Originally Posted by
tmorse
Nobody is arguing…. 42.91cm is the correct image distance. The issue is how this image distance was calculated, and to answer that we need to break this question down in terms of vergence power.
All we need to picture is an object point located on the optical axis 60cm (or 0.60m) from lens, then strikes a +4.00 lens, and then emerges some distance from the back of this lens.
This video seems to suggests that is doesn’t matter what unit of measurement is simply plugged into Lensmakers Equation as written…
1/l + 1/f = 1/l’
But this formula is derived from Vergence Power Formula = L(in D) + F(in D) = L’(in D), which holds that an incoming vergence power L in Dipoters of vergence power (from object) strikes a (thin) lens, and can exit the back of lens with only 3 possibilities… either a divergent (minus) ray, a parallel ray, or a converging (plus) ray, depending on the power of the incoming object vergences power and the F power it strikes.
Here, L + F = L’ and our object distance is 60cm = 0.60m. So here our incoming vergence L= incoming vergence in Diopters of an object 60cm (or 0.60m) away from lens 1/l(m) = 1/-0.60m = -1.66D, or 100/-60cm if we wish object distance to stay in
cm = -1.66D
However, this Georgian College video appears to calculate this same incoming object vergence L in Diopters as
1/-60 or -0.0166D
Now our incoming vergence L (in Diopters) from object then strikes our +4.00 lens power. L + F = L’ with L(incoming ray in D) + F (in D) = L’ (exiting vergence power L’ (in D) from back of lens.
L + F = L’ with object distance 60cm = 0.60m from the lens.
So incoming vergence L (in Dipoters for object vergence power) is either
a) Georgian College calculation for L=1/l = 1/-60cm or -0.0167D or,
b) BCCO calculation L= 1/l(m) = 100/l(cm) = 1/-0.60m or 100/-60cm= -1.67D
This incoming vergence power of object ray then strikes our +4.00 lens.
In Georgian College calculation…
L + F = L’
-0.0167D + (+4.00D lens) = L’(in D) = or +3.98D image vergence power which produces a real (plus) image a distance L’ = 1/l’ or 1’/+3.98D = +0.2512m or 25.13cm to right of this +4.00 lens.
In BC College of Optics calculation…
L + F = L’
1/-0.60m + (+4.00D lens) = L’(in D)
-1.67D +(+4.00) = L’ (in D) or +2.33D image vergence power which produces a real (plus) image at l’ = 1/L’ = 1/+2.33 = +0.4291m (or +42.19cm) to the right of this +4.00 lens, which is acknowledged as the correct image location l’.
Hopes this breakdown helps.
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