[Physics] Magnification of a real image formed by a concave lens

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Graph of 1/u and 1/v for Concave Lens:

Lens Formula:
$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$
Since, f is negative for a concave lens (diverging lens), re-writting the equation
$$\frac{1}{v}=\frac{1}{u}-\frac{1}{f}$$
graph

Now, we know from the graph that for a real object it's image will be virtual(formed behind the lens between f and optical center) and diminished too since $\frac{1}{v}$ is greater than $\frac{1}{u}$ i.e
v is smaller than u and $$m=\frac{v}{u}$$
Represented by the region 1.

Also, from the principle of reversibility of light, a virtual object being formed between focus and optical center will have its image formed behind the lens i.e virtual and magnified.
Represented by the region 2.

And now what about a virtual object being formed beyond focus (region 3)?
From the graph, I think it will be real but magnified for some region and diminished for some.So as to get some clarity I used Newtonian Thin Lens Equation
$$x_1\cdot x_2=f^2$$
where

$x_1$ is the distance of object from the focus

$x_2$ is the distance of the image from the focus

Now since $x_1$ is greater than 2f $x_2$ should be less than f/2 which will result in the formation of a virtual image and not a real image, which is conflicting from the result we obtained from the graph.

To summarize, my question is as follows:

  1. Whether the image of the virtual object will be magnified or diminished or
    both?
  2. The image formed will be real for sure but from Newtonian Thin Lens Equation
    I am getting a virtual image. So what am I doing wrong?

Best Answer

I note that you are using the Cartesian Convention, in which distances are +ve measured to the right of the lens. Light is travelling left to right.

I agree with region 1 (real object, virtual image, diminished). However, in region 2 I get virtual object and real image ($v>0$), magnified. As you point out, this should be the opposite of region 1.

Region 3 is virtual object ($u>0$), virtual image ($v<0$). Magnification can be $<1$ ($u>f/2$), $=1$ ($u=v=f/2$) or $>1$ ($u<f/2$). This follows quite easily from your graph.

So to answer your specific questions :

  1. In region 3 the image can be magnified or diminished, depending on the object position.

  2. The image in region 3 is virtual ($v<0$). Your calculation with Newton's formula is correct. It is your interpretation of your graph which is at fault.

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