# Homework Help: Another Index of Refraction Problem ! HELP! PLEASE

1. Jan 9, 2008

### physicsbhelp

[SOLVED] Another Index of Refraction Problem!!!!!!!!!!! HELP! PLEASE

1. The problem statement, all variables and given/known data

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!

2. Relevant equations

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

3. The attempt at a solution

180 degrees, but that didnt work

File size:
27 KB
Views:
189
2. Jan 9, 2008

### Shooting Star

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. Jan 9, 2008

### Shooting Star

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. Jan 9, 2008

### physicsbhelp

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. Jan 9, 2008

### Shooting Star

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. Jan 9, 2008

### physicsbhelp

SO;

1.13/1.83831= sin (i)

7. Jan 9, 2008

### Shooting Star

Yes. Please finish off the rest of it.

8. Jan 9, 2008

### physicsbhelp

(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. Jan 9, 2008

### Shooting Star

Quick, what is the value of r?

10. Jan 9, 2008

### Shooting Star

r not eq to i. r=90 deg. Look at the other thread.

11. Jan 9, 2008

### physicsbhelp

oooo okay so sin(37.93) / sin(90) = Nair/Nliquid

is Nliquid= 1.13?

12. Jan 9, 2008

### olgranpappy

if you could start thinking about the problem a bit on your own... THAT would be great.

13. Jan 9, 2008

### hage567

What are you using for Nair?

14. Jan 9, 2008

### Astronuc

Staff Emeritus
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. Jan 9, 2008

### physicsbhelp

sin(37.93) / sin(90) = 1/Nliquid

16. Jan 9, 2008

### hage567

Yes, but be careful with your significant figures.

17. Jan 9, 2008

thanks.