iGEM TU Delft

How can light resonate in the cell and what is the minimal size of the cavity in order to fit light inside it?

Introduction

In order to get lasing inside our bacteria we constructed an inside part of or the whole bacteria. For most normal lasers this is accomplished by two mirrors (figure 1A). In our project PHB granules are formed inside cells and by total internal reflection the light should be trapped inside them. Another way to form an optical cavity is by encapsulation of the bacteria in a material with a higher refractive index than the inside of the cell. This can be achieved by the enzyme silicatein, which polymerizes monomers such as silicic acid (Müller et al., 2008) and tin dioxide (André et al., 2011), creating the desired reflective layer (figure 1B). In this model we will investigate using both polysilicate and tin dioxide as a substrate since polysilicate is used in in our lab but tin dioxide has some potentials as it has an higher refractive index. In order to trap the light inside these cavities it is necessary that the wavelength of light ‘fits’ inside. Therefore standing waves have to be formed within the cavity. Here we will determine what the minimal required size of these cavities is to fit exactly one wavelength. This will not give us a definite answer whether we will get lasing, but a very strict lower bound of the required cavity size.

Whispering Gallery Modes

In both methods of capturing light (PHB granules and covering the cell with polysilicate and tin dioxide), the light will become trapped by whispering gallery mode resonance (Humar et al., 2015). WGM resonance is the phenomenon where waves travel around a concave surface in a closed path (figure 2). Every time the wave hits the surface, total internal reflection occurs. When light hits a interface between two materials the light gets refracted as in figure 3A depending on the angle of incidence $$\theta_i$$. If $$\theta_i$$ reaches a critical point where the refracted light is parallel with the interface (figure 3B), we call this the critical angle $$\theta_c$$. If $$\theta_i>\theta_c$$ the lights does not go through the interface and gets reflected as in figure 3C, we call this total internal reflection.

It is important that the waves follow a closed path forming a polygon so that constructive interference takes place. Constructive interference occurs when two waves travel in phase so that their amplitudes add up (figure 4). We need constructive interference so that every cycle in the resonator adds to the constructive interference. To make sure the waves travel in phase, the optical path length (OPL) is required to be an integer number of wavelengths (equation 1). In equation 1, n is the number of sides of length l, and m an integer number of wavelengths ($$\lambda$$).

$$OPL=n\cdot l = m\lambda$$

PHB Granules

For intracellular lasers, oil droplets have been used before as an optical cavity (Humar et al., 2015). In our project we use PHB granules as an optical cavity inside the E. coli instead of oil droplets. This PHB granule will then be stained with the fluorescent dye Nile Blue with an emission maximum of 580 nm (Kitamura et al., 1994) to create the gain medium.

In order to get WGM resonance, the light beam has to be reflected by total internal reflection every time it hits the surface. Therefore the light beam has to approach the surface at a minimal angle larger than the critical angle. From Snell’s law (equation 2) we can compute the critical angle (Hecht, 2001) where the incoming light is refracted to have an outgoing angle of exactly 90 degrees (figure 3C). In equation 2 and 3, $$n_1$$ and $$n_2$$ are the refractive indices of the materials before and after the interface.

$$n_1 sin(\theta_{i}) = n_2 sin(\theta_{f})$$ $$\theta_{c} = \arcsin\Big(\frac{n_2}{n_1}\sin\big(\frac{\pi}{2}\big)\Big)=\arcsin\Big(\frac{n_2}{n_1}\Big)$$

The incidence angle of light in total internal reflection requires $$\theta_i>\theta_c$$, the outgoing angle is then $$\theta_f=\theta_i$$. When we look into geometrical optics the path of the light in WGM is shown in figure 5. We want a closed optical path (polygon), therefore we have an integer number of sides on the polygon. Using equation 4 and the fact that a total round gives an angle of $$2\pi$$ we can determine the number of sides, ($$n_{sides}$$) of the polygon as in equation 5. We assume here we are at the critical point for total internal reflection so that $$\theta_i=\theta_c$$.

