geometric-opticslensesopticsvisible-light
I do not understand how to proceed in order to create the ray diagram of the case of a diverging lens where both object and image are virtual, that is $f<0$ , $p<0$ with $p<f$ (otherwhise the image wuold be real.
I did not find this particular case of ray diagram in any textbook so I would like to know where to find it or how to build it on my own.
![[1]: https://i.stack.imgur.com/p8ZMy](https://i.stack.imgur.com/JDX7L.png)
The virtual image for Interface 2 here actually means, that the image would have been there if the second interface did not alter the path of light. But due to interface 2, the position of (would have been image, l1) is changed to l. It is "would have been" image, not a virtual object. The virtual object for interface 2 is where the light seems to be coming from, means in exactly opposite direction of l1 along the ray of light. Because, it is a "would have been image", and virtual object is directly opposite, the virtual object and final image are not on the same side.
The angular magnification of a telescope $M$ is defined as the ratio of the angle subtended by the image of the object when looking through the telescope $b$ to the angle subtended by the object when looked at with the unaided eye $a$.
$$M=\dfrac ba$$
Those angles are often called visual angles and they detained the size of the image which is formed on the retina.
The bigger the visual angle, the bigger the image formed on the retina and the bobber the “object” being viewed is perceived to be.
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I have annotated your diagram which clearly shows that $b>a$ which means that the angular magnification of such a telescope is greater than one ie the Galilean telescope magnifies.
The final image can be formed at infinity as shown in the ray diagram below.
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$f_1$ if the focal length of the convex lens which converges the incoming rays and $f_2$ is the focal length of the concave hens which diverges the incoming rays.
Again the visual angle for the final image $u’$ is greater than the visual angle for the object being observed $u$.
Best Answer
The top diagram shows the formation of the virtual object where converging rays are prevented from meeting by the diverging lens.
Then those converging rays are made to diverge by the lens and so a virtual image is formed.
Update as a result of a comment from @Floris.
I included a converging lens just to check the orientation of the virtual object and the virtual image for the diverging lens.
As pointed out by @Floris the diagram that the OP had the virtual object and the virtual image both upright so I investigated whether that was possible by moving the position of the virtual object relative to the focal point of the diverging lens.
Rays labelled $1$ have the virtual object in the focal plane of the diverging lens and that produces an image at infinity.
Rays labelled $s$ have the virtual object outside the focal point of the diverging lens and that always produces an inverted and virtual image.
Rays labelled $3$ have the virtual object inside the focal point of the diverging lens and that produces an upright image which is real.