Answer:
A)6.15 cm to the left of the lens
Explanation:
We can solve the problem by using the lens equation:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}[/tex]
where
q is the distance of the image from the lens
f is the focal length
p is the distance of the object from the lens
In this problem, we have
[tex]f=-16.0 cm[/tex] (the focal length is negative for a diverging lens)
[tex]p=10.0 cm[/tex] is the distance of the object from the lens
Solvign the equation for q, we find
[tex]\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}[/tex]
[tex]q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm[/tex]
And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is
A)6.15 cm to the left of the lens
