Binoculars in Low Light

Some advertisers claim that binoculars with large objective lenses make objects appear brighter, thus making it easier to see in low light.

This sounds convincing: after all, a larger lens captures more light, and with more light entering the eye, objects appear brighter. 
I used to believe this, but surprisingly perhaps, the claim is false. In a nutshell, the wider the objective is, the narrower  is the angular range of the rays that make it to the eye. These two competing effects cancel each other exactly, as I show next. 

Referring to Figure 1, each ray is characterized by the pair \((x,y)\) where \(x=\sin\theta,\) with \(\theta\) being the angle formed by the ray with the horizontal, and \(y\) is the coordinate of the intersection with the \(y\)-axis. The same ray, when crossing the \(X\)-axis (tangent to the pupil) in the figure, is similarly characterized by coordinates \((X, Y)\) where \(Y=\sin\Theta.\) So, our binocular defines a mapping \(\varphi:(x,y)\longmapsto(X,Y)\). This map takes the left rectangle in Figure 1 to the right one. It is a fundamental fact that \(\varphi\) preserves area [1].

Now, here is a key point: Area \((R)\) = carried across the objective by the set of rays corresponding to \(R\). 

It is important to note that \(\varphi(R)\) is fixed by the size of the pupil and by the exit angle \(\beta\) between the two parallel beams in the figure, so that the area of \(\varphi{R}\) and \(R\) are independent of the size of the objective. And so by the \(\textrm{area}=\textrm{power}\) remark, the power reaching the retina does not depend on the size of the objective, as claimed. 

It remains to explain why \(\textrm{area}=\textrm{power}\) I have to come clean and mention the unstated assumptions: (i) we are looking at a homogeneous object, such as a wall, and (ii) each point of the object emits photons equally in all directions.¹ By the assumption of isotropic emission of photons, the number of photons passing per unit of time through an infinitesimal segment \(dy\) in the directions between \(\theta\) and \(\theta+d\theta\) (where \(\theta\) is the angle with the horizontal in Figure 1) is proportional to \(dy\) and to the component \(\cos\theta\) of the velocity of the photons² normal to the segment i.e., to \(dy\) \(\cos{\theta}\;d\theta=dy\;d(\sin\theta)=dy\;dx.\) And so, \(dy\;dx\) is not just the area, but also the energy flux.

A more rigorous explanation of why \(\textrm{area}=\textrm{energy flux}\) can be given by considering the phase space of photons, observing that the flow in this space is volume-preserving.  The volume measures the number of photons, i.e., the energy. The flux thus measures the energy flux. Considering the tube of trajectories corresponding to the rays in Figure 1, one can conclude that fluxes through two different sections are equal; I omit the details.

Magnification as a Ratio

The binocular in Figure 2 magnifies by the factor \(\beta/\alpha:\) indeed, \(\beta/\alpha\) is the factor by which the images on the retina spread apart when viewed through the binocular versus the naked eye. Preferring to deal with angles \(\theta\) and \(\Theta\), instead of their sines \(x\) and \(X\), we note that the error in \(\sin\alpha\approx\alpha\) with \(|\alpha|<20^{\circ}\) is about two percent, hardly noticeable in practice. With this approximation, area preservation gives

\[\textrm{Objective}\cdot\alpha=\textrm{Exit pupil}\cdot\beta,\]

and so \(\beta/\alpha\) becomes

\[\boxed{{\textrm{Magnification factor}}=\frac{\textrm{Objective diameter}}{\textrm{Exit pupil}}.}\]

This is an expression of the area-preservation of the map \(\varphi\) defined above. This area preservation can be viewed as a classical analog of the uncertainty principle: by narrowing the exit, we widen the angular range \(\beta\). 

¹ Without these assumptions, the size of the objective lens may actually matter. Indeed, imagine looking through the binocular at the moon at night. The energy reaching the retina will depend on the proportion of the dark background, and that proportion decreases as the magnification increases. But magnification increases with the size of the objective (according to the remark at the end of the article), and so the size of the objective matters in this (uninteresting) sense. However, once the moon occupies the full field of view, further increase in magnification/objective diameter will not make it brighter.

² Taking the speed of a photon to be 1. 

The figures in this article were provided by the author. 

References 
[1] Levi, M. (2014). Classical mechanics with calculus of variations and optimal control: An intuitive introduction. Providence, RI: American Mathematical Society.