﻿ ﻿OPTICS - PHYSICS TOPIC REVIEW - SAT Subject Test Physics

## SAT Subject Test Physics (2012)

### Chapter 17. OPTICS

Even though light consists of electromagnetic waves, the properties of light can be investigated by treating the wave as a straight line called a ray. This approach is known as ray approximation and will be the basis of your review of optics in this chapter.

Reflection

When light strikes a surface, it can pass through it, be absorbed by it, or bounce back from it. The process through which light bounces off a surface is known as reflection. According to the law of reflection, the angle of incidence is equal to the angle of reflection. Each angle is measured relative to the normal extending from the surface.

Smooth, shiny surfaces produce specular reflection, in which light is reflected in only one direction. For most surfaces, however, the surface is not perfectly flat. Instead, the surface is rough, even if it is only at the microscopic level. These types of surfaces produce diffuse reflection. Each incident ray still obeys the law of reflection but is reflected in a different direction according to the direction of the surface where the ray struck. Unless otherwise stated, questions on SAT Physics will assume specular reflection.

Mirrors

A mirror is an object with a reflective surface. SAT Physics will address both plane mirrors and curved mirrors.

Plane Mirrors A plane mirror is a flat mirror. When an object is placed in front of a plane mirror, light bounces off the object and is reflected by the mirror. The light rays appear to be coming from behind the mirror for an observer looking at the mirror. This type of image is a virtual image because it is formed by rays that appear to come together, but never actually do. A virtual image cannot be projected on a screen.

The image of the object is upright and the same size as the object. The distance between the object and the mirror do equals the distance between the mirror and the image di. The following diagram is an example of a ray diagram, which is a drawing that uses basic geometry to describe an image formed by a mirror.

Curved Mirrors A curved mirror appears to be part of a sphere and therefore is also known as a spherical mirror. A diverging mirror is one in which the reflective surface bulges toward the object. This type of mirror is also known as a convex mirror. A converging, or concave mirror, curves away from an object.

The point that would be at the center of that sphere is known as the center of curvature C. A line that passes from the surface of the mirror through C is known as the principal axis. The point at which the principal axis meets the mirror is the geographic center of the mirror, known as the vertex. The focal point of the mirror is halfway between the vertex and the center of curvature. Parallel light rays incident on the mirror will meet at the focal point after reflection. The distance between the focal point and the mirror is the focal length. A concave mirror has a positive focal length, whereas a convex mirror has a negative focal length value because the focal point is behind the mirror.

If an object is placed in front of a concave mirror, light rays bounce off the object and are reflected from the mirror. The reflected rays converge at a point in front of the mirror to form an image. Because the light rays do meet where the image is formed, the image is known as a real image. If a screen were placed at that point, the image would be projected on the screen.

The location of the image can be determined from the following mirror equation, where do is the object distance, diis the image distance, and f is the focal length of the mirror.

The following ray diagrams show the location of an image when do is greater than, equal to, and less than the focal length.

The magnification M of the image compares the size of the image with the size of the object. The magnification can be calculated according to the following equation.

Note that the equation includes a negative sign. Distances are considered positive if they are in front of the mirror and negative if they are behind the mirror. If the resulting magnification is positive, the image is upright and virtual. If the resulting magnification is negative, the image is inverted and real. If the magnitude of the magnification is less than 1, the image is smaller than the object. If the magnitude the magnification is greater than 1, the image is larger than the object.

If an object is placed in front of a convex mirror, light rays again bounce off the object and are reflected from the mirror. The mirror diverges the light rays so the image is formed where the rays appear to be coming from behind the mirror. The resulting image is, therefore, always virtual, and the image distance is always negative.

Refraction

If light passes through a medium rather than being reflected from it, the light can be refracted, or bent. Refraction can be explained by the fact that the speed of light changes as it travels from one medium to another. If a light ray enters straight into a medium, the entire ray speeds up or slows down together. If, however, the ray enters the medium at an angle, one part of the ray changes speed before the other. This difference causes the ray to bend.

When light travels from a less dense medium into a denser medium, such as from air to water, the ray bends toward the normal. When light travels from a denser medium into a less dense medium, such as from water to air, the ray bends away from the normal.

The amount of bending that occurs during refraction depends on the speed of light in the two mediums involved. The index of refraction n compares the speed of light in a vacuum with the speed of light in a specific medium.

According to the equation, you can see the index of refraction is dimensionless and it must always be greater than 1. The index of refraction of light in air is only slightly different than in a vacuum, so a value of 1 is used for problems involving air in SAT Physics, unless otherwise stated. In addition, the index of refraction varies with the wavelength of light. Use the information provided with each question or any tables that might be included.

Example:

The index of refraction for yellow light passing through flint glass is 1.655. At what speed does light pass through flint glass?

Example:

A glass has a refractive index of 1.5. How does the speed of light through the glass compare to the speed of light in a vacuum?

The index of refraction in a vacuum is 1.00, so . Light travels 0.67 times the speed of light in a vacuum when traveling through the glass.

Snell’s law of refraction relates the angles of incidence and refraction to the index of refraction. The relationship is summarized by the following equation, in which n1 andn2 are the indices of refraction for the two mediums, while θ1 and θ2 are the angles of incidence and refraction.

