Difference between revisions of Vertex distance

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→‎Calculation: still hadn't convinced myself with the wording
(→‎Calculation: add some words to describe what's going on in terms of virtual images.)
(→‎Calculation: still hadn't convinced myself with the wording)
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where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>
where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>


Conceptually, the location of the [[Optics_related_math#Virtual_image|Virtual image]] is changed because the lens has moved relative to the eye and source object.
Conceptually, the location of the [[Optics_related_math#Virtual_image|Virtual image]] is changed because the lens has moved relative to the eye and source object. For a distant source object, the image is formed at the focus point of the lens, and so simply moves with the lens.
* a minus lens creates the virtual image of a distant object near its focal point. When the lens is moved away from the eye, so too is the virtual image. This is equivalent to using a weaker minus lens.
* for a minus lens, moving the virtual image further from original location of the lens is equivalent to using a weaker minus lens
* a plus lens creates the virtual image of a near object on the other side of its focal point. When the lens is moved towards the object, the virtual image is pushed further away from lens, and therefore also the eye. This is equivalent to using a stronger plus lens.
* for a plus lens, moving the (real) image towards the original location to the lens is equivalent to using a stronger plus lens.


==References==
==References==
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