A 4.0-cm-tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 25.0 cm. What is the nature and location of the image?

Question

Quizplus · Accepted Answer

Using the thin lens equation for a diverging lens, we get:

1/f = 1/do - 1/di

where f = -25.0 cm (negative because it is a diverging lens), do = 50.0 cm, and di is the distance of the image from the lens.

Solving for di, we get:

1/di = 1/f - 1/do
di = -50.0 cm

The negative sign indicates that the image is virtual (i.e. cannot be projected onto a screen) and located on the same side as the object. The magnification of the image is given by:

m = -di/do
m = -(-50.0 cm)/(50.0 cm) = 1.0

This means that the image is the same size as the object (i.e. 4.0 cm tall). Therefore, the only choice that matches these criteria is D, which states that the image is virtual and 1.3 cm tall (equal to 4.0 cm divided by the magnification) located 16.7 cm same side as the object.

A 40-Cm-Tall Object Is Placed 50