Solving for Index of Refraction Using x, d, and θ

In summary, Homework Equations state that n1*sin(θ1)=n2*sin(θ2). To solve for θ2, sin(θ1-θ2) is used. When θ1 is small, an approximate solution can be found by using sinθ ≈ θ*/(1-cos2(θ2)).
  • #1
CINA
61
0

Homework Statement



Find the index of refraction n of the rectangle using x, d, and θ.

http://img338.imageshack.us/img338/6079/snellslaw.jpg [Broken]

Homework Equations



n1*sin(θ1)=n2*sin(θ2)

The Attempt at a Solution



Is it possible to solve this problem given the relevant data? What I did was try to find θ2 in terms of x, d, and θ (now called θ1). What I came up with was:

[tex]cos(\theta_{2})=\frac{x*sin(\theta_{1}-\theta_{2})}{d}[/tex]

From the two right triangles that can be constructed:

http://img11.imageshack.us/img11/6606/snellslaw2.jpg [Broken]

But either this isn't right or I don't see a simple way to proceed from here. I used wolfram to solve the above equation for θ2, but the result seems too complicated to be the answer. Can someone help, anyone with a simpler approach? Perhaps n1*sin(θ1)=n2*sin(θ2) is useless in this case?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
You do have to use Snell's law to get the relation among x, d, and θ1. Expand sin(θ1-θ2). Find sin(θ2). Write both sin(θ2) and cos(θ2) in terms of sin(θ1). Use them in the expression for d.

ehild
 
Last edited:
  • #3
What do you mean "Find sin(θ2)." Do you mean isolate it from sin(θ1-θ2)=sin(θ1)*cos(θ2)-cos(θ1)*sin(θ2)=d/a ?
 
  • #4
CINA said:
What do you mean "Find sin(θ2)." Do you mean isolate it from sin(θ1-θ2)=sin(θ1)*cos(θ2)-cos(θ1)*sin(θ2)=d/a ?

Find sin(θ2) from Snells law in terms of sin(θ1) and the refractive index of the slab relative to the surrounding medium.

ehild
 
Last edited:
  • #5
Ok, so here are the equations I'm using:
[tex](1)n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2}) [/tex]

[tex](2)cos(\theta_{2})=\frac{x}{a}[/tex]

[tex](3)sin(\theta_{1}-\theta_{2})=\frac{d}{a}[/tex]I find [tex]sin(\theta_{2})[/tex] from (1):

[tex]\frac{n_{1}}{n_{2}}sin(\theta_{1})=sin(\theta_{2})[/tex]

I expand (2):

[tex]sin(\theta_{1})cos(\theta_{2})-cos(\theta_{1})sin(\theta_{2})=\frac{d}{a}[/tex]

Plug (1) and (3) into (2):

[tex]sin(\theta_{1})\frac{x}{a}-cos(\theta_{1})\frac{n_{1}}{n_{2}}sin(\theta_{1})=\frac{d}{a}[/tex]

But..."a" is an unknown, I can't solve for n2 until a is eliminated. Am I doing something wrong?
(n1 is assumed to be 1)
 
Last edited:
  • #6
Hello CINA,

Just out of curiosity, is there anything left out of the problem statement such as θ being small (perhaps less than π/18 [which corresponds to around 10o)?

If so, approximations exist that would make solving for an approximate answer much easier.

For small θ (where θ is expressed in radians),
sinθθ
cosθ ≈ 1​

I'm not sure if assuming a small θ applies to this particular problem though.
 
  • #7
CINA said:
But..."a" is an unknown, I can't solve for n2 until a is eliminated. Am I doing something wrong?
(n1 is assumed to be 1)

cos(θ2)=x/a . Write a in terms of x and cos(θ2) and use cos(θ2)=√(1-sin2(θ2))

ehild
 
  • #8
collinsmark said:
Hello CINA,

Just out of curiosity, is there anything left out of the problem statement such as θ being small (perhaps less than π/18 [which corresponds to around 10o)?

Hello, Collinsmark,

There is no indication in the problem that θ1 is small. As the refractive index of the slab is to be obtained, θ1 and d has to be accurately measured, so they can not be two small. The problem can be solved without assuming small angle of incidence.

ehild
 

1. What is the index of refraction and why is it important?

The index of refraction is a measure of how much a material can bend light as it passes through it. It is an important concept because it helps us understand how light behaves in different materials, which is crucial in many scientific and technological applications such as optics and fiber optics.

2. How do you solve for the index of refraction using x, d, and θ?

The index of refraction (n) can be calculated using the formula n = x/(d*tan(θ)), where x is the distance between the light source and the object, d is the distance between the object and the observer, and θ is the angle of incidence of the light ray. Simply plug in the values for x, d, and θ and solve for n.

3. What are the units for the index of refraction?

The index of refraction is a unitless quantity, as it is simply a ratio of the speed of light in a vacuum to the speed of light in a given medium. However, it is often represented with the symbol "n".

4. Can the index of refraction be greater than 1?

Yes, the index of refraction can be greater than 1. In fact, most materials have an index of refraction greater than 1, meaning that they can bend light. The higher the index of refraction, the more the material can bend light.

5. How does the index of refraction change with different materials?

The index of refraction varies for different materials, as it depends on the density and optical properties of the material. Generally, materials with a higher density and stronger interactions with light have a higher index of refraction.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
855
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
8K
  • Introductory Physics Homework Help
Replies
11
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
7K
Back
Top