This is called super-saturation
hope this helps :)))))))
Answer:
q = -2 m and q = -0.5 m
Explanation:
For this exercise we must use the equation of the optical constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image, respectively
Let's start with the far vision point, in this case the power of the lens is
P = -0.5D
power is defined as the inverse of the focal length in meter
f = 1 / D
f = -1 / 0.5
f = - 2m
the object for the far vision point is at infinity p = infinity
1 / f = 1 / p + i / q
1 / q = 1 / f - 1 / p
1 / q = -1/2 - 1 / ∞
q = -2 m
The sign indicates that the image is on the same side as the object
Now let's lock the near view point
D = +2.00 D
f = 1 / D
f = 0.5m
the near mink point is p = 25 cm = 0.25 m
1 / f = 1 / p + 1 / q
1 / q = 1 / f - 1 / p
1 / q = 1 / 0.5 - 1 / 0.25
1 / q = -2
q = -0.5 m
the sign indicates that the image is on the same side as the object in front of the lens
The three main cycles of an ecosystem are the water cycle, the carbon cycle and the nitrogen cycle.
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Ahhh, there aren’t option to work with :v
Try writing them down here thnx