In this experiment, light is again assumed to be a wave (contrary to what was posted earlier) as another of its properties, Interference and diffraction, is observed.
Interference is simply the interaction of two or more waves, and works due to the principle of superposition of waves. When waves are in phase with each other, that is their crests and troughs meet at the same point, then when they interact they are said to constructively interfere which results in a wave which has a bigger crest and trough. However, when they are not in phase, or their crests and toughs do not meet at the same point, then are said to deconstructively interfere, which results to a wave with a smaller crest and trough. The figure below illustrates this effect more efficiently.
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| Figure 1: Illustration for the Interference of Waves. In this particular set up, the waves are said to either totally interfere (a) constructively, or (b) desctructively. Total constructive interference occurs when the waves are totally in phase, and total deconstructive interference occurs when the waves are totally out of phase. The red arrows mark the wave's direction of propagation. The bottom-most illustration shows the result when the waves meet or interfere with each other. (Source: Physics for Scientists and Engineers with Modern Physics 8th Edition by Serway and Jewett) |
Diffraction, as is the focus of this experiment, is the phenomena when light deviates from propagating in a straight line as it hits a boundary, or a barrier.
Many scientists before tried to observe the effects of light interfering with itself. At that time, it was already established that if two light sources were placed side by side such that the light coming from both combines, the effects of interference were not observed because the light waves from either source were being emitted independently of those from the other. Furthermore, the light emitted from the light sources are not constantly in phase due to the fact that light from either sources undergo random phase changes in a very short amount of time -- too short in fact that the human eye cannot distinguish the effects, when they happened. The light sources were said to be incoherent with each other.
However, by 1801, a British scientist by the name of Thomas Young was able to examine this phenomena in his famous Double Slit Experiment. He did it by blocking the path of light coming from a single source using a barrier with two very thin slits separated at a distance between each other, and projecting this effect on a screen. Probably his first prediction, like ours, when light passes through the slits the resulting projection would look like such:
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| Figure 2: A prediction on how the pattern on the screen will look like when Young did this experiment for the first time. The dot on the left represents the light source, The thick line to the right the screen with the white rectangle a projection of the pattern on it. |
This prediction was under the premise that the barrier blocked off most of the light except for the slits hence light would travel in a straight line after passing it -- which makes sense. When we turn this set up such that the screen and barrier are parallel to the horizontal, then pour sand over the slits, then we expect that on the screen, two piles of sand will accumulate.
To everyone's shock however, when Young did this experiment, the projection of the pattern on the screen was different, as shown:
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| Figure 3: The Main Result of Thomas Young's Double Slit Experiment. Instead of seeing just two bands, as expected, what he got was a series of bands, now called as fringes, that was projected on a screen. (Take note that corresponding intensities are not to scale) |
Instead of getting something similar as in Figure 2, Young got a series of alternating bright and dark parallel bands now called as fringes. From this experiment, Young was able to come up with the necessary conditions for interference of light (and waves in general) to be observed. One of the conditions is that the sources must be coherent, or must maintain a constant phase with respect to each other. The other necessary condition for interference of light to be observed is that the sources should be monochromatic, or that it is of a single wavelength.
Another thing to observe is how the direction of propagation of light wave changes as it passes through the two slits. From Figure 3, it is observed that at the slits, light travels as if the slits were point sources of light -- or that light deviates from its supposed path when it passes through the slits. This phenomenon, as defined earlier, is diffraction, and can be explained by Huygen's Principle. Huygen's Principle states that:
"All points on a given wave front can be considered as point sources for of a secondary wavelet that spreads at a rate equal to the speed of propagation of that wave."This simply means that a wave front is the sum of the interference of a wave generated at every point on the previous wave front.
To put it into the context of Young's Double Slit Experiment, the reason why the slits acted as 'sources' of light when light passed through it is because as the wave fronts of light approach it, as most of the light does not get through due to the barrier at some point in that wave front, a wavelet was produced which passed through the slit, and then Huygen's Principle applies again after passing through the slit. It's as if light was already interfering with itself even before we tried to manipulate it -- which is amazing!
