How Is Object Distance Calculated for Image Reduction with a Diverging Lens?

[DMOC]

Homework Statement

An object is 18 cm in front of a diverging lens that has a focal length of -12 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 2.0

Homework Equations

The thin lens equations:

(1/di)+(1/do)=1/f

The magnification equation

(hi/ho)=-(di/do)

f=focal length
di=image distance
do=object distance
hi=image height
ho=object height

The Attempt at a Solution

The first thing I do is plug in values for the thin lens equations and solve for di (do and f are given). I end up getting:

18/di=-2.5
di=-7.2 cm

So the distance of the image from the lens is -7.2 cm.

However, this is where I get confused. I could use hi/ho=-(di/do) but there are two unknowns (hi/ho).

The answer to this, by the way, is 48 cm. I don't know how to get to this, though.

 

[Gear300]

Since you know di and do, you could get the magnification at 18cm in front of lens. Adjusting the equation a bit, -(di/do) = 1 - di/f. -(di/do) is the magnification, and what they are asking for is the object distance for half the magnification; thus, set -(di/do) to half the magnification you got earlier and solve for di. Using that value of di, find the value of do.

 

[JaWiB]

Let's say your initial image height is h_i = h_0\frac{d_i}{d_0}

Now you want to move the object to a new distance so that the new image height is half of this:
h_{i2} = h_0\frac{d_{i2}}{d_{02}}=\frac{1}{2}({h_0}\frac{d_i}{d_0})

Now you have an equation with three knowns and two unknowns. But you should be able to use the thin lens equation to reduce it to one unknown. Does that help?

 

[DMOC]

Thanks for the responses! I used the method that Gear300 proposed (I tried JaWiB's but I couldn't figure it out) and I got 48 cm.