Question

When the object distance is 60 cm, a converging lens forms an image 12 cm from the lens. If the experiment is repeated with the object distance of 12 cm, what will the new image distance be? Characterize the image in each case.

Answer 1

a) Let's find the magnification of the lens in this case:

$$M=-\dfrac{d_i}{d_0}=-\dfrac{12\ cm}{60\ cm}=-0.2$$

The image is real, inverted and diminished in size.

b) Let's first find the focus length of the lens from the lens equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$f=\dfrac{1}{\dfrac{1}{d_o}+\dfrac{1}{d_i}}=\dfrac{1}{\dfrac{1}{60\ cm}+\dfrac{1}{12\ cm}}=10\ cm.$$

Then, if the object distance equals 12 cm, the image distance will be:

$$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}}=\dfrac{1}{\dfrac{1}{10\ cm}-\dfrac{1}{12\ cm}}=60\ cm.$$

The sign plus means that the image is real.

Let's find the magnification of the lens in this case:

$$M=-\dfrac{d_i}{d_0}=-\dfrac{60\ cm}{12\ cm}=-5.$$

The image is real, inverted and enlarged in size.