Question

The focal length of a diverging lens is negative. If f = −13 cm for a particular diverging lens, where will the image be formed of an object located 42 cm to the left of the lens on the optical axis?

cm to the left of the lens

What is the magnification of the image?

Answer 1

In this case, we can write

$$\frac {1} {d}+\frac {1} {d_i}=\frac {1} {f} (1)$$

Usiing (1) we got:

$$d_i=\frac {d \times f} {f-d}$$

In our case, d=42 cm, f = −13 cm

We got:

d_i=-9.93 cm

The magnification of the image is equal to

$$k=\frac {d_i} {d} (2)$$

We got: k=0.24

Answer:

d_i=-9.93 cm, to the left of the lens

0.24