Using converging and diverging lenses to understand optical propertiesAtreyi SahaPhysics 132 - Section LLMondays 8:00 AMHasbrouck 212January 25, 2016AbstractIn this experiment, we investigated the properties of light with converging and diverging lenses. We specifically focused on the ray nature of light and used geometry to measure such properties. We applied Snell's Law to a laser beam incident on the surface of water in order to calculate and compare the refractive index of water to the published value. Additionally, we used converging and diverging lenses to investigate the characteristics of real and virtual images, respectively. Questions & Answers1. Use your data to verify the Law of Reflection and then use Snell's Law to calculate the refractive index for water (nw) . Compare it with the known value given in the manual. (1 point) CalculationsLaws of reflection: θ1 = θ2Laws of refraction: nair sin θ1 = nwater sin θ3Tank 1 (1.00029) (sin 45) = nwater (sin 35) nwater = 1.233Tank 2 (1.00029) (sin 55) = nwater (sin 16) nwater = 2.97The known value for the refractive index of water is 1.33. When using Snell's law to calculate the refractive index for water in tank 1, we see that a value of 1.233 is very close to the actual value with only a 7.87% error. In contrast, when using Snell's law to calculate the refractive index for water in tank 2, we see that a value of 2.97 is not close to the actual value with a 55% error. 2. What is the focal length of the lens used in section 1.3 “Focal Length” of your manual? Can you deduce from the image seen, if the lens used was converging or diverging? (2 point) The focal length of the lens is +100mm. From the image seen, we can deduce that the lens is a converging one for two reasons. Firstly, the focal length associated with the lens is a positive value which is characteristic of converging lenses. From our measurements, we can deduce that the object was located between 2F and F (F = focal length). In this case, the object is located in front of the 2F point and the actual image is located beyond the 2F point which is confirmed by our results. The image dimensions appear to be larger. We also know that this is a real image because light rays come together at the location of the image.3. Calculate the focal length of the given converging lens based on the measurements of s and s' and thin lens equation. Tabulate your results (for all the three trials). Compare ratios of the imageand object distances and heights for the three object distances to show −(s' /s)=(h' / h) is true. Draw a ray diagram and show the different lengths for one trial. Section 1.3 “Real Images” in your manual. (2 points)Focal length for three trialsfocal length f(mm)111106105Comparing ratios - while the ratio of the image and object distances appear to be close, the relationship −(s' /s)=(h' / h), is not precisely true from the values used from our data set. This is probably due to errors in measurement.Proof, s'/sProof,h'/h−(s' /s)=(h' / h)true?-1.58333333 1.65 Close-1.26315789 1.425 Close-1.38888889 1.625 CloseRay Diagram - Trial 1 *Note: Bolded Arrows are the Light Raysh = 40mmh' = 66mmFFs = 180 mms' = 180 mmf = 111 mmf = 111 mmObjectReal image4. Use the parallax method described in the lab manual to determine the focal length of the lens. Compare this length to the previously calculated values. (1 point) object s(mm)image s'(mm)actual focal length f(mm)experimental focal lengthf (mm) (1/s) + (1/s')250 -93.75 -150 -150 -0.006666667Using the measured values, we found that the experimentally determined focal length matched the actual focal length written on the lens. In this case, the focal length is negative because the height of an inverted image is always negative. According to convention, a diverging lens has a negative sign. This is in contrast with our other focal length value of +100 mm from the converging lens. 5. Describe the virtual image seen in Section 1.3 “Virtual Image” (observing an object through a converging lens). Is the image larger/smaller than the object? Is the image upright? (2 point) In this case, the object was located in front of the focal point (inside the focal length) and therefore the image is located behind the object. The image was upright and enlarged. In this case, upon refraction, the individual light rays diverge. The point where the rays intersect is where the virtual image is located. 6. Use: s=15cm and f =−10cm to calculate s' . Show / explain why it is not possible to get a real image using your diagram and the equations above. Section 1.4 “Diverging Lens”. (1 point) (1/s) + (1/s') = 1/f(1/15) + (1/s') = 1/-10(1/s') = -0.1 - 0.0667s' = -5.998 = -6cmWe know that the lens is a diverging one because the focal length of a diverging lens is always negative by convention. Diverging lenses always generate virtual images, not real images. This isbecause the light rays that are being refracted are extended to the back to meet. When the image distance is positive, we get a real image and the focal length of the lens has a positive value. When the image distance is negative, we get a virtual image and the focal length is negative as well. In this case, the image will appear shrunken and upright. According to the equation above, since the focal length is negative, and the solution for the image distance will also be negative as confirmed by our result.7. Calculate a value for f D (the focal length of the diverging lens) using the method described in Section 1.4 “Lens Combinations” of the lab use the lenses as one lens, and thin lens equation to find f combined then use 1/ f combined=1/ f C +1 / f D to find f D . Would this experiment be possible if f C > ∣ f D ∣ ? (1 points) Fcombined = (1)/((1/s)+(1/s'))Trial F combine (mm)1 189.86842112 189.86842113 188.8157895Average of Fcombined = 189.52 mm meaning this is a converging lens overall (1/189.52) = (1/107.3) + (1/ fD)0.005276 = 0.00932 + (1/fD)876tvbn -0.004044 = 1/fDfD = -247 mmThe result is negative and smaller than the value of the converging lens. This experiment would not be possible if f C >∣f D ∣. In this case, the fcombined would be a negative value which corresponds to a diverging lens and there would be no image on the screen. For example, if fC = 200 mm and fD = -100 mm, then using the formula 1/fc + 1/fd = 1/fcombined would give us a negative value for fcombined corresponding to a diverging
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