To understand the basics of astronomical instrumentation, I find it useful to go back to the classic Young’s
double slit experiment. Draw a double slit setup as a 1-D diagram on the board. Draw it big, use straight
lines, and label features clearly. The two slits are separated by a distance D, and each slit is w wide,
where w ! D such that its transmission function is basically a delta function. There is a phosphorescent
screen placed a distance L away from the slits, where L " D. We’ll be thinking of light as plane-parallel
waves incident on the slit-plane, with a propagation direction perpendicular to the slit plane. Further,
the light is monochromatic with a wavelength λ.
This question was definitely difficult and required a significant amount of conceptual thinking about the double slit experiment and Fourier transforms. We began by drawing a diagram of the double slit experiment as described in the problem. From our previous experience with the experiment, we added a normal between the two waves of light at an angle \(\theta\), making a triangle such that sin \(\theta\) is equal to \(\frac{\delta\:r}{D}\).
a) Convince yourself that the brightness pattern of light on the screen is a cosine function. (HINT: Think about the conditions for constructive and destructive interference of the light waves emerging from each slit ).
I was able to convince myself that the brightness pattern is a cosine function because since the length \(\delta\)r determines the type of interference that will occur, as delta r changes, the brightness will change. As the selected focus point of the light moves lower on the screen, delta r intially gets shorter, causing the brightness level to decrease as less constructive interference occurs. Eventually, destructive interference takes over as delta r approaches (n+1/2) \(\lambda\). This process then works in reverse and repeats. Since it exhibits this pattern and begins at a maximum, it is a cosine function.
b) Now imagine a second set of slits placed just inward of the first set. How does the second set of slits modify the brightness pattern on the screen?
Solving this part of the problem required careful consideration of the list of Fourier transforms we were given as homework. In part a, we determined that 2 slits resulted in a cosine function. Looking at the list of FTs, it is clear that a double slit is an even pair which corresponds with a cosine function.
Adding a new set of slits will create an even pair that is slightly closer together, and thus has a slightly smaller \(x_1\), which means it will have a longer period. This will yield a second cosine function in addition to first one.
Now what are the implications of this graph of the two different brightness patterns? Well, the two cosine functions can be added together in order to see what the final modification to the brightness pattern on the screen will be. At points such as the center, the waves will constructively interfere perfectly, creating an intensity that is much brighter than the maximum with only two slits. At points where the waves perfectly interfere destructively, there will be dark spots. Since the waves have different periods, there will be less places where they are perfectly in sync and thus less maximums and dark spots.
c) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?
A continuous set of slit pairs with an ever decreasing separation would create a brightness pattern similar to the one described in part b in that it would be the summation of various cosine functions of different wavelengths. The final pattern would be similar to the brightness patter of a single slit. There would be one very intense central maximum with other bright spots, declining in intensity, spaced evenly apart.
d) Notice that this continuous set of slits forms a “top hat” transmission function. What is the Fourier transform of a top hat, and how does this compare to your sum from the previous step?
A "top hat" transmission function is formed because an infinite number of slit pairs effectively resembles a single aperture of a considerable width. This "top hat" most closely resembles the "box" on the Fourier Transmissions sheet.
This pattern is essentially what I was describing in the last step. As Andrew pointed out to us at TALC, the "ringing" maximums will be contained within the first two nulls, so the Sinc function is a good approximation.
e) For the top hat function’s FT, what is the relationship between the distance between the first nulls and the width of the top hat (HINT: it involves the wavelength of light and the width of the aperture)? Express your result as a proportionality in terms of only the wavelenght of light λ and the diameter of the top hat D.
Looking at the Fourier transform for a box, we see that the distance between the first two nulls is 2/W. To make this into a proportionality in terms of wavelength of light and the diameter of the top hat D, we can use the equation for the distance between maxima (or nulls) that we saw in the first homework reading. This equation is
\[y\:=\:\frac{n\:\times\:\lambda\:\times\:L}{d}\]
We see that since both n and L are arbitrarily fixed, we can remove them from our equation. This leaves us with \(y\:=\:\frac{\lambda}{d}\). Next plug in, \(\frac{2}{W}\) for y since we know that it is equivalent to the distance between nulls. This yields: \[\frac{2}{w}\:=\:\frac{\lambda}{d}\]
Solving for W shows us that \[W\:=\:\frac{2\:d}{\lambda}\]. Now to verify our work, we plugged this value of W into the distance between nulls, \(\frac{2}{w}\) , and found that the distance between nulls was indeed equal to \(\frac{\lambda}{d}\).
f) Take a step back and think about what I’m trying to teach you with this activity, and how it relates to a telescope primary mirror.
