Fermat's principle and laws of geometric optics

16

FARM PRINCIPLE
AND THE LAWS OF GEOMETRIC OPTICS

G. Ya. Myakishev

LAWS OF GEOMETRIC OPTICS

The propagation of light waves is a rather complex process. However, if the properties of the medium in which light propagates change little over distances approximately equal to the light wavelength, then the relatively simple laws of geometric optics turn out to be valid. There are four of them in total.

1. In a homogeneous medium, light travels in a straight line.

2. Law of reflection. The incident beam, the reflected beam and the perpendicular to the reflecting surface lie in the same plane, and the angle of reflection β is equal to the angle of incidence α (Fig. 1).

3. Law of refraction. The incident beam, the refracted beam and the perpendicular to the interface lie in the same plane, and the ratio of the sine of the angle of incidence i to the sine of the angle of refraction r (Fig. 2) is a constant, independent of the angle of incidence:

$$\frac{\sin i}{\sin r} = n.$$

The value n is called the refractive index of the second medium relative to the first.

4. Independence of light beams. Intersecting in space, light rays have no effect on each other. So when

17

Fig. 3.

read a book, the light stream coming from the window and penetrating the space between the book and your eyes does not in any way prevent you from seeing the letters.

FARM PRINCIPLE

In the middle of the 17th century, the Ilyan scientist Fermat put forward a principle from which all the laws of geometric optics flowed. It is as follows: light coming from one point to another always travels along a path that requires minimal time *).

Let’s get the laws of geometric optics the way Fermat did about 300 years ago.

1. Law of rectilinear propagation of light. Distance along a straight line is the shortest distance between two points. The speed of light in a homogeneous medium is the same at all points. Consequently, light needs the least amount of time when it travels in a straight line.

2. Law of reflection. How to get from point A to point B in the shortest time (Fig. 3)? Of course along the line AB. Light travels in a homogeneous medium. But if it is preceded by Fermat’s strict principle, it is formulated as follows: light propagates in the path that requires either minimum or maximum time. However, in most cases of light, a simpler formulation is correct.

Fig. 4.

must be reflected from the mirror and only then get to point B, it is not immediately clear which path corresponds to the minimum time. Let us draw several possible paths $AC_1B$, $AC_2B$ and $AC_3B$. Due to the fact that light, both before and after reflection, propagates in a homogeneous medium, the minimum time corresponds to a path of minimum length.

What is the shortest path? A simple trick helps to find it. Let’s construct the point $A_1$ - the image of point A. It lies on the continuation of the perpendicular $AC_0$, lowered onto the mirror from point A, at a distance $A_1C_0 = AC_0$ from the mirror. By connecting points C$_1$, C$_2$ and C$_3$ with point A$_1$, it is easy to verify that $AC_1 = A_1C_1$, $AC_2 = A_1C_2$ and $AC_3 = A_1C_3$. Therefore, when calculating the path from point A to point B, we can replace the segment $AC_1$ with the segment $A_1C_1$, the segment $AC_2$ with the segment $A_1C_2$ and the segment $AC_3$ with the segment $A_1C_3$. It is clear that the shortest path is the one in which the beam falls on the mirror at point C$_2$, lying on straight line $A_1B$: straight line $A_1B$ is shorter than any broken line passing through points A and B.

Now it remains to show that when a ray is reflected from a mirror at point C$_2$, the law of reflection is satisfied, that is, that the angle of incidence α is equal to the angle of reflection β (Fig. 4). Since triangles $AC_2C_0$ and $A_1C_2C_0$ are equal,

18

Fig. 5.

then $\angle AC_2C_0 = \angle A_1C_2C_0$. On the other hand, $\angle C_0C_2A_1 = \angle BC_2N$. Therefore, $\angle AC_2C_0 = \angle BC_2N$, and therefore $\alpha = \beta$.

It is not difficult to report that if the rays $AC_2, BC_2$ and the perpendicular $DC_2$ did not lie in the same plane, then the path from point $A$ to the mirror and then from the mirror to point $B$ would be longer.

Thus, the law of reflection follows entirely from Fermat’s principle.

Interestingly, the law of reflection was first obtained by Heron of Alexandria, who lived in the 1st century. Heron believed that light travels along the shortest path. For a single-sided medium, this coincides with Fermat’s principle of least time. But it turned out to be impossible to explain the law of refraction in this way.

3. Law of refraction. Light from point $A$ falls on the flat interface between two media (Fig. 5) and is refracted, then hitting point $B$. If the speeds of light in the first (upper) and second (lower) media are different (for definiteness, we will assume that $v_1 > v_2$), then it is clear that the straight path $AB$ requires no minimum time. The time will be shorter if the light travels a slightly longer path in the first medium,

Fig. 6.

where it propagates at a higher speed, but over a slightly shorter path in the second medium.

