Lab 10_ Reflection and refraction

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Temple University College of Science and Technology Physics Department Physics 1022 Section 052 Lab 10 Reflection, Refraction, and Lenses

Lab 10: Reflection, Refraction, and Lenses 11/9/2023 Group Members Group 65: Cynthia Klingensmith, Helen Luong Goals The objective of this experiment is to understand the ray model of light. We will be practicing and interpreting calculations using the Law of Refraction also known as Snell’s Law. We will also be representing Snell’s Law graphically. Through experimentation, we’ll observe the behavior of light in terms of reflection and refraction. Additionally, the lab will cover the application of the thin lens equation and explore the use of lenses in various optical devices. Procedure Part 1: The equipment was set up so the slit plate and the slit mask were aligned so that a single ray of light passes along the normal line of the ray table. The cylindrical lens was also aligned so that the lines will be perpendicular to each other. The ray table was rotated and the angles of incidence were observed from 10 to 80 degrees, in increments of 10 degrees. The angle of refraction was also recorded. The ray table was next rotated 180 degrees and the procedure was repeated for this as the angles of incidence were observed from 10 to 80, in increments of 10 degrees. Both data taken were graphed and the slope was estimated. Part 2: The focal length mirror was taken and held about 10 cm in front of the light source and held at an angle to observe any light in the paper. The object distance was next increased by 50 cm by moving the mirror away and the observation of how the image distance and size changes was recorded. Part 3: The equipment was set up to use the 75 mm lens and the light source was placed so that it lined up to the 10 cm mark on the optics bench. Then the lens was next placed at the 40 cm mark to allow more space on each side of the lens and the object. We started observing the object at a point between the 75 mm and 150 mm mark from the lens. The observation for this was recorded. Error and Precaution One precaution for the lab is correctly drawing the light rays on the piece of paper. I experienced difficulty connecting the rays in one spot by using a ruler. Another source of error is the placement of the viewing screen. Results

Regular: i r1 sin i sin r1 0 0 0 0 10 8 0.173 0.984 20 15 0.342 0.258 30 20 0.5 0.342 40 26 0.643 0.438 50 32 0.766 0.529 60 36 0.866 0.528 70 40 0.939 0.642 80 42 0.984 0.669 Graph: Calculated Slope N2 = 1.49 1 / N2 = 0.67

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Actual = 0.6786 Percent Error 0.6786 - 0.67 / 0.67 = 1.2% error Turned 180: i r2 sin i sin r2 0 0 0 0 10 17 0.173 0.292 20 32 0.342 0.529 30 50 0.5 0.766 40 77 0.643 0.974 50 47 0.766 0.713 60 56 0.866 0.829 70 66 0.939 0.913 80 75 0.984 0.966 Graph:

c = sin^-1 [1/1.49] c = 0.73 Percent Error N1 = 1 1 / N1 = 1 Actual = 0.8672 Percent Error 0.8672 - 1 / 1 = 13 % error Questions 1. Was all the light of the ray refracted? Was some reflected? How might you have used the Law of Reflection to test the alignment of the cylindrical lens? Yes, all the light was refracted while none of the light was reflected. The Law of Reflection could be useful when calculating the angle of reflection. 2. Is the ray bent when it passes into the lens perpendicular to the flat surface of the lens? Explain.

No, when the ray enters a lens perpendicularly to the flat surface, it does not bend. The ray will not appear bent because the incident ray lies along the same line as every other ray. 3. Is the ray bent when it passes out of the lens perpendicular to the curved surface of the lens? Explain. Yes, when a ray exits a lens at a point where the surface is curved, it bends. This bending occurs because the curvature of the lens causes the light to refract at the same angle at which it entered but in a perpendicular direction relative to the lens’s surface. 4. Where is the paper when the image is the sharpest? The image is the sharpest when it is closer to the mirror. Moving it farther away distorts the reflection and blurs the light. 5. Considering the light bulb as the object is your result approximately consistent with the mirror equation 1/d + 1/dᵢ = 1/ƒ(where d is the object distance = 10 cm, dᵢ is the image ₀ ₀ distance, and ƒ is the focal length of the mirror)? Considering the focal length of the mirror is 50 mm and d is 10 cm. We can use the ₀ equation 1/d + 1/dᵢ = 1/ƒ where 10+ 1/dᵢ = 20. 1/dᵢ is 10 which means dᵢ is 1/10 o4 .1 m ₀ or 10 cm. This means the image is sharpest when d = dᵢ ₀ 6. How does the value of di change when you do this? Is this consistent with the mirror equation? Is the approximate size of the new image consistent with the magnification equation m= -di/do? When the distance of the object increases, both d and 1/d increase. To satisfy the ₀ ₀ equation where 1/ƒ = 1/d + 1/dᵢ, dᵢ needs to decrease to keep the value of 1/ƒ. As dᵢ ₀ decreases, 1/dᵢ increases which maintains the value of 1/ƒ. For the magnification, when dᵢ decreases and d increases consequently the magnification decreases which is what we ₀ observed as the distance of the object decreases. 7. Using the thin lens equation, what should the image distance and magnification be if the object distance is 2ƒ? Should the image be inverted? (If the magnification from equation 3 is negative, the image is inverted). If the object distance is 2 f the image distance will be -1/3 and this is because of the following; 1/d i - 1 / 2 f = 1/ f 1/d i = 3 f /2

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d i = 2 f /3 2 f / 3 / 2 f = -1/3 As a result, the image shown will be inverted. 8. If dᵢ = + ∞ what does the thin lens equation simplify to? (This gives us an idea of what d₀ should be approximately equal to.) When d = +∞, then 1/d in the thin line equation becomes 0, so the equation is ᵢ ᵢ simplified to 1/ƒ = 1/d₀. This implies that d₀ should be approximately equal to the focal length ƒ, since the reciprocal of a finite number is another finite number. When the image distance is at infinity, the object is placed at the focal point of the lens. 9. Use the thin lens equation to determine the smallest possible value of d that keeps dᵢ ₀ positive. (Hint: try entering dummy values of d into the thin lens equation such as 1.1ƒ ₀ and 0.9ƒ). For an object distance d = 1.1 ƒ, the image distance dᵢ is positive at 11ƒ. Additionally, ₀ for an object distance d = 0.9 ƒ, the dᵢ is negative at approximately -9ƒ, This means that ₀ the smallest possible value of d that keeps dᵢ a positive value is slightly more than the ₀ focal length ƒ. Any object distance equal to or less than the focal length would give a negative image distance, which corresponds to a virtual image for a converging lens. d ₀ must be greater than ƒ to maintain a positive dᵢ. Discussion The ray model of light was observed and using Snell’s Law, we were able to find how that applies to the data we collected in our set up. We calculated the slope of the normal light refraction to be 0.67, and the slope we got from the experiment was 0.6786 which was a 1.2% error. We also calculated the slope of the 180 turned light refraction and we got a slope of 0.86 which was a 13% error. This error may have occurred because the ray might not have been aligned correctly which did affect the results with our numbers. There were also other light sources in the environment such as light coming from the windows and door so that may not have been dark enough to have an accurate reading. The readings we did get in our experiment were accurate and followed Snell’s Law and also allowed us to observe the lens equation. It also helped us observe how light rays bend. We are also able to observe that the distance from the object to the mirror is the same compared to the distance from the mirror to the image.

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