Locating Object and Image with Converging Lens

[sbe07phy]

Homework Statement

A converging lens with a focal length of 6.00 cm forms an image of a 4.00 mm tall real object that is to the left of the lens. The image is 1.24 cm tall and erect.
Where is the object located? (Give the object distance using the correct sign.)
wrong check mark cm (to the left of the lens)
Where is the image located? (Give the image distance using the correct sign.)
wrong check mark cm
to the right of the lens
to the left of the lens

Is the image real or virtual?
virtual
real

Homework Equations

m = q/p
(1/q) +(1/p) = (1/f)
m = (height of q)/(height of p)

The Attempt at a Solution

m = 1.24cm/.4cm
m = 3.1
q = mp
q = 3.1p
(1/3.1p) + (1/p) = (1/f)
(1/3.1p) + (1/p) = (1/6)
This is where I get stuck. I can't remember how to solve this. I know the image is located to the left of the lens and that it's virtual.

 

[dotman]

Hello,

In general:

$$\frac{1}{x} = \frac{1}{y} + \frac{1}{z} = (\frac{1}{y} \cdot \frac{z}{z}) + (\frac{1}{z} \cdot \frac{y}{y}) = \frac{z}{yz} + \frac{y}{yz} = \frac{y+z}{yz}$$

Or specifically, in your case:

$$\frac{1}{f} = \frac{1}{3.1p} + \frac{1}{p} = \frac{1}{3.1p} + (\frac{1}{p} \cdot \frac{3.1}{3.1}) = \frac{1}{3.1p} + \frac{3.1}{3.1p} = \frac{1 + 3.1}{3.1p} = \frac{4.1}{3.1p}$$

However, you need to watch your signs. You've made a sign error with regard to convention-- see http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html#c3"

Hope this helps.