Why Is Magnification Undefined for a Point Object in Ray Diagrams?

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Magnification for a point object is considered undefined because a point has no size, making it impossible to establish a meaningful image-to-object size ratio. In ray diagrams, magnification can still be discussed by considering the angles between rays rather than their sizes, particularly in systems like telescopes where rays are parallel. When drawing ray diagrams, it's essential to account for rays emanating from multiple points to define the image accurately. The concept of magnification may also vary based on context, such as when observing the apparent spacing of objects at infinity. Ultimately, the lack of size for a point object leads to ambiguity in defining magnification.
Krushnaraj Pandya
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Homework Statement


I read that the magnification for a point object is undefined, but when we draw ray diagrams we consider the rays to be emanating from a point source and also define magnification for it- what am I missing?

Homework Equations


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The Attempt at a Solution


(conceptual query)
 
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The magnification has meaning only when the image of a second point that originates at the same distance, but at a different height (a point on-axis for the second point is ok) is found. Then, the magnification is defined as ## M=\frac{h_i}{h_o} ##.
 
Krushnaraj Pandya said:

Homework Statement


I read that the magnification for a point object is undefined, but when we draw ray diagrams we consider the rays to be emanating from a point source and also define magnification for it- what am I missing?

“Magnification” can be used a couple of ways. They all can be related back to the idea of image to object ratio, but sometimes when designing an optical system it is useful to note other things that also relate to magnification.

Picture the usual single lens diagram with a magnified real image. You see the difference in the size of the object and image. However do you also notice the angle between the rays at the object and image? The ratio of the angles is also the magnification. That’s how you can show single points and still talk about magnification. If you had more than one field point the distance between them would be magnified.

Now consider an afocal telescope. The rays coming in and going out are parallel. We can see the magnification both in the decreased size of the pupil and the increased size of the angles. The more common meaning of magnification appears when you put your eye behind the telescope and the magnification translates to the apparent spacing of the objects. Note that here with the objects at infinity it becomes hard to use the object size/image size definition and the angles are the best way to define it
 
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Krushnaraj Pandya said:

Homework Statement


I read that the magnification for a point object is undefined, but when we draw ray diagrams we consider the rays to be emanating from a point source and also define magnification for it- what am I missing?

Homework Equations


--

The Attempt at a Solution


(conceptual query)
When we draw ray diagrams we consider rays from two points!
We usually draw the object as a small vertical arrow based on the axis.
We draw the "usual" rays from the head of the arrow and discovery where the image of the head of the arrow will be.
We then include the fact that the "usual" rays drawn from the base of the arrow just pass along the axis, and chose the perpendicular on the axis as the other end.
Try a ray diagram from BOTH ends of a "floating, oblique arrow" to see what you get.

Note: It is reasonable that magnification of a point object is undefined, since a single point has location, but no size. When you find the location of the image point, it will also have zero size, so we could claim any magnification we like: 2X , 3X , 12X because 2 x 0 = 0, 3 x 0 = 0, 12 x 0 = 0, so no hints as to what the answer is.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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