# What is the diameter of the image? professor gave us the steps, got it wrong though!

1. Mar 17, 2006

### mr_coffee

Hello everyone, today our professor did this hw problem for us, just different numbers. This is the question:
You produce an image of the Sun on a screen, using a thin lens whose focal length is 20.6 cm. What is the diameter of the image? (The mean radius of the Sun is 6.96x10^8 m and its mean distance from Earth is 1.50x10^11 m.)

Okay here is my work, i'm using the formula:
1/i + 1/o = 1/f
i = image distance
o = object distance;
f = focal length.

I said:
f = .206m
sove for i

I submitted .95584mm as my answer but was wrong! I did it just the professor, he does make mistakes alot when trying to help us (6 students) with homework any ideas?

2. Mar 18, 2006

### big man

I haven't really looked at your working at all, but do you know if the program accepts answers with too many significant figures?
does the program only want your answer in SI units as well??
It also looks like you'd multiply that result by 2 since you are looking for diameter and not the radius.

From the looks of it the lowest number if significant figures outlined in the problem is 3 so that will be to what precision you must state your final answer. So either submit 0.00191 m or 1.91*10^-3 m.

If this is not the case then I'm sure someone will reply with more help.

3. Mar 18, 2006

### andrevdh

The image diameter can be determined from basic geometry by realizing that the rays from the two opposite sides of the sun crosses throught the optical center of the lens and continue straight onwards to the same points in the image of the sun in the focal plane of the lens. The angle between such two rays are the same on both sides of the lens.

4. Mar 18, 2006

### mr_coffee

big man, that isn't the case but thanks for the help! he wants the answer in mm.

Andrevdh, thanks for the responce but I don't quite follow you, can you explain some more? Is my method not even close on the right track?

5. Mar 19, 2006

### andrevdh

You used the radius instead of the diameter of the sun in your calculation so the image size will be twice what you got. Notice that your image distance is equal to the focal lenght of the lens, since the object is very far away this is to be expected.

6. Mar 19, 2006

### big man

The significant figures thing was more of an additional suggestion and observation rather than the actual solution. It seems you missed this part of my post:

But no matter 'cause Andrevdh provided a more comprehensive reply to your problem.

7. Mar 19, 2006

### andrevdh

big_man
Excuse me, I am usually in too much of a hurry. I do not read everything in the thread before I reply.

mr_coffee
my suggestion for using trig will come to the exact same formula. It is just nice to know where they come from, which gives one more confidence and pleasure in applying the formulas.

You have the sun on one side and its image on the other side of the lens. The rays from the sun coming from the extreme edges of the sun cross and continue straight onwards to the image. If one works just with half of these it is a matter of simple trig to solve the problem (ratios of sides of two tringles are the same).

8. Mar 19, 2006

### big man

haha no sorry man I wasn't saying that you missed what I said. I was pointing out to mr_coffee that I didn't mean the significant figures thing to be the solution his problem and that he must have missed the radius problem I tried to point out. I just added the significant figures comment because it is usually an issue on online submission programs that I've seen.

I just gave a half-arse reply as a quick response to his question because I was in a hurry, but you took the time to explain a good thought process to follow when doing optics problems.

So yeah, I'm glad you gave those good responses 'cause mine wasn't thorough and I my previous post wasn't directed towards you. :)

9. Mar 19, 2006

### mr_coffee

Sorry about that Big man, I didn't think your suggestion was right because when the professor was doing the problem i questioned why he didn't multipy the raidus by 2. He told me because it was spherical. So i didn't want to look like a tard and say, so what does that have to do with it? But you were correct!

THanks for the explanation and also the right answer andrevdh.