How Is Magnification Calculated with Two Thin Lenses?

AI Thread Summary
The discussion revolves around calculating the magnification of an image formed by two thin lenses. The first lens, with a focal length of 50mm, creates an image at infinity when an object is placed at its focal point. The second lens, with a focal length of 400mm, then takes this virtual image as its object. The magnification for the first lens is considered infinite, while the second lens yields a magnification of 400/inf. Ultimately, the total magnification is determined to be 8 times, although caution is advised regarding the mathematical treatment of infinity.
avelon
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Homework Statement




A light source placed in the focus of a (thin) lens with a focal length of 50mm.

A second (thin) lens (with focal length = 400mm) was placed 600mm after the focal point of the first one, an image of the light source was captured at the seonds lens focal point. How many times has it been magnified?
 
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welcome to pf!

hi avelon! welcome to pf! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 


tiny-tim said:
hi avelon! welcome to pf! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

This is what I did

Magnification M = image_distance S / object_distance S'
M = 400 / 600 = 2/3
I don't think this is the right solution... any comments? :--)
 
there's two lenses …

won't you need to use the magnification equation twice?
 
tiny-tim said:
there's two lenses …

won't you need to use the magnification equation twice?

hmmm..Lens equation for lens1 would giva us
1/object_distance + 1/image_distance = 1/focal
"assuming it's a virtuall image", otherwise image_distance would be inf for lens 1
1/(-50)+1/i_d = 1/50
image_distance = 25

Magnification (lens1) = 25/50 = 1/5
Magnification (lens2) = 400/600 = 2/3

M = 1/5 * 2/3 ...still feeling I'm doing something worng here.
 
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avelon said:
hmmm..Lens equation for lens1 would giva us
1/object_distance + 1/image_distance = 1/focal
"assuming it's a virtuall image" …

hmm … i think you're misunderstanding the question :redface:

the object is a real object, so if it's placed at the focal point, then the image is at … ? :smile:
 
tiny-tim said:
hmm … i think you're misunderstanding the question :redface:

the object is a real object, so if it's placed at the focal point, then the image is at … ? :smile:

When an object is placed at the focal point, the image will be at inf :-)

For lens1:
focal = 50 mm
Object_distance = 50 mm
Image_distance = inf (this will be object for the second lins)
Magnification = (inf/50)


For lens2:
focal = 400 mm
Object_distance = inf
Image_distance = 400 mm
Magnification = (400/inf)
----

Magnification for two lenses would be:
Magnification_lens1 * Magnification_lens2 = (inf/50) * (400/inf) = 8 times (I think I got it right this time)
 
hi avelon! :smile:
avelon said:
When an object is placed at the focal point, the image will be at inf :-)

Magnification for two lenses would be:
Magnification_lens1 * Magnification_lens2 = (inf/50) * (400/inf) = 8 times (I think I got it right this time)

that's right! :smile:

(but don't let the professor catch you multiplying or dividing by ∞ ! :wink:)

i'm not sure what the official way of proving this is :redface:

i suspect you have to put the object a small distance x away from the focus, and let x -> 0
 
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