How Many Lenses Generate a Real Image?

[Mark1991]

Good afternoon,

I want to check another problem about optical physics:
There are taken N equal converging lens, ordered one by one in a distance of f, where f is the focal distance.
There is an object standing in a distance of 0,5*f of the first lens.

A Sketch:
http://img35.imageshack.us/img35/8288/aufzeichnenc.png

Question:
For which N, a real image is generatet behind the last lens? What is the factor of incresement of the image compared with the object?
_________________________________________

My solution is that a real image is generated
if 3|N (image size = object size) or 3|(N+1) (image size = 2* object size)

_________________________________________

My approach:

in general: 1/f = 1/g + 1/b

after first lens: g=f/2
-> b=-f
-> a virtual image with size A=b/g=-2

after second lens: g=2*f
->b=2*f
-> a real image with size A=1, compared with the size of the virtual image
-> size of real image = size of object * 2

after third lens: g=-f
b=f/2
-> a real image with size A=0,5, compared with the size of the real image (2)
-> size of real image = size of object


after fourth lens - nth lens:
images seems to repeat -> virtual image if 3|(N+2), real image if 3|N or 3|(N+1)
_________________________________________

Is this calculation right?

Mark

 

[Mark1991]

No opinions?

 

[Lok]

Have you tried the geometrical approach ... somehow I think that 2 lenses will be enough.

 

[Lok]

Either that or infinity.

 

[Mark1991]

Why do you think that two lenses will be enough?
Which geometrical approch do you mean?
If two lenses will be enough, where is the mistake in my consideration?

Mark

 

[Lok]

http://img34.imageshack.us/img34/5571/clipboard1n.jpg

Something like this if I remember correctly.

So the first object which is at f/2 distance get's magnified like with a usual magnifiing glass so that a virtual image appears on the same side but twice as big.

The virtual image then passes through the second lens using the same geometric method and a real but inverted image if it appears. the final real image is twice the originals size.

Again if I remember it right.

 

[Mark1991]

It seems you used the same way of solution:
Obviously, there does not arise a real image after the first lens, but after the second.
But I think that there is an image after the third lens, too.
After the fourth lens only a virtual image arises, and so on.

Mark

 

[Lok]

Yes a third lens would create a real image. And after the cycle begins.

About my infinity ... I was trying to see if the first lens does converge the light so that the second one would be useless. But it doesn't.

 

[Redbelly98]

after fourth lens - nth lens:
images seems to repeat -> virtual image if 3|(N+2), real image if 3|N or 3|(N+1)
_________________________________________

Is this calculation right?

You're solution looks right.