# Observing the image of the sun (optic)

1. Apr 10, 2010

### seifer9

1. The problem statement, all variables and given/known data

Hi everyone,

I was wondering if anyone can help me out with my project in physics. Our goal is to make the sun image appear on a panel at the end of a rail, using lens, to see if we can observe sunspots on the projected image of the sun.

This is the materials we have :

-rail of 120 cm

-a panel of 20x16 cm (so the diameter of the image of the sun would best be around 10cm if possible)

- Lens of different focal distance : +50 mm, +100 mm, +200mm, +250mm, and one -150mm

___________

2. Relevant equations

Theres the equation I got but these are from french class so might be different I am not sure.

1/p + 1/q = 1/f

m = -q/p

Sun distance : 1.5 x 10^11 m
Sun diameter 1.392 x 10^9 m

3. The attempt at a solution

From what I understand, the setup should look like this on the rail :

LENS1_____________LENS2_____________________PANEL

I believe I should place a positive focal distance lens first right? like the +250mm, and then the -150mm... but Im having a bit of trouble calculating the distance between the lens and the panel on the rail to get a clear image of the sun at the good diameter. When I calculate that I want a 10cm image of the sun, I get that I need a 10 m LENS and that my image would also be at 10m... its too far for my setup!

- Seifer

2. Apr 10, 2010

### AtticusFinch

There are multiple ways you can do this. But there are two things you should realize. One is that the Sun is so far away from your apparatus that you can say it is infinitely far from your lens. Therefore, the Sun's rays enter your lens parallel to each other. Where would such rays focus?

Secondly, due to the rays entering the lens parallel to one another you don't really need the Sun's diameter as the image formed by the first lens will be fairly small in diameter. Your real objective is to set up the system of lenses such that you achieve the largest possible magnification.

Try this: come up with a system that will allow you to use a converging lens followed by a diverging lens.

3. Apr 11, 2010

### seifer9

Thanks Atticus, I also came to that conclusion, but the problem is calculating. Since the distance of the object can be infinite, the distance of the image is the same as the focal distance of the lens for a converging (For example, my first lens have 250 mm focal distance, so my image is also at 25cm of my len). That gives me a very small image of the sun though (around 2mm diameter)

So I need to place a diverging len, which the only size I got is a -150mm focal distance... The problem is, where do I place it on the rail? My guess is that the len have to be placed BEFORE the focal distance (so before 25cm of the other lens), or else the diverging len would see the small image created by the first len and then make a virtual image which would not be projected on my panel.

So lets say I place my second len (the diverging -150mm) at 20 cm of my first len, the image would probably appear bigger and farther, but I don't know how to calculate the effect of stacking lens and the effect on the image appearing

It would look like this :

Sun ==== ___LENS1(+250mm)____LENS2(-150)_____________________Panel

I just need the right distance between them I guess

4. Apr 11, 2010

### seifer9

http://lectureonline.cl.msu.edu/~mmp/applist/optics/o.htm

This website helped me out to confirm the logic, if you pick concave to convexe, the converging-diverging lens combo allows the image to appear farther if we place the diverging len before the focal distance of the first len.

The problem remains, how to calculate the distance of that image, which depend on the distance between the converging and diverging len...

5. Apr 11, 2010

### AtticusFinch

Yes you are right. You need to place the lens before the focal length of the first lens. This makes the image a virtual object for the second lens leading to a final real image which can be focused.

To preform calculations involving systems of thin lenses simply treat each lens by itself and move down the line. So you would find where the first lens creates an image and use that image as an object for the second lens. To figure out the object distance for the second lens you simply do D - p = q, where D is the distance between lenses and p is the image distance from your first calculation. Then use the thin lens equation.

As for magnification you can use the equation you provided. You don't need to consider the magnification of your first lens (just note that the point will be small). If you use more than two lenses you can multiply the magnifications of each lens after the first to get a total magnification.

6. Apr 11, 2010

### seifer9

I believe I just got it :) However I will never get a bigger image than 2cm diameter with those lens and this rail.

I am way too tired to post my numbers right now, but I can do that tomorrow if anyone cares.

Thanks a lot AtticusFinch, greatly appreciated