UMass Amherst PHYSICS 132 - Lab Report 2-Physics 132 (1) (4 pages)
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Lab Report 2-Physics 132 (1)
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- School:
- University of Massachusetts Amherst
- Course:
- Physics 132 - Intro Physics II
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Unformatted text preview:
Thursdays 9 00 AM ILC S110 July 25 2017 Lenses and Geometric Optics Lab Abstract In this experiment we saw the difference in the properties of light with converging and diverging lenses We applied Snell s Law to a laser beam in order to calculate and compare the refractive index to the published value Additionally we used converging and diverging lenses to investigate the characteristics of real and virtual images respectively Questions Answers 1 Comment on how the measured angle of refraction would have changed if you were to use diamond ndiamond 1 6 instead of D shape lens as the medium for Part 2 2 Would the angle of refraction be larger smaller or stay the same Explain 2 point The angle of refraction would be much smaller because the combination of diamond and natural air provides one of the largest differences in the index of refraction values A small critical angle forms therefore causing most rays to enter the diamond at angle of incidence greater than the critical angle 2 A 1 75 m tall man is 4 0 m from the converging lens of a camera His image appears on a piece of film that is 40 mm behind the lens How tall is his image on the film 2 point Using the equation h i h o d i d o Height of image height of object is equal to the negative distance of the image distance from the object with the numbers from the problem we know d i 004m 40 mm converted to m d o 4m and d o 1 75 m So the equation can become h i 1 75 0040 4 so h i 0006m so the height of his image on film is 0006m 3 Use your data to verify the Law of Reflection and then use Snell s Law to calculate the refractive index for D shape lens lens Compare it with the known value of 1 52 2 points Calculations Law of Reflection 1 2 Law of Refraction n1sin 1 n2sin 2 Tank 1 Known n1 1 1 30 2 18 1sin 30 n 2 sin 18 n2 1 61 Tank 2 Known n1 1 1 20 2 12 1sin 20 n 2 sin 12 n2 1 64 The refraction index we got was 1 61 and 1 64 which is fairly close to 1 52 our expected value The known value for the refractive index is 1
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Study Guide: Midterm 2
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