Lab Report #5 (Exp. 15 Geometric Optics)

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Jan 9, 2024

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Report for Experiment #15 Geometric Optics Vaughn Montoya Lab Partner: Ricardo Macrron TA: Ji Tae 3/24/2023

Introduction Lenses that date back to 1000 years ago, were the first known lenses discovered by Viking ruins on the island of Gotland. In today’s world we live with optical devices such as CD, DVD, and blue ray players. Geometric optics refers to how light travels and relates to lenses and mirrors. This was the point of this lab to see how light travels through items we interact with and see every day such as glasses and mirrors. Under geometric optics it is assumed that light travels in a direct line which are known as rays. We used this to look at refraction and reflection. There are different types of mirrors, plane mirrors, spherical, and paraboloidal mirrors. These are used to investigate the concept of reflection. Plane mirrors in which a ray is reflected will be the same reflected back and the normal force will be perpendicular with the incidence and reflection angles being equal. Refraction is how light is altered upon entry into a transparent material. When looking at refraction we can use two formulas, the index fraction which is ratio of speed of light ( 3.00 x 10^8 m/s) which is highlighted in the following formula: n = c v Then the second equation is known as Snell’s law or the law of refraction which can be seen in the equation below: ࠵? ! ࠵?࠵?࠵?࠵? ! = ࠵? " ࠵?࠵?࠵?࠵? " Snell’s law allows light to move from a material to another and the equation above is used to manipulate and solve for the respective indices of fraction. Lenses allow for the transmission of light rays. When it comes to lenses in this lab, we looked at two, converging and diverging lenses. Converging rays refracted to a single focal point and diverging rays caused the rays to never cross paths. This lab looked at parallel light ray reflection on plane and spherical mirrors. We had to find the refractive index of a rhombic prism. Then we had to investigate refraction of parallel rays in convex and concave lens. The focal length had to be obtained using a thin lens using a form of bench optics. During Investigation 1, we examined the principles of ray tracing for concave and convex mirrors and lenses using both reflection and refraction. Then in Investigation 2 light was refracted through an acrylic trapezoid which looked at how the rays of light entered the trapezoid and exited the trapezoid at specific angles. In Investigation 3 focal length of a thin lens was looked at using an optic rail bench. The image was reflected on a metal screen through a lens and the dimeter of the refracted and original image were computed.

Investigation 1 The setup for Investigation 1 consisted of ray tracing with a concave mirror we used a triangular mirror that had both concave and convex sides during the Investigation along with using a PASCO light source. The room lights were switched off and the light source was adjusted to so that five rays came out. Our TA then gave out pieces of white paper so then we could trace it on the paper. The mirror and lenses, concave and convex were used and the focal points were determined. The table below highlights this: Focal Length, f (m) δf (m) do (m) Nominal Value (m) δNominal Value (m) Concave Mirror 0.060 0.005 0.158 0.063 0.003 Convex Mirror 0.068 0.015 0.166 0.063 0.003 Concave Lens 0.114 0.01 0.180 0.129 0.005 Convex Lens 0.142 0.15 0.14 0.129 0.005 We measured the focal length of concave/convex mirrors and lenses which were determined by measuring the distance between the focal point and the point where the incident rays converged or appeared. The error of the focal length was simply half the smallest increment. The focal length for the concave mirror was 0.060 m ± 0.005 m and the convex was 0.068 m ± 0.015 m. These values were compared to the actual, nominal values of the convex and concave mirror, 0.0630 m ± 0.003 m. The experiential values obtained were within range. Then the lens focal points were 0.114 m ± 0.01 m for the concave lens and 0.142 m ± 0.15m for the convex lens. The nominal value given by the manufacturer was 0.129 m ± 0.005 m and the experimental values are within error.

