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.

 

[ogb p]

Based on the given information, the object is located to the left of the lens and the image is located to the right of the lens. The image is also virtual, meaning it cannot be projected onto a screen. This can be determined by using the thin lens equation, (1/q) +(1/p) = (1/f), where q is the image distance, p is the object distance, and f is the focal length. By substituting the given values, we can solve for p and q.

Using the magnification equation, m = q/p, we can also determine the magnification factor, which is 3.1 in this case. This means that the image is 3.1 times larger than the object.

To solve for the object distance, we can rearrange the thin lens equation to get p = 1/(1/f - 1/q). Substituting the values, we get p = 0.517 cm. Since the object is located to the left of the lens, the object distance has a negative sign, making it -0.517 cm.

To solve for the image distance, we can use the magnification equation and substitute the calculated values. This gives us q = 1.6 cm. Since the image is located to the right of the lens, the image distance is positive, making it 1.6 cm.

Therefore, the correct answers are:
Object distance = -0.517 cm
Image distance = 1.6 cm
Image is virtual.