# Foucalt’s Method

In 1862, Foucalt designed an apparatus to determine the speed of light in the laboratory using much shorter distance and a rotating mirror. This method is called Foucalt’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 M1 is placed at A so that the light after reflection from the mirror M1, 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 M1 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 M2. The light rays after reflection from M2 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.

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 M1 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.

The displacement of the image in the Foucalt’s experiment was only 0.7 mm. The value of c found by Foucalt was 2.98×108 m/s.

Foucalt 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, Foucalt’s experiment justified the validity of the wave theory of light.