Is ‘velocity’the displacement of an object during specific unit of time?
2 answers:
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A) Image position: -19.3 cm
B) Image height: 1.5 cm, upright
Explanation:
A)
In order to calculate the image position, we can use the lens equation:
where
p is the distance of the object from the lens
q is the distance of the image from the lens
f is the focal length
In this problem, we have:
p = 13 cm (object distance)
f = 40 cm (focal length, positive for a converging lens)
So the image distance is
The negative sign means that the image is virtual.
B)
In order to calculate the image height, we use the magnification equation:
where
y' is the image height
y is the object height
In this problem, we have:
y = 1.0 cm (object height)
p = 13 cm
q = -19.3 cm
Therefore, the image heigth is
And the positive sign means the image is upright.