In the semiconductor industry, there is a continually aggravating problem of increasing communication volume. Optical interconnection technology is seen as a prime option for solving this problem.^{1, 2} In planar optical interconnects (POI) light is injected (by an in-coupling optical element) into a transparent slab (light guide) at a total internal reflection angle (
$\psi $
in Fig. 1) and propagates along it until it meets an out-coupling optical element and goes to the detector. Diffractive lenses as coupling optical elements have many advantages^{3, 4, 5} and have been suggested also in the context of quasi-optics^{6} for rapidly expanding terahertz technology.^{7} However, their intrinsic dependence of optical properties on the wavelength—chromaticity—poses difficult problems.^{8} Chromaticity leads also to temperature instability. Namely, the wavelength of semiconductor lasers is temperature-dependent. Temperature change causes wavelength shift, therefore the laser beam as deflected by the diffractive in-coupling optical element deviates from its desired path and can miss the out-coupling element. For example, for a typical 850-nm vertical-cavity surface-emitting lasers (VCSEL) source the wavelength shift is
$d\lambda \u2215dT=0.06\phantom{\rule{0.3em}{0ex}}\mathrm{nm}\u2215\mathrm{K}$
,^{9} i.e., relative wavelength shift is
$(1\u2215\lambda )(d\lambda \u2215dT)\sim 7\times {10}^{-5}\phantom{\rule{0.3em}{0ex}}{\mathrm{K}}^{-1}$
. If POI was assembeled at
$20\xb0\mathrm{C}$
, the working temperature is
$140\xb0\mathrm{C}$
(
$Max3905$
from Dallas Semiconductors, e.g.), and the POI length is
$10\phantom{\rule{0.3em}{0ex}}\mathrm{cm}$
, the resulting beam deviation is about
$1.7\phantom{\rule{0.3em}{0ex}}\mathrm{mm}$
(see below the calculation). This temperature instability is specific for diffracive POI since other thermal effects are much less: e.g., for fused silica the explicit refraction index temperature dependence is
$-3\times {10}^{-6}\phantom{\rule{0.3em}{0ex}}{\mathrm{K}}^{-1}$
.^{10} The implicit temperature dependence (due to the refraction index wavelength dependence
$dn\u2215d\lambda $
and the laser wavelength shift
$d\lambda \u2215dT$
) is even smaller:
$dn\u2215d\lambda =4\times {10}^{-5}\phantom{\rule{0.3em}{0ex}}{\mathrm{nm}}^{-1}$
,^{10} leading to
$dn\u2215dT=(dn\u2215d\lambda )(d\lambda \u2215dT)=2.4\times {10}^{-7}\phantom{\rule{0.3em}{0ex}}{\mathrm{K}}^{-1}$
.

We propose to fix this problem by bending POI in accordance to the temperature change. We suppose that POI and the laser have the same temperature since they are close to each other. A simple bending scheme is absolutely passive and implies attaching to POI a second plate, having different thermal expansion coefficient (in mass production, the attachment may be done by lamination technology^{11}). The curvature of this bending is proportional to the temperature change (as long as the thermal expansion can be considered as linear with temperature). As long as the laser wavelength shift can be also considered proportional to the temperature change, the bending curvature is proportional to the wavelength shift. Thus, with appropriate proportionality between the bending curvature and the wavelength shift, the bending will cause the diffracted beam to propagate along nearly the same path for different wavelength.

Let us consider the problem quantitatively. At some reference temperature ${t}_{0}$ , POI is strictly planar and the incident angle of the incoming beam is zero (Fig. 1). Let the propagation angle (within POI) be $\psi $ (see Fig. 1). When temperature changes $\left(t\right)$ , POI bends symmetrically in respect to perpendicular plane situated just in the middle between the input and output diffractive elements (Fig. 2).The incident angle $\alpha $ becomes nonzero since the laser beam direction does not change in the lab frame. However, the bending curvature can be tuned in such a way that the propagation angle $\psi $ retains its original value (locally in the incidence/refraction point) due to the incident angle change. We show now how this tuning is achieved.

