66sin(x/2) and y=sin(x) and see where they intersect. This method gives the answer of ~67.78 degrees. However I pose my own question: would the proctor have accepted an answer of 0 degrees? Technically, it is correct because 0/2 is 0, but my only fear is that the proctor wouldn't be the best versed in grading and may just be a volunteer and be simply going off of the answer key.
When I was playing around with the graphing calculator I also noticed that there isn't always a solution for all indexes of refraction (for angles between 0 and 90). The only indexes that this property worked for were (sqrt(2)<n<2). I wonder why specifically sqrt 2 was the limit..
Just some food for thought
Nevertheless, I am not sure how to answer this question without the graphing calculator.
I was in a bin

Rustin '19
UPenn '23

kenniky
Member
Posts: 283
Joined: January 21st, 2016, 6:16 pm
State: MA
Location: ABRHS --> UMichigan

### 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
$sin(2x)=2sin(x)cos(x)$

so

$\frac{sin(x)}{sin(x/2)} = \frac{2sin(x/2)cos(x/2)}{sin(x/2)} = 2cos(x/2)=1.66$

$cos(x/2)=0.83$

From there it's trivial to solve for x
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kenniky
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### Re: Optics B/C

Double post because editing is broken-
If I saw that problem on the test I would just take my graphing calculator and plug in y=1.66sin(x/2) and y=sin(x) and see where they intersect. This method gives the answer of ~67.78 degrees. However I pose my own question: would the proctor have accepted an answer of 0 degrees? Technically, it is correct because 0/2 is 0, but my only fear is that the proctor wouldn't be the best versed in grading and may just be a volunteer and be simply going off of the answer key.
When I was playing around with the graphing calculator I also noticed that there isn't always a solution for all indexes of refraction (for angles between 0 and 90). The only indexes that this property worked for were (sqrt(2)<n<2). I wonder why specifically sqrt 2 was the limit..
Just some food for thought
Nevertheless, I am not sure how to answer this question without the graphing calculator.
$sin(2x)=2sin(x)cos(x)$

so

$\frac{sin(x)}{sin(x/2)} = \frac{2sin(x/2)cos(x/2)}{sin(x/2)} = 2cos(x/2)=1.66$

$cos(x/2)=0.83$

From there it's trivial to solve for x
$2cos(x/2)=n$

over $0 \leq x \leq 90$, cos(x/2) goes from 1 to $\frac{\sqrt{2}}{2}$, which is why the only indices of refraction that work are $\sqrt{2}$ to 2
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jonboyage
Member
Posts: 102
Joined: December 13th, 2016, 8:32 am
State: PA

### Re: Optics B/C

Double post because editing is broken-
If I saw that problem on the test I would just take my graphing calculator and plug in y=1.66sin(x/2) and y=sin(x) and see where they intersect. This method gives the answer of ~67.78 degrees. However I pose my own question: would the proctor have accepted an answer of 0 degrees? Technically, it is correct because 0/2 is 0, but my only fear is that the proctor wouldn't be the best versed in grading and may just be a volunteer and be simply going off of the answer key.
When I was playing around with the graphing calculator I also noticed that there isn't always a solution for all indexes of refraction (for angles between 0 and 90). The only indexes that this property worked for were (sqrt(2)<n<2). I wonder why specifically sqrt 2 was the limit..
Just some food for thought
Nevertheless, I am not sure how to answer this question without the graphing calculator.
$sin(2x)=2sin(x)cos(x)$

so

$\frac{sin(x)}{sin(x/2)} = \frac{2sin(x/2)cos(x/2)}{sin(x/2)} = 2cos(x/2)=1.66$

$cos(x/2)=0.83$

From there it's trivial to solve for x
$2cos(x/2)=n$

over $0 \leq x \leq 90$, cos(x/2) goes from 1 to $\frac{\sqrt{2}}{2}$, which is why the only indices of refraction that work are $\sqrt{2}$ to 2