Interestingly, the same thing applies for a single slit, as shown below:
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| Figure 4: Projection of Light on the Screen when Light Passes through a Single Slit. Again, it does not agree with the intuitive prediction, instead looks like this. (Note that apparent intensities are not to scale) |
One would ask as to why the so-called Diffraction Pattern is still projected when light passes through a single slit when in fact there is no other 'source' for which the light would interact with. To clear this argument, one goes back to Huygen's Principle. As stated, one can consider the wave front closest to the slit (as illustrated) as a source for the successive wave fronts. In that manner, one can consider that the slit is a 'big' source of waves which then interfere with each other to create the supposed pattern on the screen. Hence, diffraction and interference is apparent even with a single slit.
To further understand the phenomena, we identify the relative positions of the bright and dark bands using mathematics. As shown below in Figure 5, let us denote $d$ as the separation distance of the two slits, $S_1$ and $S_2$ as the sources of light as it goes through the slits, $P$ is a point on the screen a height $y$ with respect to the line connecting the midpoint of $\overline{S_1S_2}$ and the screen, $L$ as the distance of the slits from the screen, and $\theta$ as the angle from the line midpoint of $\overline{S_1S_2}$ and the screen, and $P$.
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| Figure 5: Geometric Construction for the Double Slit Diffraction Pattern |
As discussed, constructive interference between waves occur when the path difference $|\overline{S_2P}|-|\overline{S_1P}|$ are integer multiples of the wavelength $\lambda$ of the light source, and deconstructive interference occur when the path difference are odd multiples of the half wavelength $\frac{1}{2}\lambda$ of the light source. By looking at the figure, one can say that it would be a bit difficult to calculate for this path difference as the lines can be considered as the hypotenuse of two different triangles. However, we physicists (who tend to be lazy), can work our way around it. To make our lives easier, we directly assume that $L\gg d$ such that locally $\overline{S_1P}\parallel\overline{S_2P}$. From this assumption, we can therefore say that the path difference between the two lines is approximately equal to the length of $\overline{S_2C}$, where $\overline{S_1C}\perp\overline{S_2C}$. Hence $|\overline{S_2P}|-|\overline{S_1P}|\approx|\overline{S_2C}|=d\sin\theta$. All we have to do now is to equate this path difference to the conditions established earlier. Hence constructive interference or bright bands occur at a point $P$ such that:
\begin{equation}
d\sin\theta=m\lambda\hspace{3cm}\forall\ m\in\mathbb{Z}
\end{equation}
and deconstructive interference or dark bands occur at point $P$ when:
\begin{equation}
d\sin\theta=\left(m+\frac{1}{2}\right)\lambda\hspace{1.4cm}\forall\ m\in\mathbb{Z}
\end{equation}
$m$ is the order number, or the number of wavelengths representing that path difference between the two slitss. The bright finge in the center is represented by $m=0$, the immediate fringes to its sides are represented by $m=\pm1$ and so on.
It should be noted however, that these are the angular positions of the bands -- to obtain expression for the linear displacements of the band with respect to the line connecting the midpoint of $\overline{S_1S_2}$, one should note that:
\begin{equation*}
\tan\theta=\dfrac{y}{L}
\end{equation*}
and hence, generally,
\begin{equation*}
y=L\tan\theta
\end{equation*}
where $\theta$ is the given $\theta$ in the previous equations. However, this again does not really look appealing to us, hence we will again assume that $\theta$ is small enough such that $\tan\theta\approx\sin\theta$, and hence putting this back our previous equations become:
\begin{equation}
y=L\,\dfrac{m\lambda}{d}\hspace{3cm}\forall\ m\in\mathbb{Z}
\end{equation}
for constructive interference, and
\begin{equation}
y=L\,\dfrac{(m+\frac{1}{2})\lambda}{d}\hspace{2cm}\forall\ m\in\mathbb{Z}
\end{equation}
for deconstructive interference.
