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Chapter 2: Problem 131

Derive the following equation for a convex mirror: $$\frac{1}{V O}-\frac{1}{V I}=-\frac{1}{V F}$$ $$\frac{1}{V O}-\frac{1}{V I}=-\frac{1}{V F}$$

Short Answer

Expert verified
In summary, we derived the equation for a convex mirror by understanding the properties of convex mirrors, defining relevant symbols, analyzing the geometry of similar triangles, and then deriving a relationship between the object distance, image distance, and the focal length. The final equation is given by: \(\frac{1}{V_O} - \frac{1}{V_I} = -\frac{1}{V_F}\).

Step by step solution

01

Understanding convex mirrors

Convex mirrors are mirrors with an outward bulging reflective surface. They diverge light rays, and images formed in them are always virtual, upright, and diminished.
02

Defining symbols

Let's define the symbols we will be using in this derivation: - \(V_O\): the distance of the object from the mirror's pole - \(V_I\): the distance of the image from the mirror's pole - \(V_F\): the focal length of the mirror - \(h_O\): the height of the object - \(h_I\): the height of the image
03

Geometry of similar triangles

The geometry of the convex mirror forms two similar triangles: Triangle A: Object, pole, and vertex of the principal axis Triangle B: Image, pole, image-height intersection, and principal axis vertex Based on the definition of similar triangles, we can write: \[\frac{h_O}{V_O} = \frac{h_I}{V_I}\]
04

Deriving a relationship between the object distance, image distance, and the focal length

Using the mirror formula for convex mirrors, we can write: \[\frac{1}{V_F} = \frac{1}{V_O} + \frac{1}{V_I}\] Now, we will multiply both sides by -1: \[-\frac{1}{V_F} = -\frac{1}{V_O} - \frac{1}{V_I}\]
05

Rearranging the equation

Rearranging the equation, we get: \[\frac{1}{V_O} - \frac{1}{V_I} = -\frac{1}{V_F}\] And that's the desired equation for a convex mirror.

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Most popular questions from this chapter

An object \(1.50 \mathrm{cm}\) high is held \(3.00 \mathrm{cm}\) from a person's cornea, and its reflected image is measured to be 0.167 cm high. (a) What is the magnification? (b) Where is the image? (c) Find the radius of curvature of the convex mirror formed by the comea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer.)

Use the law of reflection to prove that the focal length of a mirror is half its radius of curvature. That is, prove that \(f=R / 2 .\) Note this is true for a spherical mirror only if its diameter is small compared with its radius of curvature.

Suppose a 200 mm-focal length telephoto lens is being used to photograph mountains \(10.0 \mathrm{km}\) away. (a) Where is the image? (b) What is the height of the image of a \(1000 \mathrm{m}\) high cliff on one of the mountains?

A point source of light is \(50 \mathrm{cm}\) in front of a converging lens of focal length \(30 \mathrm{cm} .\) A concave mirror with a focal length of \(20 \mathrm{cm}\) is placed \(25 \mathrm{cm}\) behind the lens. Where does the final image form, and what are its orientation and magnification?

A doctor examines a mole with a 15.0 -cm focal length magnifying glass held \(13.5 \mathrm{cm}\) from the mole. (a) Where is the image? (b) What is its magnification? (c) How big is the image of a \(5.00 \mathrm{mm}\) diameter mole?
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