$$\phi = \pi - 2\theta_{f}$$ $$n_{sides} = \frac{2\pi}{\phi} = \frac{2\pi}{\pi-2\theta_{f}}$$

We can calculate the number of sides by using the reflected angle $$\theta_f$$ in equation 4 and round $$n_{sides}$$ up to an integer. This will give us the smallest polygon possible for WGM. We can then calculate the smallest possible value of $$\theta_f$$ to get total internal reflection with a closed optical path as in equation 6.

$$\theta_f = \frac{\pi}{2}-\frac{\pi}{round(n_{sides})}$$

Now that we know the minimal number of sides the polygon requires, we can compute the minimal size for the granule so that an integer number of wavelengths fit into the optical path length (equation 8). Here l is the length of one side of the polygon.

$$m\lambda = n_{sides}l$$ $$l = 2Rcos(\theta_{f})$$

We suggest that at least one wavelength should fit into the path length $$l$$ ( $$l=\lambda$$ ) so that the waves can get trapped. Therefore the minimum size of the granule can be determined as equation 9:

$$R = \frac{\lambda}{2\cos(\theta_{f})}$$

Results

For PHB granules the refractive index at a wavelength of 580 nm is $$n_{PHB} = 1.468$$ (Huglin et al., 1991) and the refractive index of cytosol is $$n_{cytosol} = 1.37$$ (Liang et al., 2007). Putting these values into our model, we find that the minimal size of the PHB granules is $$1.7\mu m$$ (figure 6). This is quite a bit larger than we expect the granules to be since the size of E. coli is only $$1-2 \mu m$$ long and the granules will not fill the entire bacterium.

Silica covered cells

Since the size of the PHB granules that can be grown inside an E. coli bacterium was expected to be smaller than the required size for lasing, we investigated the option of making a biolaser out of the entire cell. Therefore we encapsulated the cell with a layer of biosilica or tin dioxide. The gain medium in this method is provided by fluorophores which we expressed in the cytosol of the cell. To determine the minimal size in this case the calculation was a bit more tricky than for the PHB granules, since we had to take into account that the light gets refracted at the cytoplasm/shell interface but reflected at the shell/medium interface. The optical path has therefore a star-like shape (figure 7).

The critical angle can again be computed by equation 3, where $$n_1$$ is the refractive index of the layer $$n_f$$, and $$n_2$$ is the refractive index of the buffer outside the cell $$n_b$$. When the angle is slightly larger than the critical angle we have total internal reflection where the outgoing angle is equal to the incoming angle, therefore we may set $$\alpha = \theta_c$$. In this model we will neglect the curvature of the surface which means that we can also set $$\theta_i=\alpha$$. Using Snell’s Law (equation 2) again, we can compute the outgoing angle $$\theta_0$$ (equation 10) where $$n_f$$ is the refractive index of the layer and $$n_c$$ the refractive index of the cytosol.

$$\theta_0 = asin\Big(\frac{n_f}{n_c} sin(\alpha)\Big)$$

From $$\theta_0$$ we can easily determine the angle $$\phi$$ since all the angles in a triangle add up to $$\pi$$ (equation 11).

$$\phi = \pi-2 \theta_0 = \pi-2\arcsin\Big(\frac{n_f}{n_c} sin(\alpha)\Big)$$

The path of the light is star-shaped and within the optical path length (OPL) should fit an integer number ($$k_{sides}$$) of wavelengths, the $$OPL=k_{sides}\lambda$$. Furthermore we will assume that the total path length per side is minimally $$\lambda$$ and that one side is taken as the path from point A to D. The optical path length per side is given as $$OPL_{side}=2\cdot x+l\geq\lambda$$. $$X$$ and $$l$$ can be determined as in equation 12 resulting in the minimal radius for the criteria where $$OPL_{sides} = \lambda$$ (equation 13). Here t is the thickness of the layer.

$$\left.\begin{matrix} OPL_{sides} =2x+l\geq \lambda \\ x=\frac{t}{\cos\alpha}\\ l =2R\cos\theta_o \end{matrix}\right\} OPL_{sides} = \lambda = \frac{2t}{\cos\alpha}+2R \cos\theta_o =\frac{2t}{\cos\alpha}+2R \sqrt{1-\Big(\frac{n_f}{n_c}\sin\alpha\Big)^2}\geq\lambda$$ $$R_{min} = \frac{\lambda-\frac{2t}{\cos\alpha}}{2\cos\theta_o}$$

