Question

A converging lens has a focal length of 19.0 cm. Locate the images for the following object distances. A) 41 cm B) 14cm

Answer 1

We can solve both parts of the problem by using the lens equation:
$\frac{1}{f} = \frac{1}{d_o}+ \frac{1}{d_i}$
where
f is the focal length
$d_o$ is the distance of the object from the lens
$d_i$ is the distance of the image from the lens

Let's also keep in mind that for a converging lens, the focal length is taken as positive, so f=+19.0 cm.

(a) In this case, $d_o = 41 cm$. So, the lens equation becomes
$\frac{1}{d_i}= \frac{1}{f}- \frac{1}{d_o}= \frac{1}{19 cm}- \frac{1}{41 cm} =0.028 cm^{-1}$
And so the distance of the image from the lens is
$d_i = \frac{1}{0.028 cm^{-1}} =+35.4 cm$
where the positive sign means the image is real, i.e. is located behind the lens (opposite side of the real object)

(b) In this case, $d_o = 14 cm$. So, the lens equation becomes
$\frac{1}{d_i}= \frac{1}{f}- \frac{1}{d_o}= \frac{1}{19 cm}- \frac{1}{14 cm} =-0.019 cm^{-1}$
And so the distance of the image from the lens is
$d_i = \frac{1}{-0.019 cm^{-1}} =-53.2 cm$
where the negative sign means the image is virtual, i.e. is located in front of the lens (i.e. on the same side of the real object)