# Estimating image distance in lens and mirror

1. Jul 9, 2010

### ShineyPenny

I'm having difficulty understanding how to estimate the location of an image in a lens/mirror.

I tried understanding it by going off the books question/answer section:
"A spherical concave lens has a focal length of 16 cm, and an object is placed 8 cm from the lens." It says the answer is 6.0 cm.

What formula is being used, I cannot figure out how they got 6 cm? All I see in the book for formulas is f=R/2 and R=2f.

I'm confused :(

2. Jul 9, 2010

### Staff: Mentor

For a lens, it's usually called the "thin lens equation", sometimes with "Gaussian" in front of it. Your textbook surely has it. Try the index at the end of the book.

The two formulas you named, give the focal length of a mirror with radius R. There's a similar (but a bit more complicated) formula for the focal length of a lens, which is usually called the "lensmaker's equation." But you don't need that for this problem.

Oh, and either you left out a minus sign, or the book has a typographical error, or it's being deliberately sneaky. A concave lens has a negative focal length, not positive.

3. Jul 9, 2010

### ShineyPenny

OK, I believe I have it after running a search for it...it's actually not in the book :(

Is the equation:

(1/Object Distance)+(1/Focal Length) = (1/image location)
(1/45) + (1/36) = .05; then (1/.05) = 5.999

This is it? I'm just making sure :)

4. Jul 9, 2010

### Staff: Mentor

It's actually usually written as

$$\frac{1}{o} + \frac {1}{i} = \frac {1}{f}$$

where o, i and f are object distance, image distance and focal length. Some books use s and s', or p and q, instead of o and i.

Why are you using 45 and 36 for object distance and focal length, when your original problem had 8 and 16? (actually -16 because it has to be negative as I said before)

5. Jul 10, 2010

### ShineyPenny

I have no idea where 45 and 16 came from, sorry it was a long night

I input (1/8o) + (1/-16i) = .0625f, then go (1/.0625f) and come out with 16.

Either I'm missing something big, or I haven't drank enough coffee this morning. I even plugged the numbers into this website (http://www.photonics.byu.edu/Thin_Lens_Calc.phtml) and it's not coming up with 6 cm for the image location.

Sorry, I'm not a math person but I'm trying to grasp this

6. Jul 10, 2010

### ShineyPenny

OK

I have figured it out, the book is wrong and it was verified by the professor :)

I inputted: (1/16f) + (1/8o) = .1875, then (1/.1875)= 5.33.

They did not use a negative sign.

Thank you! =)