While it may not seem at first that this has anything to do with a telescope, the calculations we just made are actually very important. The distance between nulls, \(y\:=\:\frac{\lambda}{d}\) is essentially the \(\theta_{min}\) that Prof. Johnson talked about it class. By minimizing the distance between nulls, you increase the angular resolution and make it easier to distinguish far away objects. Increasing d, or the diameter of the telescope mirror, will decrease the distance between the nulls according to our equation and thus increase the angular resolution.
I would like to give a special thanks to Steve Johnson, Taylor, and Daniel Hellstrom for being a part of my group at TALC.
This question was definitely difficult and required a significant amount of conceptual thinking about the double slit experiment and Fourier transforms. We began by drawing a diagram of the double slit experiment as described in the problem. From our previous experience with the experiment, we added a normal between the two waves of light at an angle \(\theta\), making a triangle such that sin \(\theta\) is equal to \(\frac{\delta\:r}{D}\).
a) Convince yourself that the brightness pattern of light on the screen is a cosine function. (HINT: Think about the conditions for constructive and destructive interference of the light waves emerging from each slit ).
I was able to convince myself that the brightness pattern is a cosine function because since the length \(\delta\)r determines the type of interference that will occur, as delta r changes, the brightness will change. As the selected focus point of the light moves lower on the screen, delta r intially gets shorter, causing the brightness level to decrease as less constructive interference occurs. Eventually, destructive interference takes over as delta r approaches (n+1/2) \(\lambda\). This process then works in reverse and repeats. Since it exhibits this pattern and begins at a maximum, it is a cosine function.
b) Now imagine a second set of slits placed just inward of the first set. How does the second set of slits modify the brightness pattern on the screen?
Solving this part of the problem required careful consideration of the list of Fourier transforms we were given as homework. In part a, we determined that 2 slits resulted in a cosine function. Looking at the list of FTs, it is clear that a double slit is an even pair which corresponds with a cosine function.
Adding a new set of slits will create an even pair that is slightly closer together, and thus has a slightly smaller \(x_1\), which means it will have a longer period. This will yield a second cosine function in addition to first one.
c) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?
A continuous set of slit pairs with an ever decreasing separation would create a brightness pattern similar to the one described in part b in that it would be the summation of various cosine functions of different wavelengths. The final pattern would be similar to the brightness patter of a single slit. There would be one very intense central maximum with other bright spots, declining in intensity, spaced evenly apart.
d) Notice that this continuous set of slits forms a “top hat” transmission function. What is the Fourier transform of a top hat, and how does this compare to your sum from the previous step?
A "top hat" transmission function is formed because an infinite number of slit pairs effectively resembles a single aperture of a considerable width. This "top hat" most closely resembles the "box" on the Fourier Transmissions sheet.
e) For the top hat function’s FT, what is the relationship between the distance between the first nulls and the width of the top hat (HINT: it involves the wavelength of light and the width of the aperture)? Express your result as a proportionality in terms of only the wavelenght of light λ and the diameter of the top hat D.
Looking at the Fourier transform for a box, we see that the distance between the first two nulls is 2/W. To make this into a proportionality in terms of wavelength of light and the diameter of the top hat D, we can use the equation for the distance between maxima (or nulls) that we saw in the first homework reading. This equation is
\[y\:=\:\frac{n\:\times\:\lambda\:\times\:L}{d}\]
We see that since both n and L are arbitrarily fixed, we can remove them from our equation. This leaves us with \(y\:=\:\frac{\lambda}{d}\). Next plug in, \(\frac{2}{W}\) for y since we know that it is equivalent to the distance between nulls. This yields: \[\frac{2}{w}\:=\:\frac{\lambda}{d}\]
Solving for W shows us that \[W\:=\:\frac{2\:d}{\lambda}\]. Now to verify our work, we plugged this value of W into the distance between nulls, \(\frac{2}{w}\) , and found that the distance between nulls was indeed equal to \(\frac{\lambda}{d}\).
f) Take a step back and think about what I’m trying to teach you with this activity, and how it relates to a telescope primary mirror.
While it may not seem at first that this has anything to do with a telescope, the calculations we just made are actually very important. The distance between nulls, \(y\:=\:\frac{\lambda}{d}\) is essentially the \(\theta_{min}\) that Prof. Johnson talked about it class. By minimizing the distance between nulls, you increase the angular resolution and make it easier to distinguish far away objects. Increasing d, or the diameter of the telescope mirror, will decrease the distance between the nulls according to our equation and thus increase the angular resolution.
I would like to give a special thanks to Steve Johnson, Taylor, and Daniel Hellstrom for being a part of my group at TALC.
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