Example:

A ray of light passes from air to water at an angle of 64° to the normal line. What is the angle of refraction of the light ray? [ and ]

Total Internal Reflection

A combination of reflection and refraction can be used to explain a phenomenon known as total internal reflection. Consider the light shown in the following diagram with an angle of incidence equal to 0°. At this angle, most of the light passes directly across the boundary between the mediums. As the angle increases, the light experiences both reflection and refraction. You can see that the angle of refraction increases as the angle of incidence increases. The angle of refraction is 90° when the angle of incidence reaches a value known as the critical angle. As the angle of incidence increases beyond the critical angle, the light is no longer refracted. Instead, all of the light is reflected back into the medium from which it came.

Total internal reflection occurs only when the light source is in the denser medium. This principle makes it possible to transmit light over long distances with little attenuation through optical light fibers.

Lenses

A lens is a clear piece of glass or plastic through which light can be refracted. A diverging or convex lens is thicker at the ends than in the middle. A converging or concave lens is thicker in the middle than at the ends.

As in the case of curved mirrors, the principal axis connects the center of the lens to the point that would be the center of the sphere related to the lens. The focal point is the point at which the light rays meet after being refracted by the lens. Unlike mirrors, lenses have focal points on either side because light can pass through the lens in either direction.

In a convex lens, the light rays meet at the focal point after passing through the lens. A concave lens diverges the light rays so they do not meet after refraction. However, they can be traced back to the point where they intersect.

The following illustrations show the ray diagrams for a converging lens with the object at different distances from the lens. You can see that when the object is at infinity, the image is at the focal point. As the object is moved closer to the lens, but still outside 2F, a real image is formed that is smaller than the object. The image is between F and 2F. When the object is brought to 2F, a real image is formed that is the same size as the object. The object distance equals the image distance. If the object is brought between 2F and F, a real image is formed that is larger than the object. The image is formed beyond 2F.

A diverging lens always produces a virtual image. The image is upright and smaller than the object. It is located inside the focal length for any position of the object.

If the lens is assumed to be very thin, such that the thickness of the lens is much smaller than the focal length, the following thin-lens equation can be used to determine the location of the image. As you can see, the equation is identical to the mirror equation you used earlier.

As before, using the equation requires that you pay attention to sign conventions. The object and images distances are positive in front of the lens and negative behind the lens. The focal length is positive for a converging lens and negative for a diverging lens.

In addition, you can again use the magnification equation to describe the size and orientation of an image.

Test-Taking Hint

The rules for interpreting mirrors and lenses are similar, but not identical. If you see the terms converging, diverging, convex, or concave, make sure you determine whether the question involves mirrors or lenses. Once you know, it may help to draw a diagram to interpret the information provided.

REVIEW QUESTIONS

Select the choice that best answers the question or completes the statement.

1. A ray of light strikes a plane mirror. The angle formed by the incident ray and the reflected ray equals 80°. What is the angle of incidence?

(A) 20°

(B) 40°

(C) 50°

(D) 90°

(E) 110°

2. Light strikes a plane mirror as shown. If the incident ray makes an angle of 30° with the normal as shown below, what is θ?

(A) 30°

(B) 45°

(C) 60°

(D) 90°

(E) 120°

3. The diagram below shows two objects P and Q in front of a plane mirror. Each object is 0.5 m tall, but P is twice as far from the mirror as Q.

How does the size of P’s image compare to the size of Q’s image?

(A) P’s image is one-fourth the size of Q’s image.

(B) P’s image is half the size of Q’s image.

(C) P’s image is the same size as Q’s image.

(D) P’s image is twice the size of Q’s image.

(E) P’s image is four times the size of Q’s image.

4. Which characteristic do all images produced by convex mirrors have in common?

(A) they are magnified.

(B) they are at infinity.

(C) they are inverted.

(D) they are real.

(E) they are virtual.

5. The speed of light through a particular diamond is 1.24 × 108 m/s. What is the index of refraction of the diamond?

(A) 1.24

(B) 1.76

(C) 2.42

(D) 3.72

(E) 4.13

6. A ray of light in air strikes a boundary with crown glass at an angle of 38.0 degrees with the normal. What is the angle of refraction upon entering the glass? [ and ]

(A) 57.8°

(B) 40.0°

(C) 25.0°

(D) 23.9°

(E) 16.3°

7. Total internal refraction occurs when light reaches the critical angle of 34.4° in sapphire. What is the index of refraction in this material?

(A) 0.56

(B) 1.77

(C) 2.62

(D) 4.40

(E) 5.64

8. A 6.00-cm tall object is placed a distance of 22.4 cm from a diverging lens with a focal length of -14.3 cm. What is the image distance?

(A) –8.70 cm

(B) –1.04 cm

(C) –0.70 cm

(D) 6.99 cm

(E) 8.10 cm

Questions 9 and 10 relate to the diagram below, which shows a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm.

9. What is the image distance?

(A) 7.5 cm

(B) 15.0 cm

(C) 22.5 cm

(D) 30.0 cm

(E) 45.0 cm

10. The image produced by the object in the diagram is

(A) smaller than the object and erect

(B) the same size as the object and erect

(C) the same size as the object and inverted

(D) larger than the object and inverted

(E) larger than the object and erect

1. B The angle of incidence equals the angle of reflection. Therefore, the angle of incidence , which is 40°.

2. C The angle of reflection also equals 30°, so θ equals 90° – 30°, which is 60°.

3. C The size of the image produced by a plane mirror is the same as the size of the object. Because both objects are the same size, the images will also be the same size.

4. E The images formed by convex mirrors are located at the point to which the reflected rays can be traced. The image is virtual, erect, and smaller than the object.

5. C The index of refraction compares the speed of light in a vacuum to the speed in the diamond.

6.

7. B Use Snell’s law to find the solution, recognizing that the angle of refraction is 90° at the critical angle.

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