Suppose that light takes the least time if its path passes through the point $C$. Then any other path to the left and right of point $C$ will take a long time. An arbitrary beam trajectory $AFB$ can be set using the variable $x$—the distance between point $C$ and point $F$ on the interface between the media. The curve of light propagation time versus $x$ should have a minimum at point $C$, that is, at $x=0$ (Fig. 6). Near the minimum, time changes very little with $x$. Approaching at very small $x$ this time can generally be considered constant: on a very small segment in height $x=0$ the curve can be seen as a small segment of a straight line parallel to the $x$ axis.

Let us compare the propagation times of light on paths $ACB$ and $AFB$, assuming that $x=FC$ is very short. On the path $ACB$, light in the first medium passes an additional path $MC$ (FM is perpendicular to $AC$; since $FC$ is very small, then $AF=AM$). In the second medium, the additional path is already passed by the ray propagating along the path $AFB$ ($CN \perp FB$). Since the time of light propagation from point $A$ to point $B$ should not change when

20

Lemen from the sea surface. In this case, the rays of light are bent, rising into the upper, warmer layers of the atmosphere.

Another interesting consequence of Fermat’s principle is the principle of reversibility of the trajectory of light propagation. If light has passed from point A to point B along a trajectory that provides the minimum travel time, then, obviously, on the way from point B to point A (that is, if the source is located at point B), then movement along the same trajectory will require minimum time.

Another example. Everyone knows the focusing properties of a lens. Light from point A (Fig. 8) goes in all possible ways, is refracted in the glass and collected at point E. Contradiction? After all, light always follows a path that requires minimal time! Why, regardless of the direction in which the ray came out, does it arrive at the same point E?

In fact, there is no contradiction. The ADE path is shorter than the ACE path, but the distance traveled in an optically denser medium—glass—is greater in the first case. Thus, the travel time in general cases is the same and both rays will simultaneously arrive at point E. The paths ADE and ACE for light are equal!

If this were not so, we would not be able to build a focusing system.

It is precisely due to the fact that for rays of light emerging from point A in different directions and then passing through the lens, the same minimum propagation time corresponds to the trajectory induce to point E, the lens focuses the light incident on it.

Fermat’s principle allows not only to explain the operation of an arbitrarily complex optical system, but also to calculate it. However, we could calculate the optical system and use the laws of reflection and refraction, known long before Fermat’s work. What new does Fermat’s principle provide? Firstly, this law is more general, and the more general a law is, the greater its significance. Therefore, the discovery of Fermat’s principle is a big step in the development of geometric optics. Secondly, several new important consequences can be deduced from Fermat’s principle. However, one of the most important is that the refractive index is equal to the relative speed of light in the first and second media.

Refraction at the focal surface

Let us consider, as an example, the calculation of an optical system and the derivation of a lens formula known to you. To do this, it is enough to distance the refraction of light at the focal surface. All rays emanating from the exact source S, after refraction at the focal surface, will not be collected at one point S₁. This is due to the fact that only surfaces of a very complex shape can provide exactly the same time of light propagation from S to S₁ along all possible trajectories. Such a surface should be very difficult to make, and it is not even attempted to be made. Following the focal surface is much easier. Although the focal surface does not focus all the rays, the narrow beam of rays resulting from the closeness of the perpendicular to the surface (it is called the parastrong beam) is collected at almost one point.

An empty exact light source S is located at a distance a from the focal surface of radius R (Fig. 9). The refractive index of air is considered equal to unity, and the refractive index of a glass sphere is equal to n, that is, the speed of light propagation in air is n times greater.

21

Fig. 9.

speed of light propagation in glass. Let’s find at what distance $b$ from the surface of the sphere a narrow beam of rays searching from the source $S$ will be collected. According to Fermat’s principle, this distance is determined by the condition that the time of light propagation along a rectilinear path $SDS_1$ is equal to (the application of) the time of light propagation along an arbitrarily chosen path $SAS_1$, if only the point $A$ is close to the straight line $SDS_1$, perpendicular to the surface of the sphere.

From the point $A$ we drop the perpendicular $AB$ onto the line $SS_1$. Let’s denote its length by $h$. Since we are considering a paraxial ray of rays, then $h \ll a$, $h \ll b$ and $h \ll R$. We denote the distance $DB$ as $x$. On ray $SA$ we plot the segment $SD_1 = a + x$, and on ray $AS_1$ the segment $S_1B_1 = b - x$. Then, along the trajectory $SAS_1$, the light will travel only a path $\Delta_1 = D_1A$ in the air and $\Delta_2 = AB_1$ in the glass compared to the trajectory along the straight line $SS_1$. But then, on a straight path, the light will travel a path $x - \Delta_2$ longer in the glass than in the case when it follows a path $SAS_1$. The light delay in the glass on path $x - \Delta_2$ must be the same as the delay in air on path $\Delta_1$ plus the delay in the glass on path $\Delta_2$. (At $x \leq \Delta_2$ such compensation is very impossible.)

Let us now find the paths $\Delta_1$, $\Delta_2$ and $x$.

