## SAT Physics Geometric Optics - Ray Model Of Light

RAY MODEL OF LIGHT
Light can be viewed in many ways. Each model of light has its advantages in solving specific problems. In Chapter 15, “Waves,” light was encountered as a sinusoidal form. In certain problems, it is advantageous to view light as advancing wave fronts (crests). In this chapter, the angles at which light strikes a surface will be important. So a different model of light is needed. The three models of light are represented visually in Figure 16.1. Figure 16.1. Three models of light

The ray model of light shows the path of light as a straight line with an arrow indicating the direction of the light. The advantage of using the ray model of light is it allows angles of reflection and refraction to be measured.

REFLECTION
When light strikes a flat, polished surface, it is reflected in a manner similar to a ball bouncing off of a wall or the floor. In Figure 16.2, an incident ray (inbound ray) of light is shown striking a surface at an incident angle of θr Whenever surfaces are involved, angles are always measured from a normal line. The normal is a line drawn perpendicular to the surface area. The resulting reflected ray bounces off the surface with an angle of reflection of θr. Figure 16.2. Reflection

The angle of reflection is the same as the incident angle. This relationship is known as the law of reflection.
θi = θr
Specular reflection occurs when a surface is flat, smooth, and polished; all the reflected rays leave the surface parallel to one another; and the image of the reflection appears similar to the original object. Rough surfaces create diffuse reflection, where the irregularities of these surfaces cause the reflected rays to move off in random directions. Reflection allows us to see these rough surfaces, but clear images of objects cannot be reflected by rough surfaces.

Plane Mirror
The simplest mirror is a flat mirror, known as a plane mirror. When an object, such as a person, is viewed in a plane mirror, several rays of light can be visualized as starting from the object, moving toward the mirror, and reflecting off of the mirror’s surface. Each ray follows the law of reflection. Images are formed where rays intersect. However, the rays starting from the head of the person in Figure 16.3 diverge (spread out). If the reflected rays are traced backward (back trace), shown as dashed lines in Figure 16.3, then an intersection can be found. The image will appear at the intersection of reflected rays. Figure 16.3. Reflection off a plane mirror

The image formed by a plane mirror has several characteristics.
1. The image is upright. It appears right side up.
2. The image is the same size as the object, hi = h0, resulting in a magnification equal to one, M= 1.
3. The image distance and object distance are equal, di = d0.
4. The image is a virtual image. No light passes through the mirror. Therefore, intersecting light does not form the image. This means the image cannot be projected onto a screen, which is an important characteristic of a virtual image. Yet virtual images can be seen. An image is easily recognized as a virtual image since it is always upright.
5. Plane mirrors reverse front and back.
REFRACTION
When light moves into a new medium with a different density, its speed and wavelength change. If the light strikes this medium at an angle, the ray will bend at the surface of the new medium. The bending of light due to a change in density is known as refraction.

Index of Refraction
The index of refraction, n, is a ratio of the speed of light in a vacuum, c, to the speed of light in the medium, v, in which the light is moving: Since both c and v are measured in meters per second, the units cancel and the index is simply a value with no units. Light moves the fastest in a vacuum and is slower in a medium. This means the numerator, c, will either be equal to or greater than the denominator, v. As a result, the index of refraction can never be less than 1. When light is traveling in a vacuum, v and c have the same value and the index of refraction is equal to 1, nvacuum = 1. Air has a very low density. The speed of light in air is very nearly equal to the speed of light in a vacuum. This means that the accepted value for the index of refraction of air is also 1, nair = 1. Knowing that the indexes of refraction for a vacuum and for air are both equal to 1 is often important when solving problems. The index of refraction increases as optical density increases.

Snell’s Law
The amount that light refracts (bends) when it enters a new medium can be determined using Snell’s law.
n1 sin θ1 = n2 sin θ2
Snell’s law shows the relationship between the indexes of refraction of mediums 1 and 2 and the angles of the light rays when light moves from medium 1 to medium 2. Figure 16.4 shows two examples. Figure 16.4(a) shows a ray moving from medium 1 with a lower optical density into medium 2 with a higher optical density. The light ray is bent toward the dashed normal line. Figure 16.4(b) on the right shows a ray moving from medium 1 with a higher optical density into medium 2 with a lower optical density. This time the light ray bends away from the dashed normal line. In both diagrams, the larger angle is found in the medium that has the lower optical density and the lower index of refraction. The smaller angle is located in the optically denser medium with the higher index of refraction. Figure 16.4. Snell’s law of refraction

