A diverging lens of focal length -16cm projects the image of an object onto a wall. What is the object distance if the image is virtual, upright, and 40% of the object size?

Question

anonymous · Answer

You can use the magnification equation to write the image distance in terms of the object distance.  Then use that with the thin lens equation.
$$mag=\frac{ d _{i} }{ d _{o} }$$
$$\frac{ 1 }{f }=\frac{ 1 }{ d _{i} }+\frac{ 1 }{ d _{o} }$$

anonymous · Answer

yes so i did di = 0.4do then 1/-16 = 1/0.4do + 1/1do ?

anonymous · Answer

solve for do.  undistribute the do from both terms 1/-16 =do (1/0.4 + 1/1)
then divide both sides by that number and you'll have do which we expect to be negative as it's a diverging lens :)

anonymous · Answer

i got -56 and it does not work.

anonymous · Answer

I apologize I miss wrote the magnification equation.  That should read -di/do :)

anonymous · Answer

oh thank you, i got 24 and now it's right!

anonymous · Answer

Excellent.  Glad I could help.