We can give the single slit situation the same treatment, however we will deviate a bit from our previous calculation. Again, we apply Huygen's Principle and consider five evenly spaced light sources on the slit, which has a width of $a$. What makes this calculation unique to the double slit diffraction pattern is that instead of considering rays of light coming from every source, we only consider two rays -- one from the upper half of the slit and the other right in its middle as shown in Figure 6.
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| Figure 6:Geometric Construction of the Single Slit Diffraction Pattern |
We now look for the angular and linear positions of the dark band shown on the screen. Again following the same arguments as last time, we can say that position of the first dark bands the path difference between these two rays will be $\frac{1}{2}a\sin\theta$, and for deconstructive interference to occur, this path difference should be equal to the odd multiples of half the wavelength $\frac{1}{2}\lambda$ of the source. Equating these two, one gets that the first dark band is found at $P$ such that:
\begin{equation*}
a\sin\theta=\lambda
\end{equation*}
The interesting thing is that if we pick any two pairs of rays from the sources, given that the distance between them is $\frac{1}{2}a$, we would get the same result.
Looking for the second minima, we again use the same arguments, but this time, consider two rays that are equal to $\frac{1}{4}a$. Again, we know that deconstructive interference occurs when this expression is equal to even multiples of half the wavelength $\frac{1}{2}\lambda$ of the source. Equating, we gets that the second dark band is found at $P$ such that:
\begin{equation*}
a\sin\theta=2\lambda
\end{equation*}
And again, this thing works for all pairs of rays with distance $\frac{1}{4}a$.
Hence we can generalize that for a single slit, the dark bands are fount at $P$ such that:
\begin{equation}
a\sin\theta=m\lambda\hspace{3cm}\forall\ m\in\mathbb{N}
\end{equation}
Getting the linear position of $P$, with the same arguments as in the double slits we get that for small angles, the dark bands are found at $y$ such that:
\begin{equation}
y=L\,\dfrac{m\lambda}{a}\hspace{3.3cm}\forall\ m\in\mathbb{N}
\end{equation}
Take note that $m=0$ is not included. This is due to the fact that at $m=0$, we will find the central bright fringe.
One would wonder however, why there is necessarily no distinction between the bright and dark fringes for the single slit. Let us assume that at $P$, there is a bright band. If that is the case, then the path difference (of the rays from the extreme sources as in Figure 6) is equal to a multiple of the wavelength of the source (as established earlier). If that is the case, path difference would be $a\sin\theta$, equating to an integer multiple of $\lambda$, we see that we just got back to same equation as we did for a dark band. Hence there is necessarily no general formula to get the location of the bright bands.
In this experiment, Young's Double Slit Experiment, and the Single Slit Diffraction Pattern was essentially done. At first, the required materials were set up - a laser was placed on the optics bench and a slit disk (which contained many slits) around 3 cm in front of it. The screen used to project the pattern on was simply a sheet of short bond paper taped on a box, labelled with the corresponding conditions used in the experiment. The slit to screen distance and the wavelength of the laser was measured and recorded. Then, for the whole experiment, the slit width and slit separation distances were manipulated by rotating the slit disk to the desired slit width and slit separation distance.
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| Figure 7: The General Set-Up for the Whole Experiment. The black box on the far right is the laser, with the slit disk in front of it. A box was used to support the screen which was just a sheet of short bond paper. |
The whole experiment was split in two parts - one dwelling on the single slit, and another on the double slit. Furthermore, given that the slit disk was set to some slit width and separation distance, as the projection was seen on the bond paper, a mark was done on the spot where the first and second dark bands was located, and for the double slit, the number of fringes in-between the dark bands was recorded. After the whole experiment, the distance between the dark bands were measured and an experimental value for the slit widths was calculated and compared with the actual value by obtaining the percent difference between the two.
