EuPhO 2023, Теория — Задача 1. T1: Thermal lens

• b) $T_\mathrm{c} = T_\mathrm{h} + \frac{AP_\mathrm{L}}{4\pi kb}\left[1 + 2\ln\left(\frac{a}{\sigma}\right)\right] = 41^{\circ}\mathrm{C},\quad m = -\frac{AP_\mathrm{L}}{4\pi kb\sigma^2} = -1.1 \cdot 10^{7}\ \mathrm{K\,m^{-2}}$
• c) $f = \frac{1}{2\gamma b|m|} = \frac{2\pi k\sigma^2}{\gamma AP} \approx 0.94\ \mathrm{m}$

(a) Drawing a $T(r)$ graph

The graph should present or clearly infer for the four elements shown in the figure.

(b) Finding $T_\mathrm{c}$

Approach with a direct solution of the heat-transport equation.

Consider a cylindrical cut of the disk with radius $r$. Let $P_\mathrm{abs}(r)$ be the portion of laser power absorbed within the cylinder. The absorbed power is being transferred as heat towards the outer holder through the circumvent surface $2\pi br$ of the cylinder. The heat-transport equation reads:

Depending on $r$, the absorbed power is given by two different expressions. Within the illuminated area $0 \le r \le \sigma$, the incident light intensity $I = P_\mathrm{L}/(\pi\sigma^2)$ is constant, so the absorbed power is:

and the solution for $T(r)$ is quadratic in $r$:

where $T_\mathrm{c} = T(0)$ is the temperature at the center. It is clear from that expression that the parameter $m$ is:

Outside the illuminated area, i.e. for $\sigma \le r \le a$, $P_\mathrm{abs}(r) = AP_\mathrm{L}$, which does not depend on $r$. The solution for $T(r)$ is logarithmic in $r$:

where $T_\mathrm{h} = T(a)$ is the temperature of the holder, which is equal to the temperature along the outer rim of the disk. After matching the two solutions at $r = \sigma$ we obtain:

Alternative approach – direct piece-wise integration of the heat-transport equation.

After realizing that $P_\mathrm{abs}(r)$ is given by a piece-wise function:

the student may substitute the given solution $T(r) = T_\mathrm{c} + mr^2$ into heat-transport equation (1) for $0 \le r \le \sigma$. This gives directly the expression (4) for the parameter $m$. The parameter $T_\mathrm{c}$ could be easily identified with the temperature $T(0)$ at the center of the disk. On the other hand, $T_\mathrm{h} = T(a)$ due to the thermal contact between the rim of the disk and the holder. It follows from the heat-transport equation that:

The piece-wise integration of the two sides of the equation in the interval $0 \le r \le a$ gives:

which is equivalent to the expression (6) for $T_\mathrm{c}$.

(c) Finding the focal length

Approach based on the Fermat’s principle

We consider only the illuminated area of the disk ($0 \le r \le \sigma$). Due to the nonuniform temperature distribution, the index of refraction is also $r$-dependent, which leads to a bending of the light rays incident at nonzero radii $r$, as shown schematically in the figure. As a result, the light rays exiting the disk, converge toward the optical axis, and, eventually, cross it in a certain point at a distance $f$ from the disk.

Since the ray bending is relatively small, one may assume that: (i) all the rays travel approximately the same distance $b$ inside the disk; (ii) any given ray enters and exits the disk at approximately the same height $r$; (iii) $f \gg r$. Thus, the optical pathway $s(r)$ of a ray, incident at a height $r$ above the optical axis, is:

According to Fermat’s principle, in order that all the rays focus at the same point, it is necessary that $s(r)$ is constant within $0 \le r < \sigma$. In particular, $s(r) \equiv s(0)$ for any $r$, which leads to the condition:

Since:

condition (11) is satisfied if

The beam hence focuses at

Taking into account the expression for $m$ derived in part (b), we represent the answer in terms of the known parameters and calculate its numerical value:

Alternatively, the student may state that the optical pathway $s(r)$ does not depend on $r$ and use the condition $\mathrm{d}s/\mathrm{d}r \equiv 0$, which gives:

Since:

we obtain:

The beam thus focuses at:

Approach based on a direct ray/wavefront tracing

As a next approximation, the light ray inside the disk can be modeled as a circular arc of a radius $R$ ($R \gg b$, see the figure). As a result, the ray exits the disk at a smaller height $r - h$ above the optical axis ($h \ll r$). The angle of bending $\varphi$ of the ray inside the material is related to $h$ by:

From the Snell’s law it follows that:

Up to terms, linear in $h$, one may write that:

Thus, the radius of the ray inside the material, is:

and the bending angle is approximately:

Alternatively, the students may trace a small part of the wavefront, associated with two rays, incident at close distances $r$ and $r + \mathrm{d}r$ from the optical axis. By noticing that the wavefront is perpendicular to the rays, the angle $\varphi$ of deviation of the rays is equal to the angle of rotation of the wavefront. Let $v(r) = c/n(r)$ be the speed of light in the material at a distance $r$ from the axis. The time-rate $\dot{\varphi}$, i.e. the angular speed of the wavefront is:

The rays reach the opposite surface of the disk in approximately the same time $t = bn(r)/c$, so the total deflection angle of the wavefront, and of the rays, thereof, is $\varphi = \dot{\varphi}t = -n'(r)b/n(r)$

Upon exiting the disk, the ray undergoes additional refraction, and inclines at a new angle $\theta$ relative to the optical axis. The Snell’s law in the small-angle approximation ($\sin\theta \approx \theta$, $\sin\varphi \approx \varphi$) states that:

Since $n'(r) = \gamma T'(r) = 2mr$, one obtains the following expression for the angle of inclination:

It is clear from the figure that:

which reproduces the result obtained by the Fermat’s principle.

Alternatively:

Consider a ray incident at a specific height $r_0$. Let $x \in [0, b]$ be the horizontal coordinate of the ray inside the material. Our goal is to find approximately the profile $r(x)$ of the light ray inside the material. From Snell’s law:

Since:

we obtain:

Since $n(r) \approx n(r_0) + n'(r_0)(r - r_0) = n(r_0) + 2\gamma|m|r_0(r_0 - r)$, the differential equation approximates to:

which is solvable by separation of variables and gives an approximate parabolic path:

Finally, for the angle, just before exiting the disk, we obtain: