What does the intensity profile of a set of parallel rays look like after travelling through a convex lens?

I have had a discussion at my university about the following question. Imagine a 'perfect' convex lens and a set of parallel rays, falling onto that lens. The set of incoming rays has a flat intensity profile, i.e. it is equal for all positions along the axis perpendicular to the lens.

Now, after having travelled through the lens, there are two intuitive options for the intensity profile of this set of rays at a distance far away from the lens but before the focal point:

1) it is still flat, since the distance between every ray, although decreased, is still a constant.

2) it decreases with the cosine of the angle of incidence of the ray, just as the intensity of the sun on earth decreases with the cosine of it's angular position in the sky.

Which of the two is right and why is the other not? I tend to say the first one, but I'm not 100% sure.

Also, how does it relate to an actual lens and what is the main reason that an actual lens would never have a flat profile if this would theoretically be the case?

• "...just as the intensity of the Sun on Earth decreases with the cosine of it's angular position in the sky." That sounds like you are talking about the Solar irradiance on a horizontal patch of ground. (I.e., you are talking about how the irradiance changes as the Sun moves off-axis.) But when you talk about the setup with the lens, it sounds like you might be talking about the irradiance on a spot on a card that is held normal to the axis of the lens. – Solomon Slow Sep 26 '18 at 15:48
• Have you tried experimenting with a lens? Demonstration is the best way to prove a point. – sammy gerbil Sep 26 '18 at 20:59

Which of the two is right and why is the other not?

It depends.

Let's assume that the lens is perfectly shaped and does not attenuate light - just refracts it.

Let's define the intensity as a power flow per unit area normal to the direction of the flow.

If we have a flat screen parallel to the lens, the intensity in the center of the screen will be greater than the intensity at some distance from the center.

If, on the other hand, the screen is spherical, with its center at the focal point, the intensity will be the same everywhere.

To make it easier to visualize, we could reverse the direction of light, so that it originates at the focal point of the lens and the power flows out. Obviously, the intensity over any spherical surface will be uniform, while the intensity over a flat surface will change as a cosine of the angle of incidence.

The real lenses are not perfectly shaped and attenuate light unevenly due to their variable thickness, so the intensity would not be perfectly even.

Define the $$x$$ direction as parallel to the incoming rays. The flux through a small area parallel to the $$y$$-$$z$$ plane (the plane of the lens) must remain constant as the area moves radially within the beam (at one value of $$x$$). Thus, I would say that neither (1) nor (2) is correct, but that the correct answer is hidden option (3): intensity increases like the secant of the angle of the ray. Intensity is defined as the power per unit area in a plane perpendicular the propagation of the energy. In order to keep the flux in the $$x$$-direction constant, the intensity must increase away from the axis. If you draw a ray diagram with evenly spaced incoming rays and a focal point quite close to the lens, you can see that the outermost rays make smaller angles with their neighbors than do rays near the axis.

Generally when learning optics the thin lens approximation is made, in which case the cosine of the ray angle is taken to be unity, and all three options are identical.

At the focus of perfect lens there is a point, but diffraction would lead to a pattern, google Airy disk. Consider a plane at 2 times the focal distance and add another identical convex lens, you get the same output as input (except that a small diffraction has occurred twice), every ray has passed thru the focal point so the pattern is uniform.

I do not know what you mean by "intensity profile", nor by "perfect lens". I shall try to give my definitions, but it is far from a trivial thing. Pictures would help, had I time to draw them...

By perfect lens we can mean one that gives a point image of a point source (the technical term is "stigmatic"). But two things must be noted: - Stigmatic lenses exist, but only for one source/image pair. Practical solutions giving acceptable images over an extended field only approximate this ideal. - If you limit to thin lenses, the stigmatic property is unattainable. The usual treatment in elementary optics deal only with the so-called "Gauss approximation", requiring rays that are paraxial (near to the optical axis) and of small vergence (forming a small angle with optical axis).

So we must widen our search to thick lenses and restrict to a unique source/image pair. I shall discuss one simple case: the one with source at infinity on the optical axis, which means that incoming rays are all parallel to optical axis itself. Then a simple solution exists: the plano-hyperbolic lens.

Now for the intensity profile. What this is meant for the incoming rays is easy: you are assuming that if you put in a plane screen orthogonal to the rays you will find, for any area $$S$$ of the screen, an energy falling proportional to S. Not so simple is to define the intensity profile of the outgoing rays. We could mean the same, possibly with the proviso of using infinitesimal areas $$dS$$ in order to take into account that the energy falling could by not uniform.

This could be done, but is not an easy geometrical problem. I am almost certain that for a plano-hyperbolic lens the intensity profile of outgoing rays is not flat.