Answer:
1/3
Explanation:
We can solve the problem by using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
Here we have a divering lens, so the focal length must be taken as negative (-f). Moreover, we know that the object is placed at a distance of twice the focal length, so
[tex]p=2f[/tex]
So we can find q from the equation:
[tex]\frac{1}{q}=\frac{1}{(-f)}-\frac{1}{p}=-\frac{1}{f}-\frac{1}{2f}=-\frac{3}{2f}\\q=-\frac{2}{3}f[/tex]
Now we can find the magnification of the image, given by:
[tex]M=-\frac{q}{p}=-\frac{-\frac{2}{3}f}{2f}=\frac{1}{3}[/tex]