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(Figure 1: Diagrams of Investigation 1) Investigation 2 The setup for this Investigation consisted of getting an acrylic trapezoid from the kit and placing it in front of the PASCO light source. IPL manual said to rotate the ray until we got an angle between 35 ° and 70 ° my TA just wanted us to get four different angles. The incident ray was then traced exiting the trapezoid and connected via ruler. Then the refracted and incident angle were measured with respect to the normal force utilizing a protractor. The table below highlights the data gathered for this Investigation: θ incident (degrees) θ refracted (degrees) Index of refraction (n) δn average n δn avg 33.0 22.0 1.454 0.03698 1.429 0.01383 38.0 28.0 1.311 0.02603 50.0 32.0 1.446 0.02280 55.0 33.0 1.504 0.02220 Snell’s law was used to calculate the Index of refraction which was the equation used in the introduction where we solved for n and kept in mind to convert to radians while solving. Then the error of n was found by using the following equation: ࠵?࠵? = ࠵? ࠵?࠵? ࠵?ℎ࠵? ࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?ℎ࠵? ࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? √࠵? In the denominator, N refers to the 4 because that’s how many angles we measured and in the numerator that standard deviation of the values was taken. Then the average was found by adding up the index of refraction values and dividing them by four which can be seen below: ࠵?࠵?࠵? ࠵? = ࠵? ! + ࠵? " + ࠵? # + ࠵? $ 4 Then the error of n average was found by using the following equation: ࠵?࠵? %&’ = @(࠵?࠵? ! ) " + (࠵?࠵? " ) " (࠵?࠵? # ) " (࠵?࠵? $ ) " 4 Where each δn was taken and 1 through 4 and squared and divided by 4. We compared the average index of refraction to the expected value during the Investigation. The average index found was 1.429 ± 0.01383 and compared to the actual value of acrylic, 1.49 which does not fall within uncertainty. This error could have been because the light source wasn’t powerful enough which made it difficult to trace the incident and refracted rays. This could simply be improved by getting a stronger light source.

( Figure 2: Refraction through a trapezoid) Investigation 3 The setup for Investigation 3 consisted of reassembling the light source on the optic rail and adding the 100mm lens along with a metal screen. The size of the inner circle was observed and documented. Then the object was placed at the 10 cm mark on the rail and the metal screen was placed at 110 cm mark. The lens was inserted between the object and the metal screen by applying pressure to sides labeled a and b. The lens was then adjusted by sliding it until a clear, magnified image was sharply focused on the screen. The positions of everything that was setup in this experiment was noted along with the errors. This was repeated six times with the screen moving up 10 cm every time until it reached 60 cm. The table below highlights the data obtained: x Object (m) δx object (m) x Lens (m) δx lens (m) x Screen (m) δx Screen (m) do (m) δdo (m) 0.10 0.0005 0.218 0.0005 1.10 0.0005 0.12 0.00071 0.10 0.0005 0.220 0.0005 1.00 0.0005 0.12 0.00071 0.10 0.0005 0.224 0.0005 0.90 0.0005 0.12 0.00071 0.10 0.0005 0.228 0.0005 0.80 0.0005 0.13 0.00071 0.10 0.0005 0.234 0.0005 0.70 0.0005 0.13 0.00071 0.10 0.0005 0.246 0.0005 0.60 0.0005 0.15 0.00071 Then the rest of the data can be seen below: hi (m) δhi (m) ho (m) δho (m) di (m) δdi (m) dodi (m^2) δdodi (m^2) do +di (m) δdo+di (m)

0.151 0.00005 0.02 0.00005 0.88 0.000707 0.104 0.000629 1.00 0.001 0.133 0.00005 0.02 0.00005 0.78 0.000707 0.0936 0.000558 0.90 0.001 0.111 0.00005 0.02 0.00005 0.68 0.000707 0.0838 0.000486 0.80 0.001 0.090 0.00005 0.02 0.00005 0.57 0.000707 0.0732 0.000414 0.70 0.001 0.071 0.00005 0.02 0.00005 0.47 0.000707 0.0624 0.000343 0.60 0.001 0.049 0.00005 0.02 0.00005 0.35 0.000707 0.0517 0.000271 0.50 0.001 The diameter of the inner circle, ℎ ( was recorded and constant throughout the Investigation along with the error of ho was consistent throughout this Investigation. Then ࠵? ( was obtained by simply subtracting the x of the lens minus the x of the object. However, the error of ࠵? ( was obtained by using the following equation: ࠵?࠵? ( = @࠵?࠵? " + ࠵?࠵? " = √2 ࠵?࠵? = ࠵?࠵?࠵? Then the image/ lens distance was obtained by subtracting the image minus the lens. The same equation above was used to calculate the error of ࠵?࠵? . Then the hi, the inner circle diameter was obtained. Then the error of hi, which was used using a ruler was simply just half the smallest increment. Then ࠵? ( and ࠵? ) were multiplied by each other to get the respective product of image distance and object distance. Then the error was found by using the following equation: ࠵?(࠵? ( ࠵? ) ) = D E ࠵?࠵? ( ࠵? ( F " + E ࠵?࠵? ) ࠵? ) F " Then ࠵? ( + ࠵? ) was obtained by adding them together to get the sum of the image and object distance. The error of this was found by using the following equation: ࠵?(࠵? ( + ࠵? ) ) = G࠵?࠵? ( " + ࠵?࠵? ) " A graph was made using data of ࠵? ( ࠵? ) vs ࠵? ( + ࠵? ) to determine the focal length which can be seen below:

