Diverging lens problem -image should be between focus & optical centre

[s3a]

Homework Statement

Problem:
An object 2.0 cm high is placed in front of a diverging lens of focal length 1.8 cm.

The object is 2.7 cm from the optical centre of the lens.

Questions:
1) Where is the image located with respect to the focus and the optical centre of the lens, qualitatively?

2) Is the image upright or inverted?

3) Is the image real or virtual?

4) Is the image larger or smaller than the object?

Answers:
1) The image is located between the focus and the optical centre of the lens.

2) The image is upright.

3) The image is virtual.

4) The image is smaller than the object.

Homework Equations

$1/f = 1/d_o + 1/d_i$
$h_i/h_o = -d_i/d_o$

The Attempt at a Solution

This thread is about answer #1.

Modifying $1/f = 1/d_o + 1/d_i$, I get:
(1/(1.8 cm) – 1/(2.7 cm))^(-1) = 5.4 cm = $d_i$

Doesn't that mean that the image should be 5.4 cm – 2.7 cm = 2.7 cm to the left of the object instead of between the focus and the optical centre of the lens?

I'm assuming there's something wrong with the numbers because if what I am reading from this book is correct, the image of a diverging lens should always be between the primary focus and the optical centre, regardless of where the object is located.

Could someone please tell me what I am doing wrong?

Any input would be greatly appreciated!

 

[technician]

The lens is diverging...do you know what this means regarding focal length value?

 

[s3a]

Oh! Is it that the focal length is negative such that the distance of the image from the optical centre is -1.08 cm or 1.08 cm leftward of the optical centre?

 

[besulzbach]

Exactly. The focus, in this problem, is -1.8cm.

 

[technician]

Oh! Is it that the focal length is negative such that the distance of the image from the optical centre is -1.08 cm or 1.08 cm leftward of the optical centre?

You have got it. You first attempt was for a converging lens

 

[s3a]

Alright. Thank you (both) very much! :D