Tilted plane wave through a 2 lens system

[Skaiserollz89]

TL;DR Summary

How does the incident angle of a collimated beam relate to the output angle after passage through a two-lens system with f1=910mm and f2=40mm and Magnification=-f2/f1=.044?

I have 2 lenses. L1 and L2 with focal lengths f1=910mm and f2=40mm, respectively. They are separated by a distance d=f1+f2. The magnification of the system is M1=-f2/f1=.044. If I have a normally incident, collimated beam pass through my system I will have a beam, parallel to the optic axis exit the system, whose diameter is scaled by a factor of M1. I would like to know what happens if the incident beam is tilted at an angle theta? How does this incident angle relate to the output angle after passage through the second lens?

My initial thought was that the angle would also be scaled by a factor of M1, but I don't think this is right. Any thoughts would be greatly appreciated.

 

[Charles Link]

Parallel rays incident on the first lens at angle $\theta$ will converge to a point in the focal plane of the first lens at distance off-axis of $d= f_1 \theta$. When they start out in the focal plane of the second lens at a distance $d$off axis, they will emerge at angle $\theta'=d/f_2$. The result is that $\theta'=M \theta$ with a minus sign to show its direction is flipped.

 

[DaveE]

Parallel rays incident on the first lens at angle $\theta$ will converge to a point in the focal plane of the first lens at distance off-axis of $f_1 \theta$. When they start out in the focal plane of the second lens at a distance $d$off axis, they will emerge at angle $\theta'=d/f_2$. The result is that $\theta'=M \theta$ with a minus sign to show its direction is flipped.

This is for a telescope, of course, where the focal planes are the same.

 

[sophiecentaur]

TL;DR Summary: How does the incident angle of a collimated beam relate to the output angle after passage through a two-lens system with f1=910mm and f2=40mm and Magnification=-f2/f1=.044?

They are separated by a distance d=f1+f2.

This is for a telescope, of course, where the focal planes are the same.

. . . . and the angular magnification is f1/f2

TL;DR Summary: How does the incident angle of a collimated beam relate to the output angle after passage through a two-lens system with f1=910mm and f2=40mm and Magnification=-f2/f1=.044?

I will have a beam, parallel to the optic axis exit the system, whose diameter is scaled by a factor of M1.

I think the 'diameter' will relate to the the aperture of the lenses. In practice, internal apertures are used.

 

[Charles Link]

Probably outside of the scope of what the OP presently needs, but the "perfectly collimated" beam is interesting in that the light, even with perfect optics will never focus to a perfect point. Instead there is a diffraction limited focused spot size whose diameter is inversely proportional to the diameter of the collimated beam. The finite sized focused spot then becomes a source for the second lens. In the near field the beam can still be considered to be perfectly collimated, but for calculations in the far field, the beam will have a finite divergence because of the finite spot size.

 

[sophiecentaur]

but for calculations in the far field, the beam will have a finite divergence because of the finite spot size.

There is another factor here which kicks in without diffraction effects. Taking a typical arrangement of two lenses to make a telescope, the eyepiece (second) lens usually has the smaller diameter and any collimated beam, focussed at 'infinity' (parallel sides, visible through smoke), for instance, will have a diameter the same as the EP. This width I not governed by focal lengths but by a limiting 'stop', somewhere in the scope. Details like the actual focussing of the arrangement and the diffraction due to each lens edge.