Another Index of Refraction Problem HELP PLEASE

In summary, the problem involves finding the index of refraction of an unknown liquid in a rectangular metal tank. Using Snell's Law and the dimensions of the tank, the angle of the incident ray in the water can be found and used to calculate the index of refraction of the liquid. The index of refraction in air is approximately 1. The final answer should be given with the correct number of significant figures.
  • #1
physicsbhelp
298
0
[SOLVED] Another Index of Refraction Problem! HELP! PLEASE

Homework Statement



When the rectangular metal tank in Figure 33-49 is filled to the top with an unknown liquid, observer O, with eyes level with the top of the tank, can just see the corner E. A ray that refracts toward O at the top surface of the liquid is shown. If D = 1.45 m and L = 1.13 m, what is the index of refraction of the liquid?
The Figure 33-49 is attached to this thread!

Homework Equations



Sin(Theda)N1= Sin(Theda2)N2
I don't know if this equation will help or not?!

The Attempt at a Solution



180 degrees, but that didnt work
 

Attachments

  • Fig. 33-49.doc
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  • #2
The eqn to be used is Snell's law and you should use it because it describes the relationship between angles of incidence and refraction and the RIs. You are using the correct eqn.

Find i and r, the two angles mentioned above. Use geometry to find sin i and sin r.
 
  • #3
i and r are the angles of incidence and refraction respectively, what you insist on calling theda and theda2. If it is going from liquid to water, the angle i is the angle the incident ray makes with the normal at the point of incidence on the interface. I hope you know by now what angle r is.

I didn't want to say it, but doesn't r look like 90 deg to you?
 
  • #4
OOOOO now i get what you mean!
SRY it took me a while. okay so to find sin i i have to do sin(i) = 1.13m , right?
 
  • #5
Not right. Sin of an angle lies between -1 and +1.

In your diagram, sin(i) = L/diagonal.

(sin i)/(sin r) is also equal to Nair/Nliquid.

Now can you do it?
 
  • #6
SO;

1.13/1.83831= sin (i)
 
  • #7
Yes. Please finish off the rest of it.
 
  • #8
(sin i)/(sin r) is also equal to Nair/Nliquid.

so

sin(37.93) / sin(37.93) = Nair/Nliquid
would
does Nliquid= 1.13?
 
  • #9
physicsbhelp said:
(sin i)/(sin r) is also equal to Nair/Nliquid.

so

sin(37.93) / sin(37.93) = Nair/Nliquid
would
does Nliquid= 1.13?

Quick, what is the value of r?
 
  • #10
r not eq to i. r=90 deg. Look at the other thread.
 
  • #11
oooo okay so sin(37.93) / sin(90) = Nair/Nliquid

is Nliquid= 1.13?
 
  • #12
if you could start thinking about the problem a bit on your own... THAT would be great.
 
  • #13
physicsbhelp said:
oooo okay so sin(37.93) / sin(90) = Nair/Nliquid

is Nliquid= 1.13?

What are you using for Nair?
 
  • #14
Use the dimensions D and L to find the angle of the light in the water with respect to the vertical (normal to interface). The index of refraction in air is approximately 1, or 1.0003.

Use Snell's Law as you wrote it.

Refer to http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/refr.html

I think it was discussed earlier that the angle (between the ray and the normal) in air was a right angle.
 
  • #15
sin(37.93) / sin(90) = 1/Nliquid
 
  • #16
Yes, but be careful with your significant figures.
 
  • #17
thanks.
 

1. What is an index of refraction?

An index of refraction is a measure of how much a material slows down the speed of light passing through it compared to the speed of light in a vacuum. It is a unitless number typically denoted by the symbol "n".

2. How is the index of refraction calculated?

The index of refraction is calculated by dividing the speed of light in a vacuum by the speed of light in the material. The speed of light in a vacuum is approximately 299,792,458 meters per second. The speed of light in a material can vary depending on its properties.

3. Why is the index of refraction important?

The index of refraction is important because it helps determine how light will behave when passing through different materials. It is used in a variety of applications, such as designing lenses and understanding the behavior of light in different mediums.

4. What factors can affect the index of refraction?

The index of refraction can be affected by several factors, including temperature, pressure, and the wavelength of the light passing through the material. It can also vary depending on the type of material and its chemical composition.

5. How can I solve an index of refraction problem?

To solve an index of refraction problem, you will need to know the speeds of light in the vacuum and the material in question. You will also need to know the angle of incidence and refraction, and use the Snell's law equation (n1sinθ1 = n2sinθ2) to calculate the index of refraction of the material.

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