The square of $h$ of the right triangle $SAB$, according to the Pythagorean theorem, is equal to

$$h^2 = (a + x + \Delta_1)^2 - (a + x)^2,$$

or

$$h^2 = (a + x + \Delta_1 + a + x)(a + x + \Delta_1 - a - x),$$

but $a + x + \Delta_1 - a - x = \Delta_1$, and $a + x + \Delta_1 + a + x = 2a + \Delta_1 + 2x \approx 2a$ (since $\Delta_1 \ll a$ and $x \ll a$). That’s why

$$h^2 = 2a\Delta_1$$

And

$$\Delta_1 \approx \frac{h^2}{2a}.$$

Similarly, from the right triangle $ABS_1$ we find that

$$\Delta_2 \approx \frac{h^2}{2b}.$$

Finally, from the right triangle $ABC$ one can obtain the equation to determine $x$:

$$h^2 = R^2 - (R - x)^2.$$

From here

$$x \approx \frac{h^2}{2R}.$$

If the speed of light in air is $v_1$, and the speed of light in glass is $v_2$, then the light travels $\Delta_1$ in time $\tau_1 = \frac{\Delta_1}{v_1} = \frac{h^2}{2av_1}$, and the light travels $\Delta_2$ in glass in $\tau_2 = \frac{\Delta_2}{v_2} = \frac{h^2}{2bv_2}$.

On the segment $x$, the delay of light, caused by the fact that it travels in glass and not in air, is equal to the location of the time it takes light to travel through segment $x$ in glass and in air:

$$\tau = \frac{x}{v_2} - \frac{x}{v_1} = \frac{x}{v_1}\left(\frac{v_1}{v_2} - 1\right) = \frac{x}{v_1}(n - 1) = \frac{h^2(n - 1)}{2Rv_1}.$$

To focus rays, the travel time of light from point $S$ to point $S_1$ along all trajectories must be the same. Hence,

$$\tau_1 + \tau_2 = \tau.$$

Substituting here the expressions for $\tau_1$, $\tau_2$ and $\tau$ and then canceling by the common factor $\frac{h^2}{2}$, we obtain the equation

$$\frac{1}{a} + \frac{n}{b} = \frac{(n - 1)}{R}.$$

This equation allows, using the known assumption exponent $n$ and the radius of the spherical surface $R$, to find the image position (distance $b$) for a given distance from the source to the spherical surface. The fact that the distance $b$ does not depend on $h$ shows that all the rays in a narrow beam, distinguishing the values ​​$h$ from each other, converge at one point.

In the partial case when $a \to \infty$ (the beam consists of parallel rays), the position of the point at which the rays converge (focus) is determined by the formula

$$f = \frac{Rn}{n - 1}.$$

22

Fig. 10.

Let us now see what will happen at $a < \frac{R}{n-1}$. From formula (1) it follows that in this case $b < 0$. This means that point $S_1$ should not be to the right of the spherical surface, but to the left of it. But rays incident on a spherical surface cannot intersect in front of this surface. Only their continuations can intersect in front of the surface (Fig. 10). In such cases, they say that the image of the point is minimal. The orientation of the distance sign $b$ corresponds to the imaginary image of the source.

Line focal length and lens formula.

Using formula (1), obtained using Fermat’s principle, it is possible to calculate lenses, that is, find its formula connecting the distance $a$ from the object to the lens, the distance $b$ from the lens to the image and the formula for the distance of the lens (in other words, obtain the lens formula). It is possible, of course, to do the same without a chair, applying Fermat’s principle itself to all lenses, but this path is more complicated.

The lens is a solid body bounded by spherical surfaces. figure 11 shows a biconvex lens with radii of curvature of its surface equal to $R_1$ and $R_2$. The line connecting the centers $O_1$ and $O_2$ of the spherical surfaces is called the main optical axis of the lens.

Fig. 11.

We will assume that the lens is thin, that is, we will assume that the distances $a$ and $b$, as well as the radii of curvature $R_1$ and $R_2$ are much greater than the thickness of the lens. Then in calculations the thickness of the lens can be completely neglected.

After the refraction of the light ray leaving the point $S$, it appears $A S_1$. If there were no second surface, then the image of point $S$ would be at point $S_1$ at a distance $b_1$ from the lens, determined by the equation

$$\frac{1}{a} + \frac{n}{b_1} = \frac{n-1}{R_1},$$

where $n$ is the refractive index of the lens material.

But in reality, the light is refracted once again on the second spherical surface and the image produced by the lens ends up at point $S_2$ on the optical axis at a distance $b_2$ from the lens. We use the reversibility of light rays. If you place the source at point $S_2$, then after refraction on a spherical surface of radius $R_2$ the rays will appear so that their continuations intersect at point $S_1$, giving a minimum image of the source. Using formula (1) and considering that $b_1 < 0$, we can write the equation

$$\frac{1}{b} - \frac{1}{b_1} = \frac{n-1}{R_2}.$$