Ahaaa smart people. I totally forgot about the double angle formula. Not a good thing to forget :/
I was in a bin

Rustin '19
UPenn '23

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

### Re: Optics B/C

However, more attempts to isolate $x$ were unsuccessful.
If you can, invest in a calculator like the TI-89. Perfectly allowed by the rules, and has an algebra solver in it. In something like SciO you really shouldn't have the time to do this by hand, this saves so much time (and is applicable in many places outside of SciO)
UT Austin '19
Liberal Arts and Science Academy '15

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

### Re: Optics B/C

Hello,

Thanks a lot for all your help. I finally understood it after your help.
If I saw that problem on the test I would just take my graphing calculator and plug in y=1.66sin(x/2) and y=sin(x) and see where they intersect. This method gives the answer of ~67.78 degrees.
I went ahead and plugged those two equations on my TI-84 and it seemed that since that the y-scale of the equations barely ever reached -2 or 2 that I had to expand the entire scale by a large amount to see those lines. After that, I found the intersection point to be around (67.802524, 0.92588723), essentially the same answer as yours.

Going back to the original question, "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 plugged the x-value of the intersection point into Snell's Law, just to be sure that everything checks out.

$sin(67.8)=1.66*sin(x)$, $x$ being the refracted angle.

After that, I isolate $x$ to get $x=sin^-1(sin(67.8)/1.66)$.

Solving for $x$, will get you with 33.9 degrees. Multiplying this value by 2 with get you 67.8 degrees, which is exactly the same as the x-value of the intersection point.

Thanks for all your help again,
whyiamafool
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kenniky
Member
Posts: 283
Joined: January 21st, 2016, 6:16 pm
State: MA
Location: ABRHS --> UMichigan

### Re: Optics B/C

If you can, invest in a calculator like the TI-89. Perfectly allowed by the rules, and has an algebra solver in it. In something like SciO you really shouldn't have the time to do this by hand, this saves so much time (and is applicable in many places outside of SciO)
On the contrary, I would argue that being able to do these problems without your calculator is a useful skill and should be even faster than typing it into your calculator (especially if you have a bunch of trig that you have to locate in the menus)
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jkang
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Posts: 107
Joined: October 17th, 2014, 8:49 pm
State: TX

### Re: Optics B/C

On the contrary, I would argue that being able to do these problems without your calculator is a useful skill and should be even faster than typing it into your calculator (especially if you have a bunch of trig that you have to locate in the menus)
Tbh by the time I reached senior year (I got my 89 in sophomore year of hs) I knew all the shortcuts and key locations by muscle memory, and I could probs type out the equation on the solver before I could even write it. To each their own, but I believe that in terms of events where a mere seconds can lead to different placements, a calculator that could shave off quite a few of those seconds could help quite a lot.
UT Austin '19
Liberal Arts and Science Academy '15

Skink
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### Re: Optics B/C

I usually don't ask rule interpretation questions because I trust my own judgment, but I need validation here. Answers are no substitute for an official clarification, but I'm not seeking one!

4.d.xii.
The word "measurement" is used. Does this suggest to you that there might be laboratory work in addition to the box? I admittedly haven't looked closely at the S/N topics until this month.

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

### Re: Optics B/C

4.d.xii.
The word "measurement" is used. Does this suggest to you that there might be laboratory work in addition to the box? I admittedly haven't looked closely at the S/N topics until this month.
If I was writing for states/nats, I'd print a figure and either give a scale or make it to scale, then have students do analysis based on that (I'm assuming most teams at that level of competitions will have rulers/protractors/etc already prepared). Considering this is the "written test" portion, I don't think it's very likely that there will be another lab on top of the laser. But state and even national events can always be a crapshoot somehow, so there is a nonzero possibility of some event supervisor somewhere deciding they want a second lab for this event.
UT Austin '19
Liberal Arts and Science Academy '15