First, we derive the beam displacement as function of wavelength
$\lambda $
shift. For the reference radiation wave vector
${k}_{0}=2\pi \u2215{\lambda}_{0}$
, incident angle
${\alpha}_{0}=0$
, grating vector
$K=2\pi \u2215\Lambda $
(
$\Lambda $
is grating period), propagation angle
$\psi $
, and propagation vector
$\stackrel{\u20d7}{r}$
we have^{12}

## 1

$${r}_{x}=K,\phantom{\rule{1em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\psi ={r}_{x}\u2215n{k}_{0}=K\u2215n{k}_{0}={\lambda}_{0}\u2215n\Lambda $$## 3

$$\frac{\mathrm{\Delta}l}{l}\approx \frac{\mathrm{\Delta}\lambda}{\lambda \phantom{\rule{0.2em}{0ex}}{\mathrm{cos}}^{2}\psi}.$$In order to correct this deviation we want to get the same propagation angle $\psi $ at a different temperature $t$ . To achieve this we must take nonzero incident angle $\alpha $ [Fig. 3c]. Therefore ${r}_{x}^{\prime}=K+k\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\alpha $ and $\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\psi ={r}_{x}^{\prime}\u2215nk$ . We obtain

## 4

$$\frac{K}{n{k}_{0}}=\frac{K+k\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\alpha}{nk}.$$## 5

$$\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\alpha =(\delta \u2215(1+\delta ))n\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\psi .$$Let us calculate now the change of the beam trajectory in the curved element is respect to the original planar. Consider one beam reflection. In Fig. 4, $E\widehat{A}O=\pi -\psi $ . In the triangle $\mathrm{\Delta}EAO$ , $A\widehat{E}O=\pi -(\pi -\psi )-\beta =\psi -\beta $ (it should be noted here that the second incidence angle $A\widehat{E}O$ is smaller than $\psi $ and can be below the total internal reflection threshold). In $\mathrm{\Delta}EBA$ , $\mathrm{tan}(\psi -\beta )=AB\u2215EB$ . Considering circumference with origin $O$ we get $EB=EC+CB=d+R(1-\mathrm{cos}\phantom{\rule{0.2em}{0ex}}\beta )$ and $AB=R\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\beta $ . We have therefore

## 7

$$d+R(1-\mathrm{cos}\phantom{\rule{0.2em}{0ex}}\beta )=R\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\beta \bullet \mathrm{cot}(\psi -\beta )$$## 8

$$\u03f5+1-\mathrm{cos}\phantom{\rule{0.2em}{0ex}}\beta =\mathrm{sin}\phantom{\rule{0.2em}{0ex}}\beta \phantom{\rule{0.2em}{0ex}}\mathrm{cot}(\psi -\beta ).$$## 10

$${\beta}_{2}=-\frac{3+\mathrm{cos}\phantom{\rule{0.2em}{0ex}}2\psi}{2\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}2\psi}{\beta}_{1}^{2}.$$## 11

$$\frac{\mathrm{\Delta}l}{l}=\frac{{\beta}_{2}}{{\beta}_{1}}=-\u03f5\frac{3+\mathrm{cos}\phantom{\rule{0.2em}{0ex}}2\psi}{2\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}2\psi}\phantom{\rule{0.2em}{0ex}}\mathrm{tan}\phantom{\rule{0.2em}{0ex}}\psi .$$Without bending, as mentioned before [Eq. 3], the relative beam displacement is $\mathrm{\Delta}l\u2215l\sim 2\delta $ . Therefore the wavelength-shift-caused beam deviation is reduced by factor of $\sim d\u2215L$ , i.e., usually above one order of magnitude. Figure 5 presents ray tracing results of the beam deviation as function of wavelength change—with and without compensation by means of POI bending. Exact ray tracing [numerical solution of Eq. 8] results are indistinguishable from the approximation [Eq. 11].

The proposed scheme works only when the source and the detector are situated from one side of POI, but this seems to be the common case. One can realize this scheme for rectangular or circular arrays of sources/detectors. In the latter case, each pair source-detector should be situated diameterally and the bending curvature center should be at the line normal to the array circle and crossing its center. Linear behavior of curvature in respect to temperature should take place also in this case.

Finally, speaking about diffractive optical elements for VCSEL, it should be mentioned that there is an unavoidable spread of nominal VCSEL wavelengths (at a given temperature) from one laser array (chip) to another of usually up to around $10\phantom{\rule{0.3em}{0ex}}\mathrm{nm}$ or more. The problem of this “bias” may be solved at the assembling stage by off-axis adjusting of the laser array, as in Fig. 3c. The off-axis angle $\alpha $ is given by Eq. 5; its magnitude is about $\alpha \sim \mathrm{\Delta}\lambda \u2215\lambda \sim 0.01$ . As far as $\alpha \u2aa11$ , the effects are linear and this adjusting should not affect the above results regarding the temperature compensation.

## Acknowledgements

We are grateful to the anonymous referee, who made many valuable comments, leading to considerable improvement of this letter.