Angular magnification and image size

[Melawrghk]

Homework Statement

You have a lens with f=1m. You decide to make a camera to take a picture of the solar eclipse. The image of the sun and its corona have an angular size of 1.00 degrees. Find the diamter (in cm) of the resulting image.

Homework Equations

M=theta'/theta = 25cm/f

The Attempt at a Solution

I decided to find the angular magnification:
M=25cm/100cm = 1/4 (so the image the camera observes is smaller than what is in the sky?)

And since M=theta'/theta and I know theta=1.00 degree, I figured I'd find theta':
theta'=M*theta = 0.25 degrees

And I thought I'd get the height of the image formed using just a triangle where one angle is .25 degrees and one of the sides is the near point distance of 25 cm.
Which yielded image height of about 1.1mm...

Is this method right? I don't like how the height is so tiny. But I remember that when you focus sunrays with a magnifying glass you get a point (to burn whatever), so at the same time it doesn't seem unreasonable.

Any help would be greatly appreciated.

 

[Delphi51]

One property of thin lenses is that rays through their center are not bent.
Suppose a ray from the center of the sun goes on the optic axis through the center of the lens. A second ray from the rim of the sun passes through the center of the lens - it won't be bent either, so it will emerge from the lens at angle half a degree from the optic axis. Consider the triangle formed by the two rays emerging from the lens half a degree apart and a vertical line at the focal plane 1 meter away from the lens where the image is formed. This is a right triangle, so you can easily calculate the side of the triangle that is the half height of the image. It gives an answer quite different from your 1.1 mm.

 

[Melawrghk]

Ooh okay, so I got 1.75cm, does that sound more right?

Thanks!

 

[Delphi51]

Yes!
Most of those lens questions boil down to choosing two rays that are easy to work with.