Using the minimal radius of R we can determine the angle $$\psi$$ as in equation 14.

$$\psi=2\arcsin\Big(\frac{t \tan\alpha}{R_{min}}\Big)$$

Using $$\psi$$ and $$\phi$$ we can determine the minimal number of sides of the path (equation 15).

$$k_{sides} = \frac{2\pi}{\psi+\phi}$$

In equation 15, $$k_{sides}$$ has to be an integer to have a closed optical path. Taken together the constraints as in equation 16 we get an equation for $$\phi$$ and $$\psi$$ as a function of $$\alpha$$.

$$\left.\begin{matrix} \frac{2\pi}{k_{sides}}=\phi+\psi\\ \phi = \pi-2\arcsin(\frac{n_f}{n_c}\sin\alpha)\\ \psi = 2\arcsin(\frac{t\tan\alpha}{R_{min}})\\ OPL_{sides}=2x+l \rightarrow R_{min} = \frac{\lambda-\frac{2t}{\cos\alpha}}{2\sqrt{1-(\frac{n_f}{n_c}\sin\alpha)^2}} \end{matrix}\right\} \frac{2\pi}{k_{sides}}=\pi-2\arcsin\Big(\frac{n_f}{n_c}\sin\alpha\Big)+2\arcsin\Bigg(\frac{2t\sin\alpha\sqrt{1-\big(\frac{n_f}{n_c}\sin\alpha\big)^2}}{\lambda\cos\alpha-2t}\Bigg)$$

Solving equation 16 for $$\alpha$$ and using this in the equation for the radius results in the minimal radius required for WGM in a cell covered with a layer of biosilica or tin dioxide.

Results

Using the model described above we determined the minimal radius of a cell covered with polysilicate or tin dioxide to create a cavity that allows for whispering gallery modes. The refractive index of polysilicate is 1.47 (Liang et al., 2007) and the refractive index of tin dioxide is 2 (Pradyot, 2003). For both cases we calculated the minimal radius assuming the thickness of the layer to be 50 nm and a wavelength of 511 nm, 528 nm and 475 nm are the emission wavelengths of GFPmut3b, mVenus and mCerulean respectively. From our model for a cell covered with polysilicate we found a minimal diameter of $$1.3 \mu m$$ for GFPmut3b and mVenus (figure 8a) and a minimal diameter of $$1\mu m$$ for mCerulean as a gain medium. For a cell covered in tin dioxide we found a minimal diameter of $$1.6 \mu m$$ for GFPmut3b (figure 8b) and a minimal diameter of $$1.7\mu m$$ and $$1.5\mu m$$ for mVenus and mCerulean as a gain medium. These sizes are comparable to the size of a bacterium, so whispering gallery modes should in principle be possible inside a covered cell. However, we made a number of simplifications in our model that will in practice probably prevent the system from lasing at this minimum scale. First, in this model we only determined whether light waves will fit into the cavity, in Q3 we will also determine whether lasing can take place with the size of the cell and the reflectivity of polysilicate or tin dioxide. Second, the length scales we work with here are only about twice the wavelength, and the thickness of the boundary layer is much smaller than the wavelength. A more accurate description of what is happening within the cell would require us to work in the thin layer limit (known as Mie theory) as we do in Q4. There we calculate the quality factor for the cavity, which indicates how well the light can be trapped within the structure.

Conclusion

In our model we find that making intracellular lasers as PHB granules inside E. coli is physically not possible. The granules should then have a minimal diameter of $$1.7\mu m$$ which is larger than the small axis of the cell. When E. coli is encapsulated by polysilicate or tin dioxide the minimal cell size to fit light inside is $$1-1.7 \mu m$$ depending on the fluorophore used and whether the cells are encapsulated by polysilicate or tin dioxide. These sizes are comparable to the size of the cell.