Characteristics of Light Electromagnetic Waves An electromagnetic wave is a wave that consists of...

Characteristics of Light

Electromagnetic Waves

• An electromagnetic wave is a wave that consists of oscillating electric and magnetic fields, which radiate outward from the source at the speed of light.

• Light is a form of electromagnetic radiation.

• The electromagnetic spectrum includes more than visible light.

The Electromagnetic Spectrum



Electromagnetic Waves, continued

• Electromagnetic waves vary depending on frequency and wavelength.

• All electromagnetic waves move at the speed of light. The speed of light, c, equals

c = 3.00 108 m/s

• Wave Speed Equationc = f

speed of light = frequency wavelength



• Waves can be approximated as rays. This approach to analyzing waves is called Huygens’ principle.

• Lines drawn tangent to the crest (or trough) of a wave are called wave fronts.

• In the ray approximation, lines, called rays, are drawn perpendicular to the wave front.



• Illuminance decreases as the square of the distance from the source.

• The rate at which light is emitted from a source is called the luminous flux and is measured in lumens (lm).

Flat Mirrors

Reflection of Light

• Reflection is the change in direction of an electromagnetic wave at a surface that causes it to move away from the surface.

• The texture of a surface affects how it reflects light.– Diffuse reflection is reflection from a rough, texture

surface such as paper or unpolished wood.– Specular reflection is reflection from a smooth,

shiny surface such as a mirror or a water surface.

Flat Mirrors

Reflection of Light, continued

• The angle of incidence is the the angle between a ray that strikes a surface and the line perpendicular to that surface at the point of contact.

• The angle of reflection is the angle formed by the line perpendicular to a surface and the direction in which a reflected ray moves.

• The angle of incidence and the angle of reflection are always equal.

Flat Mirrors

Flat Mirrors

• Flat mirrors form virtual images that are the same distance from the mirror’s surface as the object is.

• The image formed by rays that appear to come from the image point behind the mirror—but never really do—is called a virtual image.

• A virtual image can never be displayed on a physical surface.

Image Formation by a Flat Mirror

Flat Mirrors

Curved Mirrors

Concave Spherical Mirrors

• A concave spherical mirror is a mirror whose reflecting surface is a segment of the inside of a sphere.

• Concave mirrors can be used to form real images.

• A real image is an image formed when rays of light actually pass through a point on the image. Real images can be projected onto a screen.

Image Formation by a Concave Spherical Mirror

Curved Mirrors

Curved Mirrors

Concave Spherical Mirrors, continued

• The Mirror Equation relates object distance (p), image distance (q), and focal length (f) of a spherical mirror.

1

p

1

q

1

f

1

object distance

1

image distance

1

focal length

Curved Mirrors


• The Equation for Magnification relates image height or distance to object height or distance, respectively.

'–

image height image distancemagnification = –

object height object distance

h qM

h p

Curved Mirrors


• Ray diagrams can be used for checking values calculated from the mirror and magnification equations for concave spherical mirrors.

• Concave mirrors can produce both real and virtual images.

Curved Mirrors

Sample Problem

Imaging with Concave Mirrors

A concave spherical mirror has a focal length of 10.0 cm. Locate the image of a pencil that is placed upright 30.0 cm from the mirror. Find the magnification of the image. Draw a ray diagram to confirm your answer.

Curved Mirrors

Sample Problem, continued


1. Determine the sign and magnitude of the focal length and object size.

f = +10.0 cm p = +30.0 cm

The mirror is concave, so f is positive. The object is in front of the mirror, so p is positive.

Curved Mirrors



2. Draw a ray diagram using the rules for drawing reference rays.

Curved Mirrors



3. Use the mirror equation to relate the object and image distances to the focal length.

–q

Mp

1

p

1

q

1

f

4. Use the magnification equation in terms of object and image distances.

Curved Mirrors


5. Rearrange the equation to isolate the image distance, and calculate. Subtract the reciprocal of the object distance from the reciprocal of the focal length to obtain an expression for the unknown image distance.

1 1 1

–q f p

Curved Mirrors


Substitute the values for f and p into the mirror equation and the magnification equation to find the image distance and magnification.

1 1 1 0.100 0.033 0.067– –

10.0 cm 30.0 cm cm cm cm

15 cm

15 cm– – –0.50

30.0 cm

q

q

qM

p

Curved Mirrors


6. Evaluate your answer in terms of the image location and size.

The image appears between the focal point (10.0 cm) and the center of curvature (20.0 cm), as confirmed by the ray diagram. The image is smaller than the object and inverted (–1 < M < 0), as is also confirmed by the ray diagram. The image is therefore real.

