A glow stick is placed 18.0 cm from a converging lens that has a focal length of 8.0 cm. Determine the location of the image using both a ray diagram & the lens equation.

So I'm not sure if my diagram is perfectly accurate, they just said "at least 4 squares high" but it is within reason (14cm).
Using the equation however, I'm getting 3.98 cm. I know that's not right because the image is beyond 2F & therefore the image should be between F & 2F

(1/f)=(1/do)+(1/di)
(1/8)=(1/18)+(1/di)
0.0694=1/di
di=3.98 What am I doing wrong?

Thanks a bunch :)

What you didn't do right are:

1. the reciprocal of 0.0694 is 14.4
2. the negative sign was ignored.

To interpret the lens equation correctly, we need to agree on the sign convention.
Following assumes the sign convention given here:
https://apps.spokane.edu/InternetContent/AutoWebs/AsaB/Phys103/MirrorsThinLens.pdf

f=8
do=18
then
1/f=1/do+1/di
1/di=1/f-1/do
=1/18-1/8
=8/144-18/144
=-10/144
Therefore
di=-144/10=-14.4

Your interpretation that the absolute value of the image location should be between F and 2F is correct.
However, since di is negative is negative, it is situated on the left of the object, i.e. it is a virtual image.

This is one of the drawbacks of the variety of graphing programs available on the Internet. It does not provide understanding of the concept.
I suggest you take a square paper and actually draw the image by yourself. It will help you understand how the lens equation works, and give yourself confidence for your next problem.

From your calculations, it seems that you have made a minor error in using the lens equation.

The lens equation is given by:

(1/f) = (1/do) + (1/di)

where f represents the focal length of the lens, do is the object distance, and di is the image distance.

In this case, the focal length of the lens is 8.0 cm and the object distance (do) is 18.0 cm. To find the image distance (di), we can substitute these values into the lens equation:

(1/8.0) = (1/18.0) + (1/di)

To solve for di, we need to rearrange the equation and simplify:

1/di = (1/8.0) - (1/18.0)
1/di = (9 - 4) / (8 * 18)
1/di = 5 / (8 * 18)
1/di = 5 / 144
di = 144 / 5
di ≈ 28.8 cm

Therefore, using the lens equation, the approximate location of the image is 28.8 cm.

Now, let's analyze the ray diagram:

- Draw the principal axis as a horizontal line passing through the center of the lens.
- Mark the focal points of the lens on either side of the lens along the principal axis. For a converging lens, the focal points are on the same side as the incoming light.
- Draw an incident ray from the top of the object parallel to the principal axis. This ray will pass through the focal point on the other side of the lens after refraction.
- Draw another incident ray from the top of the object that passes through the center of the lens. This ray will continue in the same direction after refraction.
- The intersection of these two refracted rays will give you the location of the image.

Based on the diagram, it looks like the image is formed between the focal point (F) and twice the focal length (2F), as you correctly mentioned. So, the image is indeed between F and 2F.

Therefore, the correct location of the image, determined both by the lens equation and the ray diagram, is between the focal point (F) and twice the focal length (2F).