40.0 cm-13.3 cm Now let\'s apply the thin-lens equation to a diverging lens, and

Question

40.0 cm-13.3 cm Now let's apply the thin-lens equation to a diverging lens, and hen we will confirm our results by constructing the image sing a ray diagram. Suppose that you are given a thin iverging lens and you find that a beam of parallel rays preads out after passing through the lens, as though all the ays came from a point 20.0 cm from the center of the lens ou want to use this lens to form an upright, virtual image that s one-third the height of a real object. (a) Where should the bject be placed? (b) Draw a principal-ray diagram di Part (b): (Figure 1) 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 Part A -Practice Problem: For a diverging lens whose focal points are 12.0 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? gure 1 of 1 Express your answer to 3 significant figures and include appropriate units do CM Submit Request Answer -13.3 cm -20.0 cm -20.0 cm-

Explanation / Answer

Part A

per lens equation

1/v - 1/u=1/f

1/v-1/u=-1/12

Also, magnivification=0.4

v=-0.4u

1/-0.4u + 1/u=-1/12

u=42 cm

v=-16.8 cm

Object should be placed 42 cm from the lens

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