a) The distance of the image from the mirror is 15 cm
b) The size of the image is -2 cm (inverted)
Explanation:
a)
We can solve this first part of the problem by applying the mirror equation:

where
f is the focal length
p is the distance of the object from the mirror
q is the distance of the image from the mirror
For a mirror, the focal length is half the radius of curvature, R:

For this mirror, R = 20 cm, so its focal length is
(positive for a concave mirror)
Here we also know:
p = 30 cm is the distance of the object from the mirror
So, by applying the equation, we can find q:

b)
We can solve this part by using the magnification equation:

where
y' is the size of the image
y is the size of the object
q is the distance of the image from the mirror
p is the distance of the object from the mirror
Here we have:
q = 15 cm
p = 30 cm
y = 4 cm
Solving for y', we find the size of the image:

and the negative sign means that the image is inverted.
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