Objects in Optics: Image Formation Explained

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In summary: I2.In summary, virtual images exist and the concept is commonly used. However, I do not know what a 'virtual object' would or could be.
  • #1
UchihaClan13
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Guys
Just wanted to clarify one doubt
When we consider image formation by objects
We always assume there's a source of light present ,right?(because all real-point and extended objects can't always be luminous )
we also know that it's diffused reflection of light or scattering of incident light from a source that allows us to perceive/see objects
Right??
Thanks :)
 
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  • #2
And there's no way virtual objects can ever be luminous as in virtual objects,the light rays "appear"to converge at a particular point
Or is there??
 
  • #3
Usually, the object is treated as a source of light. You can see an object because light scatters or reflects off the object's surface, so the light is coming from the object's surface.
 
  • #4
UchihaClan13 said:
virtual objects
Virtual Images exist and the concept is coommonly used . But I don't know what a 'virtual object' would or could be. Perhaps you are referring to an intermediate virtual image? That's something that multip[e lenses and mirrors can produce.
An object can either be luminous or illuminated by a light source. An optical system would treat both situations the same way. If an 'object' happens to be a mirror or a lens then the effect on the light falling on it could be to produce specular reflections and that could (unintentionally, perhaps) produce one or more coherent images. Ray tracing software was developed, not many decades ago, for the fraphics industry which will handle specular reflections and the formation of images in shiny objects - quite convincingly sometimes.
 
  • #5
sophiecentaur said:
But I don't know what a 'virtual object' would or could be. Perhaps you are referring to an intermediate virtual image?

Are you thinking of something like this?

lenses.gif


To me, I1 = O2 is a real image with respect to lens L1, but a virtual object with respect to lens L2.
 
  • #6
In my experience, an "object" is something you can actually get hold of and light comes from it so intermediate images in an optical system are not objects in my book. I hate to waste too much time in pointlessly classifying things and there are occasions when a projected image on a screen (like in the example of the first post) is, to all intents and purposes, an object but why change the name of an Image, formed by one lens, to an Object when its light passes through another optic in a 'coherent chain' of optics?
 
  • #7
there are recognised textbooks that refer to the image formed by a lens as being 'virtual objects' for a second lens. There is no doubt that I1 i JTBells diagram is the 'object' for L2 . As far as L2 is concerned the rays'from' this object are diverging as they strike L2.
Using the sign convention I am familiar with this counts as a 'virtual' entity...Object.
In optics it would be a strange step to describe an 'object' as something that you can actually get hold of !.
I1 is optically an 'object' and is treated as such in calculations for L2

'A collection of points which may be regarded as a source of light rays for a portion of an optical system but which does not actually have this function.'
a definition of virtual object from an online dictionary
 
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  • #8
Fair enough. :smile:
 
  • #9
By a virtual object,sophie,I mean an object towards which the light rays "appear" to converge but don't actually do so due to an intervention by an optical element such as a mirror or a lens
 
  • #10
lenses-gif.94846.gif

That's precisely what i meant :)
lenses-gif.94846.gif
 
  • #11
That's what i meant
 
  • #12
That's precisely what i meant
 
  • #13
So let me clear up the things finally
A real object is something from which light rays emanate (in case of an actual source of light) or seem to emanate
It can be a luminous or non-luminous object right?
 
  • #14
UchihaClan13 said:
A real object is something from which light rays emanate
That's the same definition as a real image but it fits both.
UchihaClan13 said:
By a virtual object,sophie,I mean an object towards which the light rays "appear" to converge but don't actually do so due to an intervention by an optical element such as a mirror or a lens
The problem with your use of the word "objet' instead of 'image' is that there is nowhere that the word 'Image' would be appropriate. A virtual image would be a location where diverging rays would meet, if produced backwards.
 
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  • #15
If the problem is the use of words then it is a problem in semantics.
Is there any doubt that I1 is an 'image' formed by L1?
Is there any doubt that the final 'image' I2 is formed by the rays forming I1 passing through L2?
The rays forming I2 come from the 'object' for L2 yes or no?
This is not a 'real' object. How better to describe it than the textbook term..'a virtual object'.
There would be no semantic problems if all of these things were just called 'things'
The thing at O1, the thing at I1, the final thing at I2.
 
  • #16
lychette said:
If the problem is the use of words then it is a problem in semantics.
Is there any doubt that I1 is an 'image' formed by L1?
Is there any doubt that the final 'image' I2 is formed by the rays forming I1 passing through L2?
The rays forming I2 come from the 'object' for L2 yes or no?
This is not a 'real' object. How better to describe it than the textbook term..'a virtual object'.
There would be no semantic problems if all of these things were just called 'things'
The thing at O1, the thing at I1, the final thing at I2.
well said
But that's how everything is!:)
 
  • #17
UchihaClan13 said:
well said
But that's how everything is!:)

So why do these quesions keep arising,...they were answered when I was a student...10, 20, 30, 40, 50...years ago
Does Physics Forums really supply an educational resource? Text books still exist, is our role to challenge the conventional wisdom ofd textbooks>?>
 
  • #18
So you seem to be implying that there is only one 'Image' in an optical system and that is the final 'thing' that is seen by an observer. I guess that could be applied consistenty in all systems - a bit like an electronic circuit in which there is a chain of inputs and outputs; it's all inputs until you get to the final output.
 
  • #19
sophiecentaur said:
So you seem to be implying that there is only one 'Image' in an optical system and that is the final 'thing' that is seen by an observer. I guess that could be applied consistenty in all systems - a bit like an electronic circuit in which there is a chain of inputs and outputs; it's all inputs until you get to the final output.

I suppose that some would argue that the final 'thing' seen by the observer is what appears on the retina but then we are a step removed from the system of lenses.
In the ray diagrams above if no arrows were marked on the rays there would be no difference to the light ray paths and so in some way we must have produced a 'convention' whereby some things are called objects and others images. Presumably this aids description and understanding. Real images and virtual images are well known terms and their formation by converging and diverging rays is understood. Real objects and virtual objects logically follow.
 
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  • #20
thankyou sir
 

Related to Objects in Optics: Image Formation Explained

1. What is the difference between real and virtual images?

The main difference between real and virtual images is that real images can be projected onto a screen and are formed when light rays actually converge at a point, while virtual images cannot be projected and are formed when light rays only appear to converge when traced back.

2. How are images formed in convex and concave lenses?

In a convex lens, images are formed when light rays converge at a point after passing through the lens. This creates a real, inverted image. In a concave lens, images are formed when light rays diverge after passing through the lens. This creates a virtual, upright image.

3. What is the relationship between object distance, image distance, and focal length?

The relationship between object distance (do), image distance (di), and focal length (f) is given by the lens equation:
1/do + 1/di = 1/f. This means that as the object distance increases, the image distance decreases and vice versa. The focal length remains constant for a given lens.

4. How does changing the shape of a lens affect image formation?

Changing the shape of a lens can affect image formation in several ways. Convex lenses, which are thicker in the middle, converge light rays and create real images. Concave lenses, which are thinner in the middle, diverge light rays and create virtual images. Additionally, changing the curvature of a lens can affect the focal length and the size of the image formed.

5. What is the difference between a magnified and a demagnified image?

A magnified image is an image that appears larger than the object, while a demagnified image appears smaller. This can occur when using convex or concave lenses, and is dependent on the distance between the object and the lens, as well as the focal length of the lens. Magnified images can also be created using mirrors, where the image is reflected and appears larger than the object.

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