Diverging lens problem? finding focal length

[snash1057]

A diverging lens is used to form a virtual image of an object. The object is 88.0 cm in front of the lens, and the image is 44.0 cm in front of the lens. Determine the focal length of the lens.

1/f = 1/do + 1/di


since do is 88cm infront of the lense, that would keep the 88 positive, and since di is 44cm in front of the lense, that would make it negative

so

1/f = 1/88 + 1/-44
1/f = -.011
f= -90.9cm


but its incorrect so if anyone could help correct me thatd be great!

 

[Stonebridge]

Check your arithmetic.
The answer to that calculation is -88cm

 

[iowagal]

f = (do*di)/(do+di) or (88.0*44.0)/(88.0+44.0) = 29-1/3 cm. Verify it by using graph paper or a ruler and drawing it to scale... Choose 4 cm for your object height.

So the real question becomes -- does a diverging lens with the object not placed between the focal point and the lens produce a smaller inverted image - or a larger upright image (magnifaction) - and is the image real or imaginary?

 

[nasu]

f = (do*di)/(do+di) or (88.0*44.0)/(88.0+44.0) = 29-1/3 cm. Verify it by using graph paper or a ruler and drawing it to scale... Choose 4 cm for your object height.

So the real question becomes -- does a diverging lens with the object not placed between the focal point and the lens produce a smaller inverted image - or a larger upright image (magnifaction) - and is the image real or imaginary?

You forgot the minus signs.

Regarding you second question, a divergent lens produces a virtual image no matter the (real) object's location.

 

[iowagal]

What would be the difference if it was a diverging mirror instead of a diverging lens?

 

[nasu]

The difference in what respect?
The light does not go through the mirror. This is one difference.