![]() The purpose of this equation is to convert resistance (what you are actually measuring) to lux (the unit of light measurement). If you bought a different photoresistor, you would need to look at its datasheet to get the slope and intercept of the line. To be clear, the numbers 375.81 and 1.084 are specific to this photoresistor. But it would also be valid to write the equation as So the slope of the line is negative 1.084, we can write that as 1/R^1.084 instead of R^(-1.084). Trying to put that in words: x to a negative number is equal to one over x to that number. The last part is also a bit of algebra your grandson may or may not have seen yet: ![]() The y-intercept of the line is 375.81, and the slope of the line is -1.084. Where a is the y-intercept of the line and k is the slope of the line (for more information and some examples, see this page ). The equation for a straight line on a log-log plot is Again, depending on what grade your grandson is in, he might not have encountered logarithms in math class yet. A plot with logarithmic scales for both axes is called a log-log plot for short. However, if you look at the graph on the datasheet you will see that it has a logarithmic scale on both axes (e.g. The equation is derived from the graph labeled "Typical Resistance vs Variable Illumination" on the photoresistor's datasheet:ĭepending on what grade your grandson is in, he is probably familiar with the equation for a line (usually expressed as y = m*x + b) on a graph with linear axes. Here's the more detailed explanation from our email: In case you didn't see it yet, there is a very brief explanation of this equation on the "Make it Your Own" tab of the project: Pasting the full reply here so others can see it as well. We sent the following message as a reply to your email - please let us know if that email didn't go through (if so, we apologize). Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License. Use the information below to generate a citation. Then you must include on every digital page view the following attribution: If you are redistributing all or part of this book in a digital format, Then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a print format, Want to cite, share, or modify this book? This book uses the This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Since the arc subtends an angle ϕ ϕ at the center of the circle, To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of Figure 4.7. In solving that problem, you will find that they are less than, but very close to, ϕ = 3 π, 5 π, 7 π, … rad. The exact values of ϕ ϕ for the maxima are investigated in Exercise 4.120. As a result, E 1 E 1 and E 2 E 2 turn out to be slightly larger for arcs that have not quite curled through 3 π 3 π rad and 5 π 5 π rad, respectively. ![]() Since the total length of the arc of the phasor diagram is always N Δ E 0, N Δ E 0, the radius of the arc decreases as ϕ ϕ increases. ![]() These two maxima actually correspond to values of ϕ ϕ slightly less than 3 π 3 π rad and 5 π 5 π rad. The proof is left as an exercise for the student ( Exercise 4.119). In part (e), the phasors have rotated through ϕ = 5 π ϕ = 5 π rad, corresponding to 2.5 rotations around a circle of diameter E 2 E 2 and arc length N Δ E 0. ![]() The amplitude of the phasor for each Huygens wavelet is Δ E 0, Δ E 0, the amplitude of the resultant phasor is E, and the phase difference between the wavelets from the first and the last sources is The phasor diagram for the waves arriving at the point whose angular position is θ θ is shown in Figure 4.7. This distance is equivalent to a phase difference of ( 2 π a / λ N ) sin θ. If we consider that there are N Huygens sources across the slit shown in Figure 4.4, with each source separated by a distance a/N from its adjacent neighbors, the path difference between waves from adjacent sources reaching the arbitrary point P on the screen is ( a / N ) sin θ. To calculate the intensity of the diffraction pattern, we follow the phasor method used for calculations with ac circuits in Alternating-Current Circuits. Calculate the intensity relative to the central maximum of an arbitrary point on the screen.Calculate the intensity relative to the central maximum of the single-slit diffraction peaks.By the end of this section, you will be able to: ![]()
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