# 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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## Lab Report 2-Physics 132 (1)

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4
School:
University of Massachusetts Amherst
Course:
Physics 132 - Intro Physics II
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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 52 When using Snell s law to calculate the refractive index in tank 1 and 3 we see that a value of 1 61 and 1 64 which are both not that different from the expected value 4 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 image and 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 2 4 Real Images in your manual 2 points The equation we used was 1 s 1 s 1 f The focal lengths were 94 3mm 100mm 98 5mm The ratios were found by s s and h h 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 from ourselves Ratio 1 s s 3 67 h h 3 75 Close Ratio 2 s s 2 0 h h 2 125 Close Ratio 3 s s 4 58 h h 4 25 Close 5 Describe the virtual image seen in Section 2 4 Virtual Image observing an object through a converging lens Is the image larger smaller than the object Is the image upright 2 point The virtual image produced was larger than the given object the target and was upright 6 Use s 15 cm and f 10 cm to calculate s Show explain why it is not possible to get a real image using your diagram and the equations above Section 2 5 Diverging Lens 2 point 1 f 1 s 1 s 1 10 1 15 1 s s 5 99cm It is impossible to have a real image because the image will never be projected It will always have a negative s and because of this it will always be visualized on the same side of the lens mirror as the object 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 2 points 1 f combined 1 f C 1 f D 1 183 1 98 0 1 fD 1 fD 1 183 1 98 0 1 fD 00474 1 00474 211 This experiment would not be possible if fc fd When using both the converging lens and the diverging lens to view the image the image that is produced is a virtual image instead of a real image 8 You have a lens with a focal length of 25 cm You need to combine this lens with another in contact to make a combination lens with a focal length of 10 cm What focal length should you choose for the second lens 2 points 1 s 1 s 1 f 1 10 1 25 1 f 1 10 1 25 1 f 06 1 f f 16 6cm The focal length of the second lens should be 16 6cm Conclusion This experiment let us look at the differences in light appearance and how different lenses distances etc could affect the image produced onto the screen We were able to see how focal length is related to lens length and how to use the law of refraction reflection and Snell s Law

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