Reading: pages 574 – 580; 593 – 602 Outline: ⇒ the ray model of light (PowerPoint) ⇒ reflection specular and diffuse reflection images from plane mirrors ⇒ image formation with spherical mirrors concave and convex mirrors principle axis and radius of curvature focal point and focal length ⇒ images from spherical mirrors (PowerPoint) ray tracing for concave mirrors ray tracing for convex mirrors ⇒ the mirror equation and the magnification equation sign conventions lots of examples

Problem Solving The problems in this chapter are relatively straightforward. Whenever a light ray hits a mirror, the angle of reflection equals the angle of incidence: θr = θi. Remember that all angles are measured with respect to the normal. The image formed by a plane mirror is a virtual image that has the following properties: 1) The image is upright. 2) The image height equals the object height. 3) The image distance equals the object distance.

You should know how to find the orientation, height and position of the image from a spherical mirror (both concave and convex) by using ray tracing. Ray tracing always involves drawing three light rays: these three light rays are listed on page 594 for a concave mirror and on page 596 for a convex mirror. To find the orientation, height, and position of the image using a more mathematical approach, the mirror and magnification equations are used. 1 1 1 + = where s is the object distance, s’ is the image distance, and f s s' f is the focal length. The focal length is given by f = ½ R for a concave mirror and f = - ½ R for a convex mirror, where R is the radius of curvature.

The mirror equation is

The magnification equation is m =

hi s' = − where hi and ho are the image and object heights. ho s

It is very important to know the sign conventions for spherical mirrors given on page 764. They are listed below for your convenience. f is + for a concave mirror f is - for a convex mirror s is + if the object is in front of the mirror (real object) s is - if the object is behind of the mirror (virtual object) s’ is + if the image is in front of the mirror (real image) s’ is - if the image is behind of the mirror (virtual image) m is + for an image that is upright with respect to the object m is - for an image that is inverted with respect to the object

Questions and Example Problems from Chapter 18 Question 1 (a) When you look at the back side of a shiny teaspoon, held at arm’s length, you see yourself upright. Why? (b) When you look at the other side of the spoon, you see yourself upside down. Why?

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CHAPTER

18

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Page 602

Ray Optics

SUMMARY The goal of Chapter 18 has been to understand and apply the ray model of light.

GENERAL PRINCIPLES

Reflection

Refraction

Law of reflection: ur = ui

Snell’s law of refraction: ui

Reflection can be specular (mirror-like) or diffuse (from rough surfaces). s

Plane mirrors: A virtual image is formed at P¿ with s¿ = s, where s is the object distance and s¿ is the image distance.

P

ur

n 1 sin u1 = n 2 sin u2

s⬘ 5 s P⬘

Normal

Incident u1 ray n1 n2

Index of refraction is n = c/v. The ray is closer to the normal on the side with the larger index of refraction.

u2

Refracted ray

If n 2 6 n 1 , total internal reflection (TIR) occurs when the angle of incidence u1 is greater than uc = sin-1(n 2 /n 1).

IMPORTANT CONCEPTS The ray model of light

Image formation

Light travels along straight lines, called light rays, at speed v = c/n.

If rays diverge from P and, after interacting with a lens or mirror, appear to diverge from P¿ without actually passing through P¿, then P¿ is a virtual image of P.

A light ray continues forever unless an interaction with matter causes it to reflect, refract, scatter, or be absorbed. Light rays come from self-luminous or reflective objects. Each point on the object sends rays in all directions. Ray diagrams represent all the rays emitted by an object by only a few select rays.

P⬘

These rays appear to have come from P⬘. P

Virtual image

If rays diverge from P and interact with a lens or mirror so that the refracted rays converge at P¿ , then P¿ is a real image of P. Rays actually pass through a real image. These rays actually do come from P⬘. Real image

P

In order for the eye to see an object (or image), rays from the object or image must enter the eye.

P⬘

APPLICATIONS Ray tracing for lenses

The thin-lens equation

Three special rays in three basic situations:

For a lens or curved mirror, the object distance s, the image distance s¿, and the focal length f are related by the thin-lens equation: 1 1 1 + = s s¿ f

Converging lens Real image

Converging lens Virtual image

Diverging lens Virtual image

The magnification of a lens or mirror is m = - s¿/s. Sign conventions for the thin-lens equation:

Ray tracing for mirrors

Three special rays in three basic situations:

Quantity Positive when

Negative when

s

Not treated here Virtual image; on same side of a lens as object, or behind a mirror Diverging lens or convex mirror Image is inverted.

s¿

Concave mirror Real image

Concave mirror Virtual image

Convex mirror Virtual image

f m

Always Real image; on opposite side of a lens from object, or in front of a mirror Converging lens or concave mirror Image is upright.

Problem 1 Two plane mirrors are facing each other. They are parallel, 3.00 cm apart, and 17.0 cm in length, as the drawing indicates. A laser beam is directed at the top mirror from the left edge of the bottom mirror. What is the smallest angle of incidence with respect to the top mirror, such that the laser beam (a) hits only one of the mirrors and (b) hits each mirror only once?

Problem 2 Two plane mirrors are separated by 120o, as the drawing illustrates. If a ray strikes mirror M1 at a 65o angle of incidence, at what angle θ does it leave mirror M2?

Problem 3 A 2.0-cm-high object is situated 15.0 cm in front of a concave mirror that has a radius of curvature of 10.0 cm. Using a ray diagram drawn to scale, measure (a) the location and (b) the height of the image. The mirror must be drawn to scale.

Problem 4 A 10.0-cm-high object is situated 25.0 cm in front of a convex mirror that has a radius of curvature of 1.00 × 102 cm. Using a ray diagram drawn to scale, measure (a) the location and (b) the height of the image. The mirror must be drawn to scale.

The ray diagram is shown below. (Note: f = –50.0 cm and s = 25.0 cm)

Scale

Reflected ray

10.0 cm

Virtual Image Object 10.0 cm

25.0 cm

6.67 cm

16.7 cm

F

The ray diagram indicates that the image is 16.7 cm behind the mirror . The ray diagram indicates that the image height is 6.67 cm .

C

Problem 5 The focal length of a concave mirror is 17 cm. An object is located 38 cm in front of this mirror. Where is the image located?

Problem 6 The image behind a convex mirror (radius of curvature = 68 cm) is located 22 cm from the mirror. (a) Where is the object located and (b) what is the magnification of the mirror? Determine whether the image is (c) upright or inverted and (d) larger or smaller than the object.

Problem 7 Convex mirrors are being used to monitor the aisles in a store. The mirrors have a radius of curvature of 4.0 m. (a) What is the image distance if a customer is 15 m in front of the mirror? (b) Is the image real or virtual? (c) If a customer is 1.6 m tall, how tall is the image?

Problem 8 A clown is using a concave makeup mirror to get ready for a show and is 27 cm in front of the mirror. The image is 65 cm behind the mirror. Find (a) the focal length of the mirror and (b) the magnification.

Problem 9 A concave mirror (f = 45 cm) produces an image whose distance from the mirror is one-third the object distance. Determine (a) the object distance and (b) the (positive) image distance.

Problem 10 An object is placed in front of a convex mirror, and the size of the image is one-fourth that of the object. What is the ratio s/f of the object distance to the focal length of the mirror?