I have a biconvex lens. f=45mm. (This is a loupe). I have an object (G). This object is G=64mm broad. When I place this object in a distance of g=42mm in front of the lens I get a virtual picture.

enter image description here

I have calculated the broad of this virtual picture (B) and the virtual distance (b) from the lens of this virtual picture:

distance from the lens: enter image description hereenter image description here -(45*42)/(45-42)=630 -> -630mm -> -0.63 Meter

broad of the virtual picture: enter image description here 630/42*64 = 960 -> 960mm -> 0.96 Meter

So my expectation was: When I make practical test (a practical experiment) I should see the virtual picture in a distance of 0.63 Meter.

So I position my smartphone in a distance of 42mm behind the loupe. Then I watch through the loupe to the smartphone display with my eyes. But I do not see my smartphone now in a distance of 0.63 Meter. It is much nearer.

Question: Can you explain me this contradiction between theory an experiment.

  • $\begingroup$ Re, "...I do not see my smartphone now in a distance of 0.63 Meter." What do you mean, exactly? The virtual image of your phone, when seen through the loupe, should not only appear to be further away from your eye (or camera lens) than the actual phone, but it also should appear to be proportionately larger than the actual phone. That's why a loupe is also sometimes called a "magnifying" glass. How did you measure the distance to the virtual image? $\endgroup$ – Solomon Slow Feb 20 at 15:01
  • $\begingroup$ Yes I agree, the virtual image should also be larger then the actual phone. But this is the other thing that is different in theorie and experiment. I have calculated that it should appear 0.96 Meter broad, but it looks max. like about 0.12 Meter. I was not able to measure the distance and the broad of my virtual image. I estimate them when looking through the loupe. And it is not what I have calculated. Maybe there is any fault in my calculation? Or there is any thinking error? Would appreciate some help. $\endgroup$ – Wogehu Feb 21 at 8:45

The problem may be that you try to know the distance without measuring it.

A clear way to know where is the geometric virtual image would be to use a projection lens that gives a real image on a screen. Then, you could use the usual relation to compute the position and size of the object (Here, your virtual image would be a real object).

Otherwise, and in a much more interesting way, the optics of virtual images are linked to the brain and vision. Your brain knows where the smartphone is and it builds a sharp image pretty much at that location. Look (without magnifying glass) a book at 5 m, 2 m, 1 m to 50 cm: you always see it about the same size while the size of the image on the retina changes by a factor of 10. It is the brain that interprets the size and position of the image in the context.


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