find object distance given only magnification(optics)

[KeepingitMoving]

An object placed in front of a lens with focal length f produces a real image that is three times as large as the object. At what distance from the lens has the object been placed?
Answer choices
A.1/3f
B. 3/4f
C. 4/3f
D. 3f

I have the right answer (C) but have no explanation. I have spent the last 2 hours trying to figure it out.
Any takers?

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[anticarbon]

1/f = 1/o + 1/i

where
f = focal length
o = object's distance
i = image's distance
*all positive numbers since real image is formed, (and therefore lens is converging lens)

if i = 3o

1/f = 1/o + 1/3o
1/f = 3/3o + 1/3o
1/f = 4/3o
f = 3/4o
o = 4/3f

 

[bear2roo]

An object placed in front of a lens with focal length f produces a real image that is three times as large as the object. At what distance from the lens has the object been placed?
Answer choices
A.1/3f
B. 3/4f
C. 4/3f
D. 3f

I have the right answer (C) but have no explanation. I have spent the last 2 hours trying to figure it out.
Any takers?

if you wanna solve this quickly, TBR has a neat method.

Ok so real image right? Must be converging lens since diverging lens always produce Small Upright Virtual

Next, A & B can be eliminated b/c if you place an object closer than the focal length for a converging lens, it will create a virtual image.

Between C & D, how can you figure it out?

If do = 3f, then the image will be much smaller. Therefore that leaves C.

This animation should help you: watch it several times--it will save so much time on these problems. I pretty much have the animation memorized, which is why I can eliminate D quickly

http://science.sbcc.edu/~physics/flash/optics/imageobject.html

 

[KeepingitMoving]

those explanations were perfect! thank you so much 😍