# Concave Lens problem

1. Apr 30, 2008

### jcpwn2004

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

http://img215.imageshack.us/img215/8457/physicsol4.jpg

2. Relevant equations

n1sin01=n2sin02 1/f = (n-1)(1/r1-1/r2)

3. The attempt at a solution

I'm really jus confused with this problem I don't really see where to start. I'm assuming the fish bowl is a converging lens? I still don't know where to go from there though.

2. May 1, 2008

### alphysicist

Hi jcpwn2004,

The problem here is not a thin lens problem; it is finding an image produced by a refracting surface. What equation would you use for these problems?

3. May 1, 2008

### jcpwn2004

ummm i really don't know lol, i thought refraction had to do with the n1sin01=n2sin02 equation...

4. May 1, 2008

### alphysicist

That equation (Snell's law) does describe the refraction of a light ray as it passes from one medium to the next.

However, in this problem we want to know how refraction at a single surface creates an image. You can use Snell's law to derive an expression that tells how an image is formed for light refracting from a surface. Your textbook probably does this for you and gives you the final result, which is something like:

$$\frac{n_1}{p}+\frac{n_2}{q} =\frac{n_2-n_1}{R}$$

Many books use many different symbols for the quantities in the denominators, so look for a section titled something like "image formed by refraction" to find out how it's written in your book. (The section is commonly between the section on spherical mirrors and thin lenses.)

Once you find it, you'll need to interpret what the symbols mean, and then apply it to your problem. What do you get?

5. May 1, 2008

### jcpwn2004

allright well I did n1/-p + n2/q = n2-n1/r and subbed in 1.33/-.2 + 1/q = (1-1.33)/.3 and got q = .18 which is the right answer but I kinda did guess and check with the numbers and i'm not sure if I but them in the right spot or if p should be negative like i made it. Also I'm not sure if di should be positive or negative, it just says inside the fish bowl, if I make p negative, shouldn't q be negative too?

6. May 1, 2008

### alphysicist

You have some errors in your equation; I guess they cancelled out somewhat to give you the right answer, but here are some notes.

There are some sign conventions that apply when using this equation. Your book may have different terminology, but it goes something like this:

The front side of the refracting surface is where the light is coming from (before it hits the surface), and the back side is where it is going to. So here the front side is the water side and the back side is the air side (because you're discussing the light that is going from the water to the air).

The sign of p: If the object is on the front side of the surface, it is a positive object length p; if it is on the back side of the surface it is a negative length p. Since the fish is in the water p is positive 0.2.

The sign of q: If the image is on the back side of the surface q is positive, and if on the front side q is negative. When you solve for q I think you should find that it turns out to be negative and so this means the image is on the front side of the lens (in the water).

The sign of R: If the light hits the surface on the concave side it is a negative R, and if it hits the surface on the convex side it is a positive R. (This is opposite the case for mirrors.) So you need to fix the sign of R in your equation.

Also, in your equation, on the left hand side you have n1=1.33 and n2=2, which is correct; but you have them switched on the right side.