Curved Mirrors

Convex Spherical Mirrors

• A convex spherical mirror is a mirror whose reflecting surface is outward-curved segment of a sphere.

• Light rays diverge upon reflection from a convex mirror, forming a virtual image that is always smaller than the object.

Image Formation by a Convex Spherical Mirror

Curved Mirrors

Curved Mirrors

Sample Problem

Convex Mirrors

An upright pencil is placed in front of a convex spherical mirror with a focal length of 8.00 cm. An erect image 2.50 cm tall is formed 4.44 cm behind the mirror. Find the position of the object, the magnification of the image, and the height of the pencil.

Curved Mirrors


Convex Mirrors

Given:

Because the mirror is convex, the focal length is negative. The image is behind the mirror, so q is also negative.

f = –8.00 cm q = –4.44 cm h’ = 2.50 cm

Unknown:

p = ? h = ?

Curved Mirrors


Convex Mirrors

Diagram:

Curved Mirrors


Convex Mirrors

2. Plan

Choose an equation or situation: Use the mirror equation and the magnification formula.

1 1 1

– and – 'p

h hp f q q

1 1 1 '

and –h q

Mp q f h p

Rearrange the equation to isolate the unknown:

Curved Mirrors


Convex Mirrors

3. Calculate

Substitute the values into the equation and solve:

1 1 1–

–8.00 cm –4.44 cm

1 –0.125 –0.225 0.100–

cm cm cm

10.0 cm

p

p

p

Curved Mirrors


Convex Mirrors3. Calculate, continued

Substitute the values for p and q to find the magnifi-cation of the image.

10.0 cm– ' – (2.50 cm)

–4.44 cm

5.63 cm

ph h

q

h

Substitute the values for p, q, and h’ to find the height of the object.

–4.44 cm

– – 0.44410.0 cm

qM M

p

Curved Mirrors

Parabolic Mirrors

• Images created by spherical mirrors suffer from spherical aberration.

• Spherical aberration occurs when parallel rays far from the principal axis converge away from the mirrors focal point.

• Parabolic mirrors eliminate spherical aberration. All parallel rays converge at the focal point of a parabolic mirror.

Spherical Aberration and Parabolic Mirrors

Curved Mirrors

Color and Polarization

Color

• Additive primary colors produce white light when combined.

• Light of different colors can be produced by adding light consisting of the primary additive colors (red, green, and blue).


Color, continued

• Subtractive primary colors filter out all light when combined.

• Pigments can be produced by combining subtractive colors (magenta, yellow, and cyan).


Polarization of Light Waves

• Linear polarization is the alignment of electro-magnetic waves in such a way that the vibrations of the electric fields in each of the waves are parallel to each other.

• Light can be linearly polarized through transmission.

• The line along which light is polarized is called the transmission axis of that substance.

Linearly Polarized Light


Aligned and Crossed Polarizing Filters


Crossed FiltersAligned Filters


Polarization of Light Waves

• Light can be polarized by reflection and scattering.

• At a particular angle, reflected light is polarized horizontally.

• The sunlight scattered by air molecules is polarized for an observer on Earth’s surface.

Multiple Choice

1. Which equation is correct for calculating the focal point of a spherical mirror?

A. 1/f = 1/p – 1/q

B. 1/f = 1/p + 1/q

C. 1/p = 1/f + 1/q

D. 1/q = 1/f + 1/p

Multiple Choice, continued

1. Which equation is correct for calculating the focal point of a spherical mirror?

A. 1/f = 1/p – 1/q

B. 1/f = 1/p + 1/q

C. 1/p = 1/f + 1/q

D. 1/q = 1/f + 1/p


2. Which of the following statements is true about the speeds of gamma rays and radio waves in a vacuum?

F. Gamma rays travel faster than radio waves.

G. Radio rays travel faster than gamma rays.

H. Gamma rays and radio waves travel at the same speed in a vacuum.

J. The speed of gamma rays and radio waves in a vacuum depends on their frequencies.

Multiple Choice, continued3. Which of the following correctly states the law of

reflection?A. The angle between an incident ray of light and the normal to the mirror’s surface equals the angle between the mirror’s surface and the reflected light ray.B. The angle between an incident ray of light and the mirror’s surface equals the angle between the normal to the mirror’s surface and the reflected light ray.C. The angle between an incident ray of light and the normal to the mirror’s surface equals the angle between the normal and the reflected light ray.D. The angle between an incident ray of light and the normal to the mirror’s surface is complementary to the angle between the normal and the reflected light ray.


