(II) You look at yourself in a shiny 9.2 -cm-diameter...

(II) You look at yourself in a shiny 9.2 -cm-diameter
Christmas tree ball. If your face is 25.0 $\mathrm{cm}$ away from the
ball's front surface, where is your image? Is it real or
virtual? Is it upright or inverted?

Key Concepts

Image Orientation

The orientation of an image (upright or inverted) is determined by the type of mirror and the object distance. Concave mirrors can produce either inverted or upright images depending on object distance, whereas convex mirrors always produce images that are upright relative to the object.

Virtual and Real Images

Images formed by spherical mirrors can be either real or virtual. Real images are formed when light rays actually converge and can be projected on a screen, while virtual images are formed when rays appear to diverge from a point behind the mirror. Convex mirrors, in particular, always produce virtual images since the reflected rays diverge.

Focal Length and Radius Relationship

For spherical mirrors, the focal length is directly related to the radius of curvature (R) by f = R/2 for concave mirrors. In the case of convex mirrors, the focal length is negative and still follows the relationship in magnitude, allowing us to calculate image properties from the mirror’s geometry.

Mirror Equation

The mirror equation, given by 1/f = 1/d_o + 1/d_i, relates the object distance (d_o), image distance (d_i), and the mirror's focal length (f). This fundamental equation is used to determine the position and nature of images formed by spherical mirrors, whether concave or convex.

Transcript

00:01 Okay, in this problem we have light deflecting off a spherical reflector, which we can model as a convex mirror.

00:10 And we know the diameter of this spherical deflector, this sphere.

00:17 Thus, we know its radius.

00:20 If its diameter is 9 .2 centimeters, we know its radius to be 4 .6 centimeters.

00:25 We also know that the object that we're interested in its image is located some distance 25 centimeters away.

00:35 From the surface of the sphere.

00:38 And so we're asked to find the distance to the image and essentially say everything about it.

00:43 Is it virtual? is it upright? is it magnified? what is it? so it starts off, we're going to be using the lens maker equation for ideal lenses and mirrors.

00:57 And that equation writes 1 over d0, or the distance to the object, plus 1 over d to the image is equal to 1.

01:09 One over the focal length of the mirror.

01:15 So what we need to supplement this is the fact that for a mirror, the focal length is always equal to one half the radius of curvature for that mirror.

01:25 And then we can take this as a given.

01:27 So if we combine these two results, we end up with d .i., the image distance, equal to d .0.

01:38 F or d .0 minus f.

01:45 Just a little bit of algebra.

01:49 And we use the fact that f, in this case, has to be equal to negative 2 .3 centimeters.

02:06 Okay? so when we plug in these results, we have negative 2 .106 centimeters, which is our main result.

02:20 Or approximately 2 .1 is fine...

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