At the end of the single slit diffraction part of the experiment, there were two main observations. It was observed that as the slit width increased, the distance between the dark bands decreased, and vice versa. Furthermore, as the distance from the center increased, the intensity of the fringes decreased as well. Figures 8 and 9 show these observations explicitly. The given slit width for Figures 8 and 9 are 0.02 mm and 0.04 mm respectively.
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| Figure 8: The diffraction pattern when the slit width $a=0.02\ \text{mm}$. It was observed that the central bright band is the brightest of all the bright bands and that from the center, everything is dimmer. Also, it was observed that the bands have a larger width. |
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| Figure 9: The diffraction pattern when the slit width $a=0.04\ \text{mm}$. It was observed that compared to when the slit width $a=0.02\ \text{mm}$ the bright bands have a smaller width. |
This can be explained by referring to Equation (6). From this equation, it is seen that the distance of the black bands from the center is inversely proportional to the slit width - hence this part of the experiment essentially shows this relationship at least from a general perspective.
Quantitatively, the data gathered from this part of the experiment can be summarized into Table 1.
| $a=0.02\ \text{mm}$, $m=1$ | $a=0.04\ \text{mm}$, $m=1$ | |
|---|---|---|
| Distance between the side orders, $\Delta y_1$ | $26.50$ mm | $11.50$ mm |
| Distance from center to side, $y_1$ | $13.25$ mm | $5.75$ mm |
| Calculated Slit Width | $0.0174$ mm | $0.0401$ mm |
| Percent Difference | $12.924\ \%$ | $0.326\ \%$ |
| Slit-to-Screen Distance ($L$) | $35.50$ cm | |
As seen in table one, the distance of the dark band from the center does increase with a smaller slit, which further proves the relation established in Equation 6.
Furthermore, it was observed that even if we get the distance between the second dark band and the center, using Equation 6, we will still get the same magnitude for the slit width, as shown in Table 2.
| $m=1$ | $m=2$ | |
|---|---|---|
| Distance between the side orders, $\Delta y_m$ | $11.50$ mm | $24.00$ mm |
| Distance from center to side, $y_m$ | $5.75$ mm | $12.00$ mm |
| Calculated Slit Width | $0.0401$ mm | $0.0385$ mm |
| Percent Difference | $0.326\ \%$ | $3.854\ \%$ |
The increase in percent error may be due to the fact that the dark and bright bands were not that clear with respect to each other as in these parts the intensity of the second bright band is much less than that of the first bright band.
Things got more interesting when the Double Slit Experiment part was done. Similar to the single slit, the intensity of the bright bands does decrease as their distance from the center increased, and at the same time, as the slit width increased, the distance of the dark bands from the center decreased. However this time, there were more bands! Furthermore, the intensity of these bands follow the same pattern as in the single slit part, however, when the distance between the slits increased, the width of these bands decreased and the number of bright fringes within that central bright band increased. Figures 10 and 11 show these results more explicitly. For these figures, the slit width is 0.08 mm and the slit separation distance for figures 10 and 11 is 0.25 mm and 0.50 mm respectively. Furthermore, in Figure 11, the fringes are very thin that the camera cannot finely distinguish it.
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| Figure 10: The diffraction pattern for when the slit width $a=0.08\ \text{mm}$ and slit separation distance $d=0.25\ \text{mm}$. As observed, the fringes are relatively thick and the whole bright band has a relatively few number of bright fringes. |
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| Figure 11: The diffraction pattern for when the slit width $a=0.08\ \text{mm}$ and slit separation distance $d=0.50\ \text{mm}$. Compared to when $d=0.25\ \text{mm}$ there are more fringes within the central bright band here to the point the camera cannot finely distinguish it. One can say there are more fringes here than in the last figure. |
This phenomena can be explained by referring to either Equation 3 or 4. In either equation, it is seen that the distance of the bright fringes from the center is inversely proportional to the distance between the two slits. Since the experiment shows this relationship as well, then the equation is, more or less, generally correct.
