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.

Question

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 :)

Answer 1

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.