BRAVO!!! That's correct.

Consider an object placed 50 cm in front of a concave mirror with a focal length of f = + 30 cm.

Where is the image located? Describe the image.


We can solve for the image distance di, numerically, using the "image equation",

1/di + 1/do = 1/f

1/di + 1/50 cm = 1/30 cm

1/di = 1/30 cm - 1/50 cm

1/di = (0.033 - 0.020) (1/cm)

1 /di = 0.013 (1/cm)

di = (1/0.013) cm

di = 75 cm

Since the image distance is positive, this is a real image.

While we can solve this numerically -- and have done just that -- there is usually more to learn by looking at the ray diagram for any image formation question. Start with a reasonably good sketch -- drawn to scale. Here we show the object 50 cm from the mirror and indicate the focal point with F, 30 cm from the mirror. Place the "tail" of the object on the "optic axis", the axis of symmetry of the mirror.

Look back at the principle rays if you need to.

The first principle ray is a ray of light that leaves the tip of the object parallel to the optic axis:

That ray of light is bent by the mirror so it passes through the focal point.

The second principle ray is a ray of light that leaves the tip of the object and passes through the focal point on its way to the mirror:

After being reflected by the mirror, this ray leaves parallel to the optic axis. Notice, again, that this is essentially the "reversed" version of the previous principle ray.

The third principle ray leaves the tip of the object and strikes the mirror at the optic axis:

This ray of light is bent, according to the Law of Reflection, so that the angle of reflection is equal to the angle of incidence.

All of these rays of light come together at the tip of the image,

From this ray diagram, you can see that the image is real (that means it can be projected). This real image is upside-down or inverted and it it larger than the original object. You might well estimate the image distance to be di = 75 cm, just as we calculated.

(c) 2002, Doug Davis; all rights reserved.