Chapter 24 Geometric Optics

Outline of Chapter Ray Model of Light Reflection: Image Formation by Plane Mirrors Image formation by Spherical Mirrors Index of Refraction Refraction: Snell’s Law Visible Spectrum and Dispersion Total Internal Reflection Fiber Optics Refraction at a Spherical Surface

The ray approximation is used to represent beams of light. The Ray Model of Light Ray Optics (sometimes called geometric optics) involves the study of the propagation of light. Uses the assumption that light travels in a straight-line path in a uniform medium & changes its direction when it meets the surface of a different medium or if the optical properties of the medium are nonuniform. The ray approximation is used to represent beams of light.

The Ray Model of Light Often, light travels in straight lines, so we can represent it using Rays. Rays are straight lines coming from an object. This is a model or an idealization, but it is very useful for geometric optics. Light Rays Figure 32-1. Light rays come from each single point on an object. A small bundle of rays leaving one point is shown entering a person’s eye.

Reflection: Image Formation-Plane Mirrors Experiments show that rays always obey The Law of Reflection: Angle of reflection (angle that the ray makes WITH THE NORMAL to the mirror) = angle of incidence. θi = θr Figure 32-2. Law of reflection: (a) Shows a 3-D view of an incident ray being reflected at the top of a flat surface; (b) shows a side or “end-on” view, which we will usually use because of its clarity.

Image Formation by a Plane Mirror When light reflects from a rough surface, the law of reflection still holds, but the angle of incidence varies. This is called diffuse reflection. Figure 32-3. Diffuse reflection from a rough surface.

With diffuse reflection, the eye sees reflected light at all angles. With specular reflection (from a mirror), the eye must be in the correct position. Figure 32-4. A narrow beam of light shines on (a) white paper, and (b) a mirror. In part (a), you can see with your eye the white light reflected at various positions because of diffuse reflection. But in part (b), you see the reflected light only when your eye is placed correctly mirror reflection is also known as specular reflection. (Galileo, using similar arguments, showed that the Moon must have a rough surface rather than a highly polished surface like a mirror, as some people thought.)

Image Formation by a Plane Mirror Example: Reflection from Flat Mirrors Two flat mirrors are perpendicular to each other. An incoming beam of light makes an angle of 15° with the first mirror as shown. What angle θ5 will the outgoing beam make with the second mirror? Note that the angles in the Law of Reflection are with respect to mirror normal, NOT with respect to the mirror itself!) Solution. The rays are drawn in figure 32-5b. The outgoing ray from the first mirror makes an angle of 15° with it, and an angle of 75° with the second mirror. The outgoing beam then makes an angle of 75° with the second mirror (and is parallel to the incoming beam).

Example: Reflection from Flat Mirrors Solution: Geometry  θ1 + 15º = 90º, so θ1 = 75º. Law of reflection: θ2 = 75º. Also, θ2 + θ3 + 90º = 180º. So θ3 = 15º. Law of reflection: θ4 = 15º. So, θ4 = 75º. So, the outgoing ray is parallel to the incoming ray! Solution. The rays are drawn in figure 32-5b. The outgoing ray from the first mirror makes an angle of 15° with it, and an angle of 75° with the second mirror. The outgoing beam then makes an angle of 75° with the second mirror (and is parallel to the incoming beam).

When you look into a plane (flat) mirror, you see an image appearing to be behind the mirror. This is called a Virtual Image, because the light does not go through it. The distance of the image from the mirror is equal to the distance of the object from the mirror. Figure 32-7. Formation of a virtual image by a plane mirror.

Example: How tall must a full-length mirror be? A woman, 1.60 m tall, is in front of a vertical plane mirror. Find the minimum mirror height & the distance its lower edge must be from the floor, so that she can see her whole body. Assume that her eyes are 0.1 m below the top of her head. Solution: At the minimum height, light from her feet strikes the bottom edge of the mirror and reflects into her eyes; light from the top of her head strikes the top edge of the mirror and reflects into her eyes. Geometry then shows that the mirror must be 80 cm high, with its bottom 75 cm off the floor.

Example: How tall must a full-length mirror be? A woman, 1.60 m tall, is in front of a vertical plane mirror. Find the minimum mirror height & the distance its lower edge must be from the floor, so that she can see her whole body. Assume that her eyes are 0.1 m below the top of her head. Solution: At the minimum height, light from her feet strikes the bottom edge of the mirror and reflects into her eyes; light from the top of her head strikes the top edge of the mirror and reflects into her eyes. Geometry then shows that the mirror must be 80 cm high, with its bottom 75 cm off the floor. Solution: Consider the ray that leaves her foot at A. If it comes to her eye at E, this means a reflection at B, with θi = θr. So, in the figure, BD = (½)AE. We know that AE = 1.6 – 0.1 = 1.50 m. So, BD = 0.75 m. Similar analyses can be made for other rays leaving the woman, reflecting & going to her eye. For example, from the top of her head at G to get to her eye at E, the top edge of the mirror needs to be at F, which is 0.05 m below G. So, DF = 1.55 m & the mirror needs to have height FB = 1.55 – 0.75 = 0.8 m. Also, distance BD = 0.75 m = distance of the lower mirror edge above the floor.

Conceptual Example: Is the photo upside down? Close examination of this photograph reveals that in the top portion, the image of the Sun is seen clearly, whereas in the lower portion, the image of the Sun is partially blocked by the tree branches. Show why the reflection is not the same as the real scene by drawing a sketch of this situation, showing the Sun, the camera, the branch, and two rays going from the Sun to the camera (one direct and one reflected). Is the photograph right side up? Solution: Draw a sketch of the situation – the camera, sun, and branch are all above the water. Light goes directly from the sun to the camera, and also bounces off the water and to the camera. In order for the sun to appear below the branches in one image and through them in the other, the branches must be between the sun and the camera. The picture is upside down.

Solution: Solution: Draw a sketch of the situation – the camera, sun, and branch are all above the water. Light goes directly from the sun to the camera, and also bounces off the water and to the camera. In order for the sun to appear below the branches in one image and through them in the other, the branches must be between the sun and the camera. The picture is upside down.