Magnification of a real image formed by a concave lens

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 $$\frac{1}{v}=\frac{1}{u}-\frac{1}{f}$$ 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?
• -1. Have you searched the internet? eg using your title I get this as #2 hit : physicsclassroom.com/class/refrn/Lesson-5/… – sammy gerbil Apr 20 '17 at 22:39
• It doesn't talk about the image formation of a VIRTUAL object. – Alu Apr 21 '17 at 3:11
• youtube.com/watch?v=9kyD3aiO70c. This tutorial uses a convex lens, but the principles of ray-tracing for a virtual object are the same. – sammy gerbil Apr 21 '17 at 13:00
• Do you have a particular situation in mind? If so, please can you post the details and show your attempt. It is easiest to explain using the problem you are struggling with, and the step you are confused about. – sammy gerbil Apr 21 '17 at 13:03
• Yes I am aware you are asking about a concave lens. That is why I said that the principles are the same as those discussed in the youtube video. – sammy gerbil Apr 21 '17 at 13:25

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.

• In Cartesian Convention distances are measured +ve in the direction of incident light and not in the right.Also, I have pointed out the principle of reversibility of light from which this case will be similar to region 1.Can you explain the region 3 with the help of Newtonian Thin Lens Equation? – Alu Apr 22 '17 at 17:35
• Yes, I am assuming light is travelling left to right, which is usual. Your calculation using Newton's formula appears to give the correct answer. – sammy gerbil Apr 22 '17 at 17:40
• Since you are assuming light to travel from left to right, in the region 2 our object is in the direction of the light i.e behind the lens.Similarly, our image to is in the direction of light i.e behind the lens.So the image formed will be virtual and not real.And similarly, in the region 3 as visible from the graph the image will be real and no virtual.And that's what I am asking? The image should be real which is not coming from the Newtonian Lens equation. – Alu Apr 22 '17 at 17:51
• In the Cartesian Convention all distances to the right of the lens are +ve, all distances to the left are -ve. If u>0 the object is virtual, if v>0 the image is real. – sammy gerbil Apr 22 '17 at 18:00
• Here v>0 signifies that the direction of image from the optical center is in the direction of the incident light as is the case with the our virtual object.So the image formed will be on the same side as our virtual object is being formed and hence it will appear to be virtual. – Alu Apr 22 '17 at 18:30