amsikking: SOLS optimum tilt

Description

In a single-objective light-sheet (SOLS) style microscope the light-sheet consumes some of the numerical aperture (NA) from the emission path. i.e. more NA for the light-sheet gives better sectioning but less NA for collecting the emission. The trade between light-sheet and emission NA is governed by the tilt \(\theta_t\) of the 3rd microscope (\(M_3\) below). For example, some builders have used 30 degrees for \(\theta_t\). So the question is, what is the optimal* tilt?

Here's a simple analytical estimate for matching the light-sheet thickness to the axial point spread function of the emission. This may be the maximum tilt that can be effectively used, i.e. more tilting reduces emission path collection with no obvious advantage.

To quickly see the optimum* tilt for a particular primary objective lens (\(O_1\) below) simply edit and run this Python script: SOLS_optimum_tilt.py

*Note: what is optimal really depends on the sample and application. So the point of this calculation is not to find a 'true optimum', but to highlight that this choice is important and should be considered before building a system.

Angular constraints

The SOLS architecture forces constraints on the angular ranges of the system, which can be seen in the space of objective 1 (\(O_1\)). With no tilt (\(\theta_t=0\)), microscope 3 (\(M_3\)) would simply re-image the traditional focal plane. As \(\theta_t\) increases, at some point it crosses the boundary of the \(O_1\) half angle \(\theta_1\), thereby allowing a finite amount of excitation via \(\theta_x\), but in doing so reducing the collection angle for the emission \(\theta_e\) (albeit only on one side): \[\theta_e = \theta_1 - \theta_x \tag{1}\] and, \[\theta_t = \frac{\pi}{2} - \theta_e \tag{2}\]

Matching excitation with emission

One possible criteria for optimizing the tilt angle is to match the light-sheet thickness to the axial point spread function of the emission: \[\sin\theta_e = \frac{w_0}{z_0} \tag{3}\] where, \[w_0 = \frac{\lambda_x}{\pi NA_x} \tag{4}\] and, \[z_0 = 2 \frac{n\lambda_e}{NA_e^2} \tag{5}\]

Find the angles

So we can now substitute equations (4) and (5) into (3) to give: \[2\frac{n\lambda_e}{NA_e^2}\sin\theta_e = \frac{\lambda_x}{\pi NA_x} \tag{6}\] which using \(NA = n \sin\theta\) simplifies to: \[\frac{\sin\theta_x}{\sin\theta_e} = \frac{\lambda_x}{2\pi \lambda_e} \tag{7}\] We can now use the angular constraint of equation (1) to get: \[\frac{\sin(\theta_1 - \theta_e)}{\sin\theta_e} = \frac{\lambda_x}{2\pi \lambda_e} \tag{8}\] and using the identity \(\sin(\theta_1-\theta_e) = \sin\theta_1\cos\theta_e - \cos\theta_1\sin\theta_e\) we can convert (8) into: \[\frac{\sin\theta_1\cos\theta_e - \cos\theta_1\sin\theta_e}{\sin\theta_e} = \frac{\lambda_x}{2\pi \lambda_e} \tag{9}\] which simplifies to: \[\frac{\sin\theta_1}{\tan\theta_e} = \frac{\lambda_x}{2\pi \lambda_e} + \cos\theta_1 \tag{10}\] or if we allow \(\lambda_e \approx \lambda_x\) more conveniently to: \[\tan\theta_e \approx \frac{\sin\theta_1}{\frac{1}{2\pi} + \cos\theta_1} \tag{11}\]

Results

Applying equation (11) to a 1.35 silicone objective, which has very high angular collection, returns an tilt angle of about 25 degrees. On the other end of the spectrum a 1.0 water objective has a much lower angular range, which forces a higher tilt angle of about 47 degrees: