Now let\'s apply the thin-lens equation to a diverging lens, and then we will co

Question

Now let's apply the thin-lens equation to a diverging lens, and then we will confirm our results by constructing the image using a ray diagram. Suppose that you are given a thin diverging lens and you find that a beam of parallel rays spreads out after passing through the lens, as though all the rays came from a point 20.0 cm from the center of the lens. You want to use thin lens to form an upright, virtual image that is one-third the height of a real object. (a) Where should the object be placed? (b) Draw a principal-ray diagram. SET UP AND SOLVE: The behavior of the parallel incident rays indicates that the focal length is negative: f = -20.0 cm. We want the interval magnification to be m = +1/3 (positive because the image is to be upright). From the distribution of magnification, m = 1/3 -d_i/d_0. So we use d_i = -d_0/3 in the thin lens equation: 1/d_0 + 1/d_i = 1/f 1/d_0 + 1/-d_0/3 = 1/-20.0 cm d_0 = 40.0 cm d_i = - 40.0 cm/3 = -13.3 cm shows our principal-ray diagram for this problem REFLECT In part(a), the image distance is negative, so the real object and the virtual image are on the same side of the lens. For a diverging lens whose focal points are 8.0 cm away from the lens, where should an object be placed to form an upright, virtual image that is 0.4 of the size of the object. Express your answer to two significant figures and include appropriate units.

Explanation / Answer

hi/ho = v/u

0.4 = v/u

v (image disatnce) = 0.4 u

using the lens equation,

1/f = 1/v - 1/u

1/ (-8) = 1/0.4 u - 1/u

-0.125 = 1/u ( 1.5)

u = - 1.5/ 0.125 = -12 cm

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