Question

An object is placed a distance of twice the focal length away from a diverging lens. What is the magnification of the image?

Answer 1

Answer:

1/3

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

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

$p=2f$

So we can find q from the equation:

$\frac{1}{q}=\frac{1}{(-f)}-\frac{1}{p}=-\frac{1}{f}-\frac{1}{2f}=-\frac{3}{2f}\\q=-\frac{2}{3}f$

Now we can find the magnification of the image, given by:

$M=-\frac{q}{p}=-\frac{-\frac{2}{3}f}{2f}=\frac{1}{3}$