Image formation (lenses) I have a question regarding image formation by lenses. It is the following:
A lense with a focal length of $-48.0cm$ forms an image $17.0cm$ to the right of the lense. Where is the object positioned?
My attempt at a solution:
Well, since the focal length is negative, we know that the lense is divergent (concave). The problem statement says that the image is to the right of the lense, so $s_i=17cm (>0)$
We can compute the position of the object by using the formula below:
$$\dfrac{1}{d_0}+\dfrac{1}{d_i}=\dfrac{1}{f}$$
Changing what we know for the values, we get:
$$\dfrac{1}{d_0}+\dfrac{1}{17}=- \dfrac{1}{48}$$
And doing the calculations, it comes down to:
$$s_0=-12.55$$
However, the solutions say the solution is $s_0=26.3$. One can get that if we take $s_i$ to be negative, which is wrong, in my opinion. (?)
So, why am I wrong? Or am I correct?
 A: I think you have some problems with the sign convention. At first, ask yourself the  question "why did I need the sign convention?"
For the moment, forget the sign convention and let's derive the formula related to lenses from the scratch using only geometry.
Let us consider:


*

*p= distance of the object from the optical centre

*q= distance of the object from the optical centre

*f= distance of the principal focus from the optical centre.


Now let's work out the case of convex lens:

so in the case above figure, $\rm OA=p, OA'=q,  OF_2=f$
Now, using similarity of $\triangle \rm AOB \ \  and\ \  \triangle \rm A'OB'$ , we have $$\frac{AB}{OA}=\frac{A'B'}{OA'}   \ \ \ --------(1)$$  . 
Again, using similarity of $\triangle \rm OF_2C \ \  and\ \  \triangle \rm A'F_2B'$ , we have $$ \rm \frac{OC}{OF_2}=\frac{A'B'}{A'F'_2}$$ $$\rm \frac{AB}{OF_2}=\frac{A'B'}{OA'-OF_2}\ \ \ --------(2)$$.$\ $ (note C is not shown in the figure, but it is the point above O within the lens. AB=OC may help you which point is C.)
using (1) and (2), one will get$$ \rm \frac{OA'}{OA}=\frac{OA'-OF_2}{OF_2}$$
$$ \rm \frac{q}{p}=\frac{q-f}{f}$$
$$ \rm \frac{1}{p}=\frac{q-f}{qf}$$
$$ \rm \frac{1}{p}=\frac{1}{f}-\frac{1}{q}$$
$$\large\boxed{ \rm \frac{1}{p}+\frac{1}{q}=\frac{1}{f}} \ \ \ ---------(3)$$ (REMEMBER NO SIGN CONVENTION HAS BEEN USED. ALL THE QUANTITIES(p,q,f) ARE POSITIVE.)
Now let's work out the case of concave lens:

Working out in the same way as above, you will get
$$ \rm \frac{OA'}{OA}=\frac{OF_1-OA'}{OF_1}$$
$$ \rm \frac{q}{p}=\frac{f-q}{f}$$
$$ \rm \frac{1}{p}=\frac{1}{q}-\frac{1}{f}$$
$$ \large\boxed{\rm \frac{1}{q}-\frac{1}{p}=\frac{1}{f}} \ \ \ \ --------(4)$$
(REMEMBER NO SIGN CONVENTION HAS BEEN USED. ALL THE QUANTITIES(p,q,f) ARE POSITIVE.)
The equation(4) is the formula you are looking for. Put q=17 cm , f=48 cm
then 
$$\frac{1}{17}-\frac{1}{48}=\frac{1}{p}$$
p= 26.32
Now to find where the object is, you may see the above figure. (Now you may be confused as in your problem you told the image is at the right of the lens. But that really doesn't matter. That depends on your current position. You just take a print out of the above figure and see the figure from the back of the page.)
Now let's get back to the question:" Why do we need sign convention?"
look at the equ.3 and equ.4. They are different. Then how to remember a lens formula irrespective of whether the lens is convex or concave.
That's why (I think, maybe there are other reasons) the sign convention is used.
Now the convention is such :


