Constants REFLECT In part (a), the image distance is negative, so the real objec
ID: 2032280 • Letter: C
Question
Constants 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. 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 this 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. Part A - Practice Problem: For a diverging lens whose focal points are 7.50 cm away from the lens, where should an object be placed to form an upright, virtual image that is 0.3 of the size of the object? Express your answer to 3 significant figures and include appropriate units. 1G + 2 0 ? ??? FigureExplanation / Answer
M = 0.3 = - di / do
di = 0.3 do
Applying lens equation,
1/f = 1/ do + 1/ di
1/(-7.50) = 1/do - 1/0.3do
do = 17.5 cm