Optics and Image Determination

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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...ind the image distance:

1/s + 1/i = 1/f

Where s is the distance to the object (source) and i is the distance to the image. In our case we have a source distance of 3.17 m. Thus:

1/3.17 + 1/i = 1/.194
i = .21 m
Don't forget to convert the focal length into the units being used (meters)!

Since i is positive, that means our image will be produced .21 m to the right of the lens.

2. This is two-stage question. First, we are given the position of the grasshopper and its image. From this, we can find the focal length of the lens it is standing beside. With that information and the knowledge of the new position of the grasshopper, we can find its new image location.

First, find the focal length from the lens equation:

1/31.4 + 1/68.6 = ...