Wednesday, March 28, 2012

Young's Double-Slit Experiment

Young's Double-Slit Experiment

The first serious challenge to the particle theory of light was made by the English scientist Thomas Young in 1803. Young possessed one of the most brilliant minds in the history of science. A physician by training, he was the first to describe how the lens of the human eye changes shape in order to focus on objects at differing distances. He also studied Physics, and, amongst other things, definitely established the wave theory of light, as described below. Finally, he also studied Egyptology, and helped decipher the Rosetta Stone.
Young knew that sound was a wave phenomenon, and, hence, that if two sound waves of equal intensity, but $180^\circ$ out of phase, reach the ear then they cancel one another out, and no sound is heard. This phenomenon is called interference. Young reasoned that if light were actually a wave phenomenon, as he suspected, then a similar interference effect should occur for light. This line of reasoning lead Young to perform an experiment which is nowadays referred to as Young's double-slit experiment.
In Young's experiment, two very narrow parallel slits, separated by a distance $d$, are cut into a thin sheet of metal. Monochromatic light, from a distant light-source, passes through the slits and eventually hits a screen a comparatively large distance $L$ from the slits. The experimental setup is sketched in Fig. 86.

Figure 86: Young's double-slit experiment.
According to Huygens' principle, each slit radiates spherical light waves. The light waves emanating from each slit are superposed on the screen. If the waves are $180^\circ$ out of phase then destructive interference occurs, resulting in a dark patch on the screen. On the other hand, if the waves are completely in phase then constructive interference occurs, resulting in a light patch on the screen.
The point $P$ on the screen which lies exactly opposite to the centre point of the two slits, as shown in Fig. 87, is obviously associated with a bright patch. This follows because the path-lengths from each slit to this point are the same. The waves emanating from each slit are initially in phase, since all points on the incident wave-front are in phase (i.e., the wave-front is straight and parallel to the metal sheet). The waves are still in phase at point $P$ since they have traveled equal distances in order to reach that point.

Figure 87: Interference of light in Young's double-slit experiment.
From the above discussion, the general condition for constructive interference on the screen is simply that the difference in path-length ${\mit\Delta}$ between the two waves be an integer number of wavelengths. In other words,
\begin{displaymath} {\mit\Delta} = m\,\lambda, \end{displaymath}
where $m=0,1,2,\cdots$. Of course, the point $P$ corresponds to the special case where $m=0$. It follows, from Fig. 87, that the angular location of the $m$th bright patch on the screen is given by
\begin{displaymath} \sin\theta_m = \frac{\mit\Delta}{d} = \frac{m\,\lambda}{d}. \end{displaymath}
Likewise, the general condition for destructive interference on the screen is that the difference in path-length between the two waves be a half-integer number of wavelengths. In other words,
\begin{displaymath} {\mit\Delta} = (m+1/2)\,\lambda, \end{displaymath}
where $m=1,2,3,\cdots$. It follows that the angular coordinate of the $m$th dark patch on the screen is given by
\begin{displaymath} \sin\theta_m' = \frac{\mit\Delta}{d} = \frac{(m+1/2)\,\lambda}{d}. \end{displaymath}
Usually, we expect the wavelength $\lambda$ of the incident light to be much less than the perpendicular distance $L$ to the screen. Thus,
\begin{displaymath} \sin\theta_m \simeq \frac{y_m}{L}, \end{displaymath}
where $y_m$ measures position on the screen relative to the point $P$.
It is clear that the interference pattern on the screen consists of alternating light and dark bands, running parallel to the slits. The distances of the centers of the various light bands from the point $P$ are given by
\begin{displaymath} y_m = \frac{m\,\lambda\,L}{d}, \end{displaymath}
where $m=0,1,2,\cdots$. Likewise, the distances of the centres of the various dark bands from the point $P$ are given by
\begin{displaymath} y_m' = \frac{(m+1/2)\,\lambda\,L}{d}, \end{displaymath}

          then
             $\lambda$ in  bright bands   $\lambda$=(Y*d)/(mD)

where $m=1,2,3,\cdots$. The bands are equally spaced, and of thickness $\lambda\,L/d$. Note that if the distance from the screen $L$ is much larger than the spacing $d$ between the two slits then the thickness of the bands on the screen greatly exceeds the wavelength $\lambda$ of the light. Thus, given a sufficiently large ratio $L/d$, it should be possible to observe a banded interference pattern on the screen, despite the fact that the wavelength of visible light is only of order 1 micron. Indeed, when Young performed this experiment in 1803 he observed an interference pattern of the type described above. Of course, this pattern is a direct consequence of the wave nature of light, and is completely inexplicable on the basis of geometric optics.
It is interesting to note that when Young first presented his findings to the Royal Society of London he was ridiculed. His work only achieved widespread acceptance when it was confirmed, and greatly extended, by the French physicists Augustin Fresnel and Francois Argo in the 1820s. The particle theory of light was dealt its final death-blow in 1849 when the French physicists Fizeau and Foucault independently demonstrated that light propagates more slowly though water than though air. Recall, that the particle theory of light can only account for the law of refraction on the assumption that light propagates faster through dense media, such as water, than through rarefied media, such as air.
Posted by at
Labels:

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)