Cornea/Refractive Surgery Quiz 22: Optics


Figure 1
An object is located 1/3 of a meter to the left of a +4 D lens. A +3 D lens is 2 meters to the right of the +4 D lens. 

Questions and Answers  
1. What formula is the most useful in determining the location of the final image? Answer: The vergence formula U + D = V is the most useful formula for solving this type of optics problem. U = object vergence = 1/object distance, D = power of the lens in diopters, V = image vergence = 1/image distance. 2. Where is the final image located in this lens system? Answer: 0.5 meters to the right of the +3 D lens. See diagram 3. What is the relative size of the final image compared to the object? Answer: The relative size of image1 to the object is 3:1 because the image1 distance is 3 times the object distance i.e. +1/(1/3) = 3. Image1 serves as the object for the +3 D lens and the image2 distance is 1/2 of the object distance i.e. (1/2)/(1)= 1, therefore image2 is 1/2 the size of image1. The relative size of the final image (image2) relative to the original object is 3 x (1/2) = 1.5 times. 4. Is the object real or virtual? Answer: The object is real, because light rays move away FROM the object and through the lenses. By convention, a real object is usually depicted as being left of the lenses with the path of light moving toward the right. 5. Is the final image real or virtual? Answer: The final image is real because it is located on the opposite side of the lens system as the object. 6. What is the practical difference between a real and virtual image? Answer: A real image may be projected onto a screen, whereas a virtual image cannot. 7. Is the final image upright or inverted relative to the object? Answer: The final image is upright relative to the object because the magnification (+1.5) is a positive number. If the magnification were a negative number the image would be inverted. The location and orientation of the image can also be determined by ray tracing where one ray is drawn FROM the top of the object through the center of lens and another ray is drawn parallel FROM the top of the object to the lens and then through the focal point of the lens. The intersection of these two rays will identify the location of the top of the image.  
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