# Foucault’s Method

In 1862, Foucault designed an apparatus to determine the speed of light in the laboratory using much shorter distance and a rotating mirror. This Foucault’s method is called Foucault’s rotating mirror method. The experimental arrangement of this method is shown in the following figure:

The rays of light from a bright source $S$ converge at a point $I$ after passing through the glass plate $P$ and a convex lens $L$. But, a rotating plane mirror $M_1$ is placed at $A$ so that the light after reflection from the mirror $M_1$, converges at the pole $Q$ of the concave mirror $C$, whose distance from $A$ is adjusted equal to its radius of curvature $AQ=a$. The light is reflected back from $C$ along its original path and finally, an image is formed coincident with $S$.

A half silvered glass plate $P$ is placed at $45°$ to the axis of the lens, so that the returning light is reflected from it and form the real image at $E$ which can be viewed with the help of a micrometre eyepiece.

Now, the plane mirror $M_1$ is rotated rapidly at a uniform angular speed about an axis passing through $A$ and light reflected from $C$ will find the plane mirror displaced by an angle $θ$ to a new position $M_2$. The light rays after reflection from $M_2$ appear to come from $I’$ and the image is now observed at $E’$ due to reflection and image $S’$ due to transmission by $P$. The displacement $SS’=EE’=d$ is measured by the micrometre eyepiece.

## Expression to Determine the Speed of Light by Foucault’s Method

If $c$ is the speed of light, then the time taken $t$ by the light to cover distance $2a$ i.e. from $A$ to $Q$ and back to $A$ is given by, $t=\frac{2a}{c}\text{ ___(1)}$

If the plane mirror $M_1$ is rotating at $f$ number of revolutions per second, then in the same time $t$, the plane mirror rotates through an angle $θ$. So,

$\omega=\frac{\theta}{t}$

$2πf=\frac{\theta}{t}$

$\therefore t=\frac{\theta}{2πf} \text{ ___(2)}$

From $(1)$ and $(2)$, $\frac{2a}{c}=\frac{\theta}{2πf}$

$\therefore \theta = \frac{4πfa}{c} \text{ ___(3)}$

According to the laws of rotation of light, if the mirror is turned through an angle $θ$, the reflected ray is turned through an angle $2θ$. So, from the figure,

$∠IAI’=2\theta$

$\frac{\text{(arc) } II’} {\text{(radius) }IA}=2\theta$

$\frac{II’}{a}=2\theta$

$II’=a×2\theta \text{ ___(4)}$

Now, let $b$ be the distance between the plane mirror and the lens $L$ and $k$ be the distance between the lens $L$ and the source $S$. Since $S$ and $S’$ are conjugate points with respect to $I$ and $I’$ for the lens, we have the following relation,

$m=\frac{\text{image size}}{\text{object size}}=\frac{\text{image distance}}{\text{object distance}}$

$\therefore \frac{II’}{SS’}=\frac{(a+b)}{k}$

$\frac{II’}{d}=\frac{(a+b)}{k}$

$II’=\frac{(a+b)d}{k} \text{ ___(5)}$

From $(4)$ and $(5)$, $a×2\theta=\frac{(a+b)d}{k}$

$\theta=\frac{(a+b)d}{2ak} \text{ ___(6)}$

From $(3)$ and $(6)$, $\frac{4πfa}{c}=\frac{(a+b)d}{2ak}$

$\therefore c=\frac{8πfa^2k}{(a+b)d}$

As $f$, $a$, $b$, $k$ and $d$ are measurable quantities, the speed of light $c$ can be calculated.

## Result Obtained From Foucault’s Experiment

The displacement of the image in the Foucault’s experiment was only $0.7$ $\text{mm}$. The value of $c$ found by Foucault was $2.98×10^8$ $\text{m/s}$.

Foucault also found the speed of light in water by placing a tube filled with water between rotating plane mirror and concave mirror, but the speed of light in water was found to be less than that in air which was against the Newton’s corpuscular theory of light according to which the speed of light in water should be more than that in air. But according to wave theory of light, speed of light in water should be less than that in air. So, Foucault’s experiment justified the validity of the wave theory of light.