4. Which of the following processes does not linearly polarize light?

F. scattering

G. transmission

H. refraction

J. reflection


5. Which kind of mirror is shown in the ray diagram?

A. flat

B. convex

C. concave

D. Not enough information is available to draw a conclusion.

Use the ray diagram below to answer questions 5–7.


6. What is true of the image formed by the mirror?F. virtual, upright, and diminishedG. real, inverted, and diminishedH. virtual, upright, and enlargedJ. real, inverted, and enlarged



7. What is the focal length of the mirror?

A. –10.0 cm

B. –4.30 cm

C. 4.30 cm

D. 10.0 cm



8. Which combination of primary additive colors will produce magenta-colored light?

F. green and blue

G. red and blue

H. green and red

J. cyan and yellow


9. What is the frequency of an infrared wave that has a vacuum wavelength of 5.5 µm?

A. 165 Hz

B. 5.5 1010 Hz

C. 5.5 1013 Hz

D. 5.5 1016 Hz


10. If the distance from a light source is increased by a factor of 5, by how many times brighter does the light appear?

F. 25

G. 5

H. 1/5

J. 1/25

Short Response

11. White light is passed through a filter that allows only yellow, green, and blue light to pass through it. This light is then shone on a piece of blue fabric and on a piece of red fabric. Which colors do the two pieces of fabric appear to have under this light?

Short Response, continued

11. White light is passed through a filter that allows only yellow, green, and blue light to pass through it. This light is then shone on a piece of blue fabric and on a piece of red fabric. Which colors do the two pieces of fabric appear to have under this light?

Answer:

The blue fabric appears blue. The red fabric appears black.


12. The clothing department of a store has a mirror that consists of three flat mirrors, each arranged so that a person standing before the mirrors can see how an article of clothing looks from the side and back. Suppose a ray from a flashlight is shined on the mirror on the left. If the incident ray makes an angle of 65º with respect to the normal to the mirror’s surface, what will be the angle q of the ray reflected from the mirror on the right?


12. The clothing department of a store has a mirror that consists of three flat mirrors, each arranged so that a person standing before the mirrors can see how an article of clothing looks from the side and back. Suppose a ray from a flashlight is shined on the mirror on the left. If the incident ray makes an angle of 65º with respect to the normal to the mirror’s surface, what will be the angle q of the ray reflected from the mirror on the right?

Answer: 65º


13. X rays emitted from material around compact massive stars, such as neutron stars or black holes, serve to help locate and identify such objects. What would be the wavelength of the X rays emitted from material around such an object if the X rays have a frequency of 5.0 1019 Hz?


13. X rays emitted from material around compact massive stars, such as neutron stars or black holes, serve to help locate and identify such objects. What would be the wavelength of the X rays emitted from material around such an object if the X rays have a frequency of 5.0 1019 Hz?

Answer: 6.0 10–12 m = 6.0 pm

Extended Response

14. Explain how you can use a piece of polarizing plastic to determine if light is linearly polarized.

Extended Response, continued

14. Explain how you can use a piece of polarizing plastic to determine if light is linearly polarized.

Answer: Polarized light will pass through the plastic when the transmission axis of the plastic is parallel with the light’s plane of polarization. Rotating the plastic 90º will prevent the polarized light from passing through the plastic, so the plastic appears dark. If light is not linearly polarized, rotating the plastic 90º will have no effect on the light’s intensity.


15. What is the distance between the surface of the mirror and the image?

Use the ray diagram below to answer questions 15–19.A candle is placed 30.0 cm from the reflecting surface of a concave mirror. The radius of curvature of the mirror is 20.0 cm.


15. What is the distance between the surface of the mirror and the image?

Answer: 15.0 cm




Answer: 10.0 cm



17. What is the magnification of the image?



17. What is the magnification of the image?

Answer: –0.500



18. If the candle is 12 cm tall, what is the image height?



18. If the candle is 12 cm tall, what is the image height?

Answer: –6.0 cm



19. Is the image real or virtual? Is it upright or inverted?



19. Is the image real or virtual? Is it upright or inverted?

Answer: real; inverted


Refraction

Refraction of Light

• The bending of light as it travels from one medium to another is call refraction.

• As a light ray travels from one medium into another medium where its speed is different, the light ray will change its direction unless it travels along the normal.

Refraction

Refraction of Light, continued

• Refraction can be explained in terms of the wave model of light.

• The speed of light in a vacuum, c, is an important constant used by physicists.

• Inside of other mediums, such as air, glass, or water, the speed of light is different and is usually less than c.