Quantitatively, Table 3 shows the magnitude of the width of each fringe, together with the number of fringes and width of the central bright band given a certain slit width and slit separation distance.
| $a=0.04\ \text{mm}$ $d=0.25\ \text{mm}$ |
$a=0.04\ \text{mm}$ $d=0.50\ \text{mm}$ |
$a=0.08\ \text{mm}$ $d=0.25\ \text{mm}$ |
$a=0.08\ \text{mm}$ $d=0.50\ \text{mm}$ |
|
|---|---|---|---|---|
| Number of fringes | $13$ | $25$ | $7$ | $13$ |
| Width of Central Maximum | $12$ mm | $12$ mm | $5$ mm | $5$ mm |
| Fringe Width | $0.923$ mm | $0.480$ mm | $0.714$ mm | $0.385$ mm |
Interestingly, while it is observed that the slit separation distance affects the width of one fringe projected on the screen, the width of the central maximum is not affected. This is because, as in Equations 3 and 4, the width of the central maximum is independent of the slit separation distance - it is dependent on the slit width. Again, behaving in the same behavior as in the first part of the experiment, the width of the central maximum is inversely proportional to the slit width, and is still observed in the diffraction pattern between two slits. Hence one can infer that the diffraction pattern for a double slit is generally similar to that of the single slit except there are more fringes -- and that is because the interference of light is more evident as there are distinct sources of light. In the single slit, all of these sources are cramped up in one space and hence light interferes with itself even more - but in the double slit, the light source is more defined so there are more fringes observed.
With this, one can still get the slit width using the distance between the first order and second order minimums (dark band) and get the same value. Table 4 explicitly shows this, given that the set-up has a slit width $a=0.04$ mm and slit separation distance $d=0.25$ mm.
| $m=1$ | $m=2$ | |
|---|---|---|
| Distance between the side orders, $\Delta y_m$ | $12.00$ mm | $25.00$ mm |
| Distance from center to side, $y_m$ | $6.00$ mm | $12.50$ mm |
| Calculated Slit Width | $0.0384$ mm | $0.0369$ mm |
| Percent Difference | $3.854\ \%$ | $7.700\ \%$ |
| Slit-to-Screen Distance $L$ | $35.50$ cm | |
Lastly, it is observed that when the screen is farther away from the slits, the whole diffraction is scaled up, or that everything is larger. The distance between a bright band and a dark band is bigger, however keeping the same number of fringes in between the bright bands. Figure 12 shows the effect of distance of the screen to the slit given that the slit width is 0.04 mm. This was done during the single slit part of the experiment and was not done again until the number of fringes enclosed in one bright band was to be counted.
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| Figure 12: The effect of distance of the screen to the slit. As observed the central bright band has a very large width while retaining intensity. Also, it is seen that the second bright bands are farther from the center and fainter. |
Again, this can be explained by referring to Equations 3 through 6. From any of those equations one sees that the distance of the bright and dark bands from the center is directly proportional to the distance of the screen to the slits. Since this experiment manifests this relationship, then the equations are, more or less, generally correct.
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| Figure 13: Picture of the Diffraction Pattern I got when I tried doing the double slit experiment one night. |
On the other hand though, I was happy because it was always my 'dream' per-say to be able to do this type of experiment. Before, I was very determined to do it because I was curious as to how it actually occurs and why it happens. That's why one night, when there was a brownout in our village, I made use of a ruler, pencil, folders, cutters, and our flashlight to make two slits and shine the flashlight between the slits and project in onto the wall. To my dismay, as the light I had was only white, the effect was not evident, furthermore, the light was too bright that the patterns was not really obvious. However it was fun anyway. At least I was able to do it more properly in the lab!
It was fun reading your blog, Chris. Good job! :-)
ReplyDelete- Ma'am Anj