(II) A 4.5 -cm tall object is placed 26 cm in front of a spherical mirror. It is desired to prod

(II) A 4.5 -cm tall object is placed 26 cm in front of a...

(II) A 4.5 -cm tall object is placed 26 $\mathrm{cm}$ in front of a spherical mirror. It is desired to produce a virtual image that is upright and 3.5 $\mathrm{cm}$ tall. (a) What type of mirror should be used? (b) Where is the image located? (c) What is the focal length of the mirror? (d) What is the radius of curvature of the mirror?

(II) A 4.5 -cm tall object is placed 26 $\mathrm{cm}$ in front of a spherical
mirror. It is desired to produce a virtual image that is upright
and 3.5 $\mathrm{cm}$ tall. (a) What type of mirror should be used?
(b) Where is the image located? (c) What is the focal length of
the mirror? (d) What is the radius of curvature of the mirror?

Key Concepts

Virtual Image

A virtual image is formed when the reflected rays appear to diverge from a point behind the mirror. Unlike real images, virtual images cannot be projected onto a screen and are often upright. This concept is essential in understanding how mirrors, particularly convex mirrors or concave mirrors under specific conditions, produce images based on the ray diagram.

Mirror Equation

The mirror equation (1/f = 1/do + 1/di) provides a mathematical relationship between the object distance (do), the image distance (di), and the focal length (f) of a spherical mirror. This equation is central to the analysis of image formation and is used to determine unknown parameters in optical systems.

Magnification

Magnification is defined as the ratio of the image height to the object height and can also be expressed as minus the ratio of the image distance to the object distance (m = -di/do). It gives a measure of how much larger or smaller the image is compared to the object and indicates whether the image is upright (positive magnification) or inverted (negative magnification).

Focal Length and Radius of Curvature

The focal length of a spherical mirror is related to its radius of curvature by the equation f = R/2. This relationship is fundamental to optics, as it links the geometric shape of the mirror to its focusing properties. The sign of the focal length provides insight into the type of mirror—positive for concave mirrors and negative for convex mirrors.

Sign Conventions in Mirror Problems

Consistent use of sign conventions is critical when applying the mirror equation. Typically, distances measured in the direction of the incident light are considered positive, while those measured against it are negative. These conventions help in correctly assigning the object distance, image distance, and focal length, thereby ensuring accurate calculations in mirror problems.

Transcript

00:01 Okay, in this problem we have an object that's 4 .5 centimeters tall, located 26 centimeters from some kind of spherical lens, some kind of mirror.

00:09 And we want to ultimately produce a virtual image that's 3 .5 centimeters tall.

00:14 We're first asked what kind of lens, what kind of mirror would we need to use and what are everything else we can learn from this problem? essentially the image distance, the focal length, and the radius curvature of the mirror.

00:24 So we're asked a number of things.

00:26 These all can be accomplished with a little bit of ray tracing and the main lens equation, one over d not, plus 1 over d .i equals 1 over raff.

00:35 Raff's the focal length and d, i, and d .nod, or the distances from the lens or mirror for the image and object respectively.

00:42 We also know the fact that the radius curvature is two times the focal length.

00:46 That's always the case for lenses and mirrors.

00:49 And we may need to use the definition.

00:51 The fact that the ratio of heights is also equal to the negative ratio of distances, which has been proved in a previous problem.

00:59 So it starts off, in order to produce a virtual image, that is smaller than the original object, we need to use a convex mirror.

01:10 So that can be answered immediately.

01:13 There's always the case that to produce a smaller image, that's virtual, we need a convex.

01:20 Now the rest of these are quantitative, so we're going to have to make use of our equations.

01:24 So we can do that.

01:26 We can start by using m, as i invoked earlier.

01:32 It's a ratio of heights.

01:34 It's also the ratio of distances.

01:39 This is looking a little messy.

01:40 We can deal with that.

01:51 There.

01:51 Ratio heights or a negative ratio of distances.

01:55 We know the height...

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