What Is the Index of Refraction of the Liquid in the Tank?

AI Thread Summary
To determine the index of refraction of the unknown liquid in the tank, Snell's Law is applied, which relates the angles of incidence and refraction to their respective indices of refraction. The observer can see the corner of the tank, indicating a specific angle of refraction. The calculations involve using the dimensions provided (D = 1.45 m and L = 1.13 m) to find the angles and subsequently the index of refraction. The index of refraction for air is approximately 1, and the necessary calculations lead to the conclusion that the index of refraction of the liquid is derived from the ratio of the sine of the angles. The final value for the index of refraction of the liquid is confirmed through these calculations.
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[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
 

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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.
 
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?
 
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?
 
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?
 
SO;

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

so

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