### Optics Tutorial 6 HW solutions

 1.A single refractive surface is used to image an object. The object is 100 mm away and is 7.5 mm tall. The refractive surface has a 100 mm convex radius of curvature and an index of 3.2. Use a YNU ray trace to determine the image location and height.  – Extra credit: Use the lens makers equations to check image location and similar triangles to check your image height. Solution: Using the blank spreadsheet provided at the homework page and a solver to find the final image distance, the height is 6.25 mm Spreadsheet below Thin lens solution: Have to find image distance using: n'/s' = ((n'-1)/R) + (1/s), solve for s' s' = n'[((n'-1)/R) + (1/s)]-1 = 3.2[((3.2-1)/100) + (1/{-100})]-1= 3.2[((2.2)/100) - (1/100)]-1= 3.2[(1.2)/100]-1= 3.2*100/1.2 = 266.7mmThe chief ray will refract at the lens. Since the chief ray height at the lens is zero, the surfaces optical power does not come into the equation.To first order u = n' * u', and we don't need full snell's law. The refracted angle is simply u' = u / n'.The input chief ray angle = object_height/object_distance or u = h / sOr u' = h / (s*n')Finally the object_height = u' * image_distance orObject_height = h * image_distance / (s * n') = 7.5 * 100 / (266.7 * 3.2)Image_height = 6.25 mm Zemax Solution (attached below) shows an image of ~10 and the errors are likely due to the small angle approximation 2.A biconvex singlet is used to visual observe a image. Conduct a YNU ray trace for the eye’s fovea. Assume:  – Eye has a pupil diameter of 5 mm – Eye relief of 25 (eye is 25 mm from lens) – Edmund lens #32624 • Radius of 50.8 mm • Lens thickness of 5.0 mm • N-BK7 (index of 1.5168 at 587 nm) Fovea has a 1° half angle (chief ray angle Solution: Spreadsheet belowZemax model below.
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Scott Sparrold,
Mar 7, 2013, 11:25 AM
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Scott Sparrold,
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Scott Sparrold,
Mar 7, 2013, 11:30 AM
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Scott Sparrold,
Mar 7, 2013, 1:04 PM
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Problem 1.ZMX
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Scott Sparrold,
Mar 7, 2013, 1:05 PM
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Problem 2.ZMX
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Scott Sparrold,
Mar 7, 2013, 1:05 PM