A screen is placed a distance d = 35.5 cm to the right of a small object. At wha

Question

A screen is placed a distance d = 35.5 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 +5.15 cmbe 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' =

,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

s'(s-f) = sf

s'*s - s'f = sf..............1

s' = d-s.............(2)

2 in 1

(d-s)*s -(d-s)f = sf

35.5s - s^2 - 182.825 + 5.15s = 5.15s

s^2 -35.5s + 182.825 = 0

s = , 6.25 cm ,   29.25 cm

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