*

*All the quantities are measured by starting from the optical center. And quantities will be assigned sign  + if you need to go along the direction of the incident light and - if you need to go against the direction of the incident light.


let 


*

*u= object distance

*v=image distance

*F= focal length


(all u,v,F are with convention)
then for convex lens, u= -p, v= q, F=f
so equ.3 becomes:
$$ \rm \frac{1}{-u}+\frac{1}{v}=\frac{1}{F} $$
$$ \large\boxed{\rm \frac{1}{v}-\frac{1}{u}=\frac{1}{F}} \ \ \ ---------(5)$$
then for concave lens, u= -p, v= -q, F=-f
so equ.4 becomes:
$$ \rm \frac{1}{-v}-\frac{1}{-u}=\frac{1}{-F} $$
$$ \large\boxed{\rm \frac{1}{v}-\frac{1}{u}=\frac{1}{F}} \ \ \ ---------(6)$$
Now see equ.(5) and equ.(6) are same. Now you don't need to remember the both the formulas equ.(3) and equ.(4). Rather you just need to remember$$ \rm \frac{1}{v}-\frac{1}{u}=\frac{1}{F} $$ with the sign convention stated above.
Now let's again solve your problem, with this generalized formula( which we have got using sign convention.)( Now again we have to use the sign convention when solving the problem, so that we can cancel the ad-hoc sign convention used to bring the formula equ.(5) or equ.(6) )
F=-48 cm,
v=-17 cm,
u=?
$$\frac{1}{-17}-\frac{1}{u}=\frac{1}{-48}$$
$$-\frac{1}{17}+\frac{1}{48}=\frac{1}{u}$$
u=-26.32
(SEE, in the convention there is no mention of right or left. What is important is the direction of the incident light).
A: There are many different forms of the mirror and lens formulas, each varying slightly in appearance, and each using a different set of rules as to what distances, focal lengths, and radii are positive or negative! The most important part of any formula is what I call the "wheres". Mixing one formula with another set of "wheres" leads to total chaos...
In the Wikipedia formula cited above, the "wheres" say that real objects/images have positive distances, virtual objects/images have negative distances, and converging lenses have positive focal lengths.
The question refers to a negative focal length, so the lens is diverging.
The question does not state whether the image is real or virtual, so that the question is ambiguous.  The value placed in the formula for the image distance can be +17 or -17.Each value leads to a different object position;  if the object position is found to be negative, the object is a virtual object, formed by some other optical element....
BTW, a diverging lens can form a real image;  it just needs a virtual object! 
EDIT:
There is another set of "wheres", that I recall using many years ago.  In this, the lens was placed in a standard position, with the light coming always from the left, passing thru the lens, and exiting to the right.  The sign of the distances for objects and distances followed different rules.  It worked, but was harder to use...
A: Something may be wrong with the statement of the problem. A diverging lens cannot form a real image (see this SE answer).  
If the problem meant to specify a virtual image, try this:  turn the system around so that the virtual image is on the left.  This presents a less confusing situation w.r.t. the sign convention.   
In any event, I'm not sure which side of the lens the given solution places the object.  For a virtual image the object distance and image distance have opposite signs, and, well ... one could argue that the sign should be negative.   I think the best thing to do in a case like this, where the statement of the problem leads to ambiguity, is to say the object is on the right side of the lens, and that the absolute value of the object distance is such-and-such. Don't even specify a sign. 
update
Upon reflection, I think there's a additional condition on the usual sign convention that is  sometimes not stated (and easily neglected).  That is, in all cases the object distance is always taken as positive.  With that, and the geometry as given, the image distance would have to be taken as negative (same side as object --> opposite sign):  $d_i = -17$ cm.  Carrying on with the lens equation then gives $d_o = +26.3$ cm.
