# Biconvex Lens

1. Mar 29, 2005

### Jacob87411

An object is located 36 cm to the left of a biconvex lens of index of regraction 1.5. The left surface of the lens has a radius of curvature of 20 cm. The right surface of the lens is to be shaped so that a real image will be formed 72 cm to the right of the lens. What is the required radius of curvature for the second surface?

I asked earlier on this, but I still got it wrong: this is what I did -

I used 1/F=(n-1)(1/R1 + 1/R2)

1/72 = (1.5 - 1) (1/.2 + 1/R2)
Solved it and got R2 to equal 45 cm. I am pretty sure 1/F shouldnt be 72, but if its unknown that would make it unsolvable unless I am missing something?

2. Mar 29, 2005

### Staff: Mentor

First use the thin lens equation to solve for the focal length of the lens:
$1/o + 1/i = 1/f$​

3. Mar 29, 2005

### Jacob87411

Yeah I thought of that.

1/f = 1/56 + 1/72

Is it 56 because there is the 20 cm radius of the left lens and then 36 cm to the object?

4. Mar 29, 2005

### Staff: Mentor

No. 20 cm is the radius of curvature of the left surface of the lens, not the thickness of the lens. Assume the lens is thin enough to use the thin lens formula:
$1/36 + 1/72 = 1/f$​

5. Mar 29, 2005

### Jacob87411

Ah alright, that gives focal length of 4.167 cm, then just plug that into the

1/f=(n-1)(1/R1 + 1/R2)

Last edited: Mar 29, 2005
6. Mar 29, 2005

### Staff: Mentor

Careful not to mix up f with 1/f.

7. Mar 29, 2005

### Jacob87411

Right, thanks for the help