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Students allow a narrow beam of laser light to strike a water surface. They arrange to measure the angle of refraction for selected angles of incidence and record the data shown in the following table:

$$\begin{array}{ll}\hline{\text { Angle of Incidence }} & {\text { Angle of Refraction }} \\ {\text { (degrees) }} & {\text { (degrees) }} \\ \hline \quad\quad {10.0} &\quad {7.5} \\ \quad\quad {20.0} & \quad {15.1} \\ \quad\quad {30.0} & \quad {22.3} \\ \quad\quad {40.0} & \quad {28.7} \\ \quad\quad {50.0} & \quad {35.2} \\ \quad\quad {60.0} & \quad {40.3} \\ \quad\quad {70.0} & \quad {45.3} \\ \quad\quad {80.0} & \quad {47.7}\\ \hline \end{array}$$

Use the data to verify Snell’s law of refraction by plotting the sine of the angle of incidence versus the sine of the angle of refraction. From the resulting plot, deduce the index of refraction of water.

$\approx 1.334$

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University of Washington

Hope College

University of Sheffield

McMaster University

So for this question, were given a bunch of, uh, data, Right. So we're given the data for the angle of incident in the ring angle of refraction for light going into water and were asked a plot that data and then use that data to find the index of refraction of water. Okay, so I just used a simple graphing calculator to do this plot. I plugged all the data in and then just asked it to give me the plot of the relationship between the two data. Um, you could do it that way. You can use excel, but or you could just do it by hand. But using that, I found that this came out to be a straight line where the X axis here is sine of the angle of refraction. And the Y axis is the sine of the angle of incident. So the ex access goes from about 0.1, where it starts in 0.8, where it ends and 0.2 for the Y and one where it ends. It's just a straight line, so I'll do my best here to just draw a straight line. Something like that. Okay, so then the index of refraction would just be the slope of this line. So that's rise over. Run that sign of data I over scientific data are. But I also just ask my graphing calculator to give me the slope. And it came out to be just from this rough data about 1.334 which is pretty close to the index of refraction of water so we can box that in as our solution to the question.