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From this we plugged the data into IPL straight line of fit calculator and the slope obtained was 0.1052 m ± 0.001070 m which does not fall within the expected focal length of the lens, 0.1 m. The slope of the graph should be a positive linear relationship which it is, but some potential sources of error could have been measuring the image and object distance incorrectly or just human error in conducting the experiment. Then M ( h , /h - ) vs ࠵? , ࠵? - was plotted and can be seen below: Using IPL straight line of fit to calculate the slope it was 0.1018 ± 0.006912. When comparing this with the magnification and that it should be 1, it was a little off, so the experimental value does not fall within the range of uncertainty. This could have been because of human error, not reading the correct measurements on the inner circle and the ruler. y = 0.1046x - 0.0003 0.000 0.020 0.040 0.060 0.080 0.100 0.120 0.00 0.20 0.40 0.60 0.80 1.00 1.20 dodi (m^2) do+di (m) dodi versus do+di y = 1.0157x + 0.0024 -8 -7 -6 -5 -4 -3 -2 -1 0 -8.00 -7.00 -6.00 -5.00 -4.00 -3.00 -2.00 -1.00 0.00 (hi/ho) di/do (hi/ho vs di/do)

Conclusion For Investigation 1 we looked at Rays and how they reflected off convex and concave mirrors and lenses. This was done by getting five light rays and out of the PASCO light source and tracing the reflected and refracted rays on a piece of paper. The focal length for the concave lens was 0.114 m ± 0.01 m and the cave were 0.142 m ± 0.15m. Both fell within the nominal range. Then the mirrors focal lengths were 0.060 m ± 0.005 m for the concave and 0.068 m ± 0.015 m for the convex. These also fell within the nominal range. For Investigation 2 we looked an acrylic trapezoid and how light entered and excited which then gave the refracted and incident angles when we traced everything out. We found four different angles and used Snell’s law to find the index of refraction. It was 1.429 ± 0.01383 which did not fall within the expected value of 1.49. This could have been because of human error in measuring the incident and refracted angles. Investigation 3 consisted of setting up the optic rail system and getting the lens and mirror set up. A graph of ࠵? ( ࠵? ) vs ࠵? ( + ࠵? ) was made which showed a positive linear relationship but the slope obtained from IPL, 0.1052 m ± 0.001070 m did not fall within error. This could have been because of incorrectly measuring the object and image distances or also just humor error. Then a graph of ( h , /h - ) vs ࠵? , ࠵? - was made to look at magnification which was supposed to be 1. The slope obtained was 0.1018 ± 0.00691which did not fall within error. Also, for this di/do was multiplied by negative 1 because if it wasn’t a negative slope value would have been obtained for instance, – 1.018. But some errors for this Investigation and just in general for this lab was the light sources power. If it was more powerful the incident and refracted rays would have been easier to trace causing the focal lengths to be more accurate. Also, a way to make the image more focused would be better and give more precise results for the last Investigation. Questions 1. The white screen displays the reflected image of the object which light is generated. This also is what happens when light passes through are retina which makes a reflected image. 2. To obtain a clear image on the screen, the object would need to be positioned 30 cm away from the screen, given that the screen was 15 cm away from the lens, Solving below: 1 ࠵? ( = 1 f - 1 ࠵? , = 1 10 - 1 15 = 30 cm Then the magnification would be 0.5, half the size of original image, 15/30= 0.5 3. If the ray deflected is 2.37 cm by an Acrylic, the thickness is as follows using Snell’s law to calculate: 1࠵?࠵?࠵?(50.5º) = (1.49) ࠵?࠵?࠵?࠵? " = 31 º Then,

t = 2.37 cos(31°) sin (19.5°) = 6.10 cm so, the thickness t of the Acrylic is 6.10 cm at 50.5 º. 4. Using the value we obtained, 1.429 can be plugged into the following equation to solve for speed of light in the Acrylic: ࠵? = c n = 3.0࠵?10 . 1.429 = 2.09 x 10 . m s 5. The brain flips the image even though the eyes might initially take in the image as inverted, so the brain makes the image upright.

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Lab Report #5 (Exp. 15 Geometric Optics)

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