Refraction

The Law of Refraction

• The index of refraction for a substance is the ratio of the speed of light in a vacuum to the speed of light in that substance.

speed of light in a vacuum

index of refractionspeed of light in medium

cn

v

Indices of Refraction for Various Substances

Refraction

Refraction

The Law of Refraction, continued

• When light passes from a medium with a smaller index of refraction to one with a larger index of refraction (like from air to glass), the ray bends toward the normal.

• When light passes from a medium with a larger index of refraction to one with a smaller index of refraction (like from glass to air), the ray bends away from the normal.

Refraction

Refraction

Refraction

The Law of Refraction, continued

• Objects appear to be in different positions due to refraction.

• Snell’s Law determines the angle of refraction.

index of refraction of first medium sine of the angle of incidence =

index of refraction of second medium sine of the angle of refraction

sin sini i r rn n

Image Position for Objects in Different Media

Refraction

Refraction

Sample Problem

Snell’s Law

A light ray of wavelength 589 nm (produced by a sodium lamp) traveling through air strikes a smooth, flat slab of crown glass at an angle of 30.0º to the normal. Find the angle of refraction, r.

Refraction


Snell’s Law

Given: i = 30.0º ni = 1.00 nr = 1.52

Unknown: r = ?

Use the equation for Snell’s law.

–1 –1

sin sin

1.00sin sin sin sin30.0º

1.52

19.2º

i i r r

ir i

r

r

n n

n

n

Thin Lenses

Types of Lenses

• A lens is a transparent object that refracts light rays such that they converge or diverge to create an image.

• A lens that is thicker in the middle than it is at the rim is an example of a converging lens.

• A lens that is thinner in the middle than at the rim is an example of a diverging lens.

Thin Lenses

Types of Lenses, continued

• The focal point is the location where the image of an object at an infinite distance from a converging lens if focused.

• Lenses have a focal point on each side of the lens.

• The distance from the focal point to the center of the lens is called the focal length, f.

Thin Lenses

Lenses and Focal Length

Thin Lenses

Types of Lenses, continued

• Ray diagrams of thin-lens systems help identify image height and location.

• Rules for drawing reference rays

Thin Lenses

Characteristics of Lenses

• Converging lenses can produce real or virtual images of real objects.

• The image produced by a converging lens is real and inverted when the object is outside the focal point.

• The image produced by a converging lens is virtual and upright when the object is inside the focal point.

Thin Lenses

Characteristics of Lenses, continued

• Diverging lenses produce virtual images from real objects.

• The image created by a diverging lens is always a virtual, smaller image.

Thin Lenses

The Thin-Lens Equation and Magnification

• The equation that relates object and image distances for a lens is call the thin-lens equation.

• It is derived using the assumption that the lens is very thin.

1 1 1

distance from object to lens distance from image to lens focal length

1 1 1

p q f

Thin Lenses

The Thin-Lens Equation and Magnification, continued

• Magnification of a lens depends on object and image distances.

image height distance from image to lensmagnification = –

object height distance from object to lens

'–

h qM

h p

Thin Lenses

The Thin-Lens Equation and Magnification, continued• If close attention is

given to the sign conventions defined in the table, then the magnification will describe the image’s size and orientation.

Thin Lenses

Sample Problem

Lenses

An object is placed 30.0 cm in front of a converging lens and then 12.5 cm in front of a diverging lens. Both lenses have a focal length of 10.0 cm. For both cases, find the image distance and the magnification. Describe the images.

Thin Lenses


Lenses

1. Define

Given: fconverging = 10.0 cm fdiverging = –10.0 cm

pconverging = 30.0 cm pdiverging = 12.5 cm

Unknown: qconverging = ? qdiverging = ? Mconverging = ? Mdiverging = ?

Thin Lenses


Lenses1. Define, continued

Diagrams:

Thin Lenses


Lenses

2. Plan

Choose an equation or situation: The thin-lens equation can be used to find the image distance, and the equation for magnification will serve to describe the size and orientation of the image.

1 1 1 –

qM

p q f p

Thin Lenses


Lenses

2. Plan, continued


1 1 1–

q f p

Thin Lenses

Sample Problem, continuedLenses

3. Calculate

For the converging lens:1 1 1 1 1 2

– –10.0 cm 30.0 cm 30.0 cm

15.0 cm

15.0 cm– –

30.0 cm

–0.500

q f p

q

qM

p

M

Thin Lenses


3. Calculate, continued

For the diverging lens:1 1 1 1 1 22.5

– ––10.0 cm 12.5 cm 125 cm

–5.56 cm

–5.56 cm– –

12.5 cm

0.445

q f p

q

qM

p

M

Thin Lenses


4. Evaluate

These values and signs for the converging lens indicate a real, inverted, smaller image. This is expected because the object distance is longer than twice the focal length of the converging lens. The values and signs for the diverging lens indicate a virtual, upright, smaller image formed inside the focal point. This is the only kind of image diverging lenses form.

