UMass Amherst PHYSICS 132 - COMPL Physics 132 Lab 3 - Human Eye (6 pages)

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COMPL Physics 132 Lab 3 - Human Eye



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COMPL Physics 132 Lab 3 - Human Eye

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Pages:
6
School:
University of Massachusetts Amherst
Course:
Physics 132 - Intro Physics II
Intro Physics II Documents
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Physics 132 Lab 3 Geometric Optics and the Human Eye Reflection and Refraction 1 2 Repeat this procedure for a total of 6 sets of angles Record your measurements in the spreadsheet Incident Reflection Refraction 10 11 7 15 15 5 10 5 20 21 13 75 25 25 15 5 30 31 19 35 36 5 22 5 40 41 5 25 5 45 46 28 Measure the size of the image on the screen with a ruler and record in the spreadsheet Lenses Image Object Distance Distance 3 4 O 1 do I 1 di Image Size cm 20 19 0 05 0 05263157895 4 19 19 5 0 05263157895 0 05128205128 4 3 18 20 5 0 05555555556 0 0487804878 4 8 17 21 7 0 05882352941 0 04608294931 5 3 16 23 9 0 0625 0 04184100418 6 3 15 26 7 0 06666666667 0 03745318352 7 8 14 31 9 0 07142857143 0 03134796238 10 13 37 5 0 07692307692 0 02666666667 12 7 12 48 5 0 08333333333 0 0206185567 17 4 11 77 5 0 09090909091 0 01290322581 29 2 Place a piece of paper over half the lens What happens to the image The obscured half of the image is out of focus but still visible It appears as if the left half is obscured when in fact the right half is If the top half of the object is covered it appears that the bottom half of the image is covered This is because the image is inverted Is the image upside down or right side up The image is upside down because the converging lens refracts the rays so that the image is inverted 5 6 Repeat this procedure of measuring object and image distance in spreadsheet Move the lens closer to the light box in 1 cm steps until you can no longer form an image on the screen What is the distance between the object light box and the lens when you can no longer form an image on the screen At 8 cm the image no longer appears on the screen The image distance would be 48 cm The image would be virtual because the object is less than the focal length away from the lense meaning the light rays would converge on the same side of the lense as the object Law of Reflection 7 Make a graph of the angle of reflection vs angle of incident What pattern would you describe between the two angles It seems that the angle of refraction is always a little larger than the angle of incident Though the ratio appears to be basically 1 1 8 What are the parameters slope and intercept predicted in the relationship between and The y intercept of this chart is 20 We selected a linear plot for our data points in Google Sheets It automatically calculated the most accurate measure of slope to be 0 981 Law of Refraction 9 Make a graph of the angle of refraction vs angle of incident How would you describe the pattern between the two angles W e again selected a linear scatter plot this time with the slope being 3 The y intercept is 7 10 In the spreadsheet for each value of and and Incident Rad 0 1745329252 calculate Reflection Rad Refraction Rad Sin Inc Radians Sin Refrac Rad Index of refraction 0 1919862177 0 1221730476 0 1736481777 0 1218693434 1 424871693 0 2617993878 0 2705260341 0 1832595715 0 2588190451 0 1822355255 1 420244732 0 3490658504 0 3665191429 0 2399827721 0 3420201433 0 2376858923 1 438958535 0 436332313 0 436332313 0 2705260341 0 4226182617 0 2672383761 1 581428042 0 5235987756 0 5410520681 0 3316125579 0 5 0 3255681545 1 535776743 0 6108652382 0 637045177 0 3926990817 0 5735764364 0 3826834324 1 498827459 0 6981317008 0 7243116396 0 4450589593 0 6427876097 0 4305110968 1 493080235 0 7853981634 0 8028514559 0 4886921906 0 7071067812 0 4694715628 1 506175959 11 Make a graph of sine of the angles vs How would you describe the pattern between the 12 What are the parameters slope and intercept predicted in the relationship between sine of the angles Go back and look at the equation of law of refraction to help you IMPORTANT assume the index of refraction for air is 1 Slope 1 54 Y intercept 0 012 13 What is the index of refraction of the D shape lens Use the function LINEST to find the value of the index of refraction of the D shape lens 1 487 is the average value of index of refraction The slope of sin incident vs sin refracted is 1 54 meaning that the index of refraction of the lens is 1 54 The index of refraction in air is about 1 Based on the average index of refraction the error is about 4 Thin Lens Equation 14 Make a graph of vs How would you describe the pattern between the two distances The trend in the data does not look linear but rather parabolic indicating that the relationship between image distance and object distance is not linear The line represents the average slope and aids in demonstrating the function itself is not linear 15 In the spreadsheet for each value of O 1 do and calculate and I 1 di 0 05 0 05263157895 0 05263157895 0 05128205128 0 05555555556 0 0487804878 0 05882352941 0 04608294931 0 0625 0 04184100418 0 06666666667 0 03745318352 0 07142857143 0 03134796238 0 07692307692 0 02666666667 0 08333333333 0 0206185567 0 09090909091 0 01290322581 16 Make a graph of vs How would you describe the the pattern between 17 What are the parameters slope and intercept predicted in the relationship between 0 995 y intercept 0 104 Slope 18 What is the focal length of your lens Use the function LINEST to find the value of the focal length 9 63291864229223 What is the error in the value of the focal length The focal length of the lens is 9 63 cm The error is about 0 07 Part II Vision Correction For nearsighted myopic people the image forms in front of the retina For farsighted people hyperopic people the image forms behind the retina The shorter the focal length the higher the focal power In the case of nearsightedness the lens of the eye is too strongly focusing high power so the corrective lens need to decrease the eye s focus For farsightedness the lens is too weakly focusing low power the corrective lens needs to increase the eye s focus Converging lenses have a positive focal length and therefore a positive focal power Diverging lenses have negative focal lengths and negative focal power Lightbox cm 0 Lens cm 70 Screen cm 90 19 Using the focal length of the lens and the object distance calculate the image distance Where is the image of your model eye located relative to the screen retina The image forms in front of the screen The image distance is 11 66 cm 20 Is your model of the eye nearsighted myopic or farsighted hyperopic T he model of the eye is nearsighted myopic The focal power is 21 Calculate …


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