Refraction has some specific characteristics that are worth noting.
1. Refraction occurs only if the two mediums have different optical densities and different indexes of refraction. If the indexes of refraction are equal, no refraction occurs even though the light travels through two different mediums.
2. In order for light to refract, it must strike the boundary between the two mediums at an angle that is not perpendicular to the surface. If light hits the boundary perfectly perpendicular to the surface (the light ray follows the normal), then θ1 = 0° and no bending due to refraction occurs. In this case, the light will continue along the normal line without bending, θ2 - 0°. The speed of the light will change upon entering the new medium. However, the effect will not be visibly noticeable as the angle with reference to the normal line is zero.
3. The optically denser medium will have the larger index of refraction, n, the slower speed of light, v, the shorter wavelength, A, and the smaller angle, θ.
Index of Refraction and Snell’s Law A beam of light moving in air strikes the surface of a lake at an angle of 45°, as shown in the diagram above. Which statement below is true?
(A) nwater  <  nair and θ < 45°
(B) nwater  =  nair and θ = 45°
(C) nwater  >  nair and θ > 45°
(D) nwater  <  nair and θ > 45°
(E) nwater  <  nair and θ < 45°

WHAT'S THE TRICK?

The denser medium has the greater index of refraction, n, the slower speed of light, v, the shorter wavelength λ, and the smaller angle, θ. This problem addresses only the index of refraction and the angle of refraction. Since water is denser than air, it will have a larger n and a smaller θ. The answer is E.

Total Internal Reflection
A special case involves light moving from a medium of higher density into a medium of lower density. Since the second medium has a lower density, the ray of light traveling through it will refract through a larger angle as measured from the normal. If the angle in the first medium is steadily increased, then the larger angle in the second medium will also increase, as shown Figure 16.5. This figure depicts a light source under water shining upward into air. Several rays with different incident angles are shown for comparison. Figure 16.5. Total internal reflection

The light ray labeled A strikes the interface between the mediums perpendicular to the surface and does not refract. Ray B shows traditional refraction. In ray C, the angle in the first medium has become so large that it results in a refracted angle equal to 90°. The angle in the first medium that creates this 90° refraction is known as the critical angle,θC. Snell’s law still applies, but the critical angle, θC, replaces θ1 and 90° replaces θ2.
n1 sin θ1 = n2 sin θ2
n1 sin θc = n2 sin 90°
n1 sin θc = n2(l)
In the scenario described above, the light ray bent at 90° never enters the second medium. No refraction occurs. So where does the light go? As mentioned earlier, many things are possible when light strikes a new medium. Not only can light bend due to refraction, but it can simply reflect. Usually a little of both is happening simultaneously. However, if no light is refracting, then all of the light must now be reflecting. This phenomenon is known as total internal reflection. All of the light (total) stays in medium one (internal), where it has been reflected (reflection). Total internal reflection occurs at the critical angle and at any incident angle greater than the critical angle.

Critical Angle and Total Internal Reflection
When a light ray is shined from a dense medium into air and the incident angle is slowly increased, total internal reflection is first noticed when the incident angle reaches 53°. Determine the index of refraction of the initial dense medium.

WHAT'S THE TRICK?

Total internal reflection uses an adaptation of Snell’s law, where θ1 = θc and θ2 = 90°.
n1 sin θc = n2 sin 90°
The second medium is air, nair = 1. Remember that, sin 90° = 1. The critical angle,
θc, is 53°, and this is the larger angle in a 3-4-5 triangle, sin 53° =  PINHOLE CAMERA
A pinhole camera is a simple device consisting of a closed box with a tiny pinhole punctured on one side and a piece of film inside the box opposite the pinhole. If the camera is aimed at an object such as a person and the pinhole is uncovered, an image of the person is projected onto the film. In Figure 16.6, two rays of light are shown leaving the object, a person, and passing through the pinhole. One ray originates at the head of the person and the other from the feet of the person. The only way a ray of light reflected from the person’s head can enter the pinhole is at a downward angle. The only way a ray of light reflected from the person’s foot can enter the pinhole is at an upward angle. When these rays pass through the pinhole, they cause the image of the person to be inverted on the film. Figure 16.6. pinhole camera

The object distance, d0, is the distance from the object to the pinhole. The image distance, dt is the distance from the image to the pinhole. Object height, h0, and image height, h;, are also shown in the diagram above. The magnification, M, of the image can be determined using either the ratio of image height to object height or the ratio of image distance to object distance: The' signs on the variable are an important aspect when solving problems in geometric optics. In Figure 16.6, the object is upright i (+h0) while the image is inverted (-hi). The object distance will always be positive (+d0). However, the sign on the image distance depends on whether the image is real or virtual. Real images can be projected onto a screen, in this case the film. They are inverted and have a positive image distance (+di). Virtual images cannot be projected onto a screen. They are upright and have a negative image distance (-di). The image formed by a pinhole camera is real. It is inverted (-hi) and projected onto a screen (+di). The signs on image height and image distance are always opposite each other. Therefore, a negative sign has been added to the formula so that the ratios of image height and image distance are equal. The above magnification formula, along with these sign conventions, works for this simple camera and for all lens and mirrors encountered in the remainder of this chapter.
A pinhole camera works very similarly to the eye. The black pupil at the front of the eye is a small pinhole. Light rays enter the eye and are then projected as an inverted real image on the back of the eye. The brain then inverts the image so it is perceived as right side up.

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