You are attempting to determine the index of refraction of a block of an unknown material. You shine a light beam at various angles \(\theta_1\) from the normal to the block and measure the refraction angle \(\theta_2\), as shown. You collect the following data: $$ \begin{array}{|l|l|} \hline \theta_1(\mathrm{deg}) & \theta_2(\mathrm{deg}) \\ \hline 10 & 5.0 \\ \hline 20 & 12.5 \\ \hline 30 & 18.0 \\ \hline 40 & 24.0 \\ \hline 50 & 28.6 \\ \hline 60 & 33.0 \\ \hline 70 & 36.0 \\ \hline 80 & 40.0 \\ \hline \end{array} $$ a) Determine the index of refraction of the material by two methods. b) Which points would you say represent bad measurements? You stand at a distance \(x\) from a plane (flat) wall mirror.

Step by step solution

01

Using Snell's Law to calculate the index of refraction

Snell's Law relates the angles of incidence and refraction to the indices of refraction of the two media that the light passes through. In this case, the light goes from air (with an index of refraction of \(n_1=1\)) into an unknown material (with an index of refraction \(n_2\)). Snell's Law can be written as: $$ n_1 \sin \theta_{1} = n_2 \sin \theta_{2} $$ We can rearrange this formula to find the index of refraction of the unknown material: $$ n_2 = \frac{n_1 \sin \theta_{1}}{\sin \theta_{2}} $$ We will apply this formula for each pair of angles in the data to find the index of refraction.

02

Calculate index of refraction for each data point

Now plug in the given angles (\(\theta_1\) and \(\theta_2\)) into the formula and calculate \(n_2\) for each data point: \(n_2(10, 5) = \frac{1 \times \sin(10°)}{\sin(5°)} \approx 1.92\) \(n_2(20, 12.5) = \frac{1 \times \sin(20°)}{\sin(12.5°)} \approx 1.56\) \(n_2(30, 18) = \frac{1 \times \sin(30°)}{\sin(18°)} \approx 1.53\) \(n_2(40, 24) = \frac{1 \times \sin(40°)}{\sin(24°)} \approx 1.56\) \(n_2(50, 28.6) = \frac{1 \times \sin(50°)}{\sin(28.6°)} \approx 1.60\) \(n_2(60, 33) = \frac{1 \times \sin(60°)}{\sin(33°)} \approx 1.72\) \(n_2(70, 36) = \frac{1 \times \sin(70°)}{\sin(36°)} \approx 1.78\) \(n_2(80, 40) = \frac{1 \times \sin(80°)}{\sin(40°)} \approx 1.84\)

03

Use the least-squares method to find the best-fit line

Now we will use the least-squares method to find the best-fit line that connects the data points. This method minimizes the sum of squared residuals and provides a linear approximation of the relationship between \(\sin \theta_1\) and \(\sin \theta_2\). We can rewrite Snell's law as \(y = mx\), where \(y = \sin \theta_1\), \(x = \sin \theta_2\), and the slope \(m = n_2\). In this case, we want to find the slope of the best-fit line that connects the data points. We will have to graph the data and perform the least-squares method calculations, which should give us the slope of the best-fit line \(m\), which represents the approximate index of refraction \(n_2\).

04

Calculate the approximate index of refraction

After performing the least-squares method to find the best-fit line, we calculate the slope of the line to be approximately \(m = 1.6\). Therefore, the approximate index of refraction \(n_2\) is \(1.6\).

05

Identify bad measurements

Comparing the indices of refraction calculated using Snell's law for each data point to the approximate index of refraction found using the least-squares method, we can identify the bad measurements. Data points with a significantly different index of refraction from the approximate value (1.6) are considered bad measurements. Bad measurements: - \((10, 5)\) with \(n_2 \approx 1.92\) - \((60, 33)\) with \(n_2 \approx 1.72\) - \((70, 36)\) with \(n_2 \approx 1.78\) - \((80, 40)\) with \(n_2 \approx 1.84\) In conclusion, the approximate index of refraction of the unknown material is \(1.6\). The bad measurements in the dataset are the points \((10, 5)\), \((60, 33)\), \((70, 36)\), and \((80, 40)\).

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