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1. A point source of light lies on the central axis of a bi-convex spherical lends who radii of curvature are both measured to be 19.4 cm. The lens is made up of glass with an index of refraction of 1.50 and the source is 3.17 meters to the left of it. What is the image distance (in meters)?
2. A grasshopper sitting 31.4 cm to the left of a convex lends sees its image on a screen 68.6 cm to the right of the lens. It then jumps 16.2 cm toward the lens. Where will its image be now?
3. A point source is moved along the central axis of a thin spherical glass (n=1.5) lens, and it is found that when the source is at 10.1 cm from the lens the emerging light forms a perfectly cylindrical beam travelling along the axis. Where will the image be when the source is moved out to 150 cm from the lens?
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This job emphasizes optics problems. The image determinations are provided for different distances.
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1. In order to find the image distance, first we need to know the focal length of the lens. That is given by the lensmaker's equation (for thin lenses):
1/f = (n-1)*(1/R1 - 1/R2)
Where R1 and R2 are the radii of curvature of the two surfaces of the lens. Note that in a biconvex lens such as this one, one radius is taken to be negative - this is because one surface is concave towards the incident light.
So, in our case, we have R1 = |R2| = 19.4 cm, and n = 1.5.
This gives us a focal length of f = 19.4 cm.
Now, we resort to the familiar lens equation to ...
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