A screen is placed a distance d = 45.0 cm to the right of a small object. At wha
ID: 1447965 • Letter: A
Question
A screen is placed a distance d = 45.0 cm to the right of a small object. At what two distances to the right of the object can a converging lens of focal length +5.75 cm be placed if the real image of the object is at the location of the screen? (Hint: Use the thin lens equation in the form s' = sf s f , so s'(s f) = sf. Write s' = d s and solve for s in the resulting quadratic equation.)
shorter distance cm
longer distance cm
For this lens, what is the smallest value d can have and an image be formed on the screen?
Explanation / Answer
di = distance of image , do = distance of object
di + do = 45 cm
do = 50.5 - di
1/di + 1/do = 1/f
1/di + 1/(45-di) = 1/5.75
di = 6.77 cm
di =38.23 cm
shorter distance =6.77cm
longer distance = 38.23 cm
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