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Here’s an image from my textbook

  • enter image description here

It shows, how an image is obtained from a convex lens. The second and the third images, shows in depth , that how a convex lens behaves.

They say, that suppose ( in figure 2 ) the light ray enters the lens from the interface N1 B C. This forms a virtual image I1. Now this virtual image acts like an object for the second interface, and thus forms a real image at I ( figure 3 ) .

What is this virtual object? How can rays come from this virtual object towards the lens, from the right , and again the image is formed at the right hand side. This is possible only in concave lenses and in convex lenses when the object is between P and F. But neither of the cases apply here. What is it actually?

WHY do we treat this as an ‘object’ even when it does not exist?

PS- I know we do this with plane mirrors and others too, but thats reasonable there but here, the object and the image are on the same side! So whats it?

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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.

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  • $\begingroup$ I am sorry, but i didn’t understand that if the virtual object is on the same side, how can a real image be produced on the same side? $\endgroup$ – Aaryan Dewan Mar 4 '16 at 9:05
  • $\begingroup$ @AaryanDewan: You need to differentiate between the two - "Virtual object" - is some where in the direction where the light ray is coming FROM, not where, it is going TO. l1 is not virtual object. If it is called so, it is wrong. You may possibly call it a "virtual image", or a "would have been image" because interface 2 moved this image to l. $\endgroup$ – kpv Mar 4 '16 at 15:46
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The idea is that a virtual image is an image that is not real image. To put it more concretely, a true image can be put on a screen. For example, if there is real image behind a lens, putting a white screen on that location will show the image.

A virtual image is an image because it is obtained the same way, but virtual in the sense that putting a white screen on the location will show no image. Hence the term "virtual" image. This definition is very physical.

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  • $\begingroup$ What are speaking about? Please, read my question again sir. $\endgroup$ – Aaryan Dewan Mar 4 '16 at 9:00
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I will answer to the best of my capacity bro Look bro i will try to explain how does the virtual object tactic work We know that if rhe ray was allowed to move straight after ist refeaction image at I1 will have been formed But it suffers another refraction and image i is formed Now what will happen if we place an object at I Light retraces its path so it will emerge anti parallel to that of The path towards i1 Thus a virtual image will be formed at i1 So if an object is placed at i Image is formed at i1 Since the refraction formulae for curved surfaces does not changes anyhow if u&vare replaced (N1/u__N2/v=N1_N2÷R) as it is symmetric therefore We can equally say that if object is asumed at I1 image at I will be formed

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