Question

hello there again!

you guys were very helpful and made my last question clear, so thank you for that :)

I have one more question, if someone could patiently read through.

Currently, in this lab I am trying to find a relationship between the angle of incidence and the angle of transmission.

My professor is using one of these:

70 DO 80 70 SO PASCO sentific COMPONENT 60 RAY TABLE DEGREE SCALE - NORM CYLINDRICAL LENS NORMAL or COMPONENT 08 od 09

my positive values for the angle of incidence and its corresponding angle of transmission.

my negative values the angle of incidence and its corresponding angle of transmission.

My professor told me that by looking at my data I might be able to figure out the relationship between the two, however, this may not help me work out what the relationship is specificaly.

He suggests making a plot: angle of transmission (y-axis) vs. angle of incidence (x-axis).

He states when graphed there should be one line total (my positive set of data in quadrant one, and my negative set of data in quadrant 3).

He states that a curve will likely be observed (being linear or being non-linear).

He said that if the curve is straight to find the equation of the line (which would be the realtionship).

however,

if the curve is not straight, he told us to turn the line into a straight line using the following "tricks" I have pictured below.

Once we find the correct "trick" we have to work out that equation (which is the realtionship).

* The relationship must exist between the angle of incidence and the angle of transmission. Not the relationship between x and y.

I appreciate whomever helps, thank you very much.

Answer 1

First, let us make a combined table of positive and negative values of angle of incidence and angle of transmission

Angle of incidence(θi )	Angle of transmission (θt)
80	41.5
70	40
60	35.5
50	30
40	25
30	20
20	13.5
10	6.5
0	0
-10	-6
-20	-13
-30	-20
-40	-25
-50	-31
-60	-35
-70	-40
-80	-41.5

Now as suggested by the professor let us first plot the graph of angle of incidence on x axis vs angle of transmission on y axis

By doing that we get the graph as follows:

O, Vs ; 60 y = 0.5729x + 0.0294 40 20 0 o -100 -80 -60 -40 -20 20 40 60 80 100 -20 -40 -60

From the above graph, we can see that the line is not a straight line but is a curve.We have also tried linear fitting on the line and the equation is displayed on the chart. Here y is angle of transmission and x is angle of incidence. As the line is not exact straight, we try other methods mentioned in the problem and check if we obtain a straight line.

1) Plotting y² Vs x, that is (θ_T)² on vertical axis and θ_i on horizontal axis

Here from the values of y, we get the hint that on squaring the terms all values will become positive and there will be no line in third quadrant. So this plot will not give the required relation between the two angles. The graph we obtain is as follows:

0,2 Vs ; 2000 1800 012 1600 1400 1200 1000 800 600 400 200 -100 -80 -60 -40 -20 0 20 40 60 80 100

Which is indeed not a straight line

2) plot of ln y Vs x, that is ln θ_T on y axis and θ_i on x axis

The natural log of negative numbers is not possible so we can't take natural log of angle of incidence for negative values. Thus this plot does not give the required relationship between the two angles. The plot we get is as follows:

In @ Vs 4 3.5 In 3 2.5 2 1.5 1 0.5 e -100 -80 -60 -40 -20 0 20 40 60 80 100 -0.5

3) Plot of e^y Vs x, that is e^θ_TVs θ_i

Here taking exponential of the values of angle of transmission results in all positive values on the vertical axis. So there will be no straight line extending from the negative quadrant. Thus this plot will not provide the required relation between the two angles. The plot we get is as follows:

ey Vs ; 1.2E+18 ey 1E+18 8E+17 6E+17 4E+17 2E+17 -100 -50 50 100 -2E+17

4) Plot of sin x Vs sin y. This gives the plot as follows:

sin e Vs sin 1 sin 01 y = 0.8849x - 1017 0.8 0.6 0.4 0.2 sin 0; o -1.5 -1 -0.5 0.5 1 1.5 -0.2 -0.4 -0.6 -0.8 -1

Here we can see that the equation is almost a straight line. And the equation of linear fitting is y=0.8849x-10¹⁷. So we get the relationship between angle of incidence and angle of transmission as sin(θ_T)=0.8849sin(θ_i) - 10¹⁷

5) Plot of 1/y Vs x. Here we observe that while plotting 1/y that 1/angle of transmission, the value where angle of transmission is 0 will give a not defined value and hence this plot does not provide the required relation between the two angles

The plot obtained is as follows:

1/0r Vs 0. 0.2 1/07 0.15 0.1 0.05 0. 2. 6 10 12. 14 16. 18 20 -0.05 -0.1 -0.15 -0.2

6) Plot of 1/y Vs 1/x. Here again while taking inverse of both the angles at their value equal to 0, we get a value equal to not defined, due to which the line is discontinuous and hence this plot does not provide us with the required relation

The plot obtained is as follows:

1/0 Vs 1/0 0.2 1/0 0.15 0.1 0.05 1/0 -0.15 -0.1 -0.05 0.05 0.1 0.15 -0.1 -0.15 -0.2

from above we see that plot number 4 is the best fitting and provides us with the reqired relation