Thin Lenses

Eyeglasses and Contact Lenses

• The transparent front of the eye, called the cornea, acts like a lens.

• The eye also contains a crystalline lens, that further refracts light toward the light-sensitive back of the eye, called the retina.

• Two conditions, myopia and hyperopia, occur when light is not focused properly retina. Converging and diverging lenses can be used to correct these conditions.

Thin Lenses

Farsighted and Nearsighted

Thin Lenses

Combination of Thin Lenses

• An image formed by a lens can be used as the object for a second lens.

• Compound microscopes use two converging lenses. Greater magnification can be achieved by combining two or more lenses.

• Refracting telescopes also use two converging lenses.

Optical Phenomena

Total Internal Reflection

• Total internal reflection can occur when light moves along a path from a medium with a higher index of refraction to one with a lower index of refraction.

• At the critical angle, refracted light makes an angle of 90º with the normal.

• Above the critical angle, total internal reflection occurs and light is completely reflected within a substance.

Optical Phenomena

Total Internal Reflection, continued

• Snell’s law can be used to find the critical angle.

sin for

index of refraction of second mediumsine critical angle

index of refraction of first medium

rC i r

i

nn n

n

• Total internal reflection occurs only if the index of refraction of the first medium is greater than the index of refraction of the second medium.

Total Internal Reflection

Optical Phenomena

Optical Phenomena

Atmospheric Refraction

• Refracted light can create a mirage.

• A mirage is produced by the bending of light rays in the atmosphere where there are large temperature differences between the ground and the air.

Optical Phenomena

Dispersion

• Dispersion is the process of separating polychromatic light into its component wavelengths.

• White light passed through a prism produces a visible spectrum through dispersion.

Optical Phenomena

Rainbows

Optical Phenomena

Lens Aberrations

• Chromatic aberration is the focusing of different colors of light at different distances behind a lens.

• Chromatic aberration occurs because the index of refraction varies for different wavelengths of light.

Multiple Choice

1. How is light affected by an increase in the index of refraction?

A. Its frequency increases.

B. Its frequency decreases.

C. Its speed increases.

D. Its speed decreases.


1. How is light affected by an increase in the index of refraction?

A. Its frequency increases.

B. Its frequency decreases.

C. Its speed increases.

D. Its speed decreases.


2. Which of the following conditions is not necessary for refraction to occur?

F. Both the incident and refracting substances must be transparent.

G. Both substances must have different indices of refraction.

H. The light must have only one wavelength.

J. The light must enter at an angle greater than 0° with respect to the normal.



3. What is the focal length of the lens?

A. -12.5 cm

B. -8.33 cm

C. 8.33 cm

D. 12.5 cm



4. What is true of the image formed by the lens?

F. real, inverted, and enlarged

G. real, inverted, and diminished

H. virtual, upright, and enlarged

J. virtual, upright, and diminished


5. A block of flint glass with an index of refraction of 1.66 is immersed in oil with an index of refraction of 1.33. How does the critical angle for a refracted light ray in the glass vary from when the glass is surrounded by air?

A. It remains unchanged.

B. It increases.

C. It decreases.

D. No total internal reflection takes place when the glass is placed in the oil.


6. Which color of light is most refracted during dispersion by a prism?

F. red

G. yellow

H. green

J. violet


7. If an object in air is viewed from beneath the surface of water below, where does the object appear to be?

A. The object appears above its true position.

B. The object appears exactly at its true position.

C. The object appears below its true position.

D. The object cannot be viewed from beneath the water’s surface.


8. The phenomenon called “looming” is similar to a mirage, except that the inverted image appears above the object instead of below it.What must be true if looming is to occur?F. The temperature of the air must increase with distance above the surface.G. The temperature of the air must decrease with distance above the surface.H. The mass of the air must increase with distance above the surface.J. The mass of the air must increase with distance above the surface.


9. Light with a vacuum wavelength of 500.0 nm passes into benzene, which has an index of refraction of 1.5. What is the wavelength of the light within the benzene?

A. 0.0013 nm

B. 0.0030 nm

C. 330 nm

D. 750 nm


10. Which of the following is not a necessary condition for seeing a magnified image with a lens?

F. The object and image are on the same side of the lens.

G. The lens must be converging.

H. The observer must be placed within the focal length of the lens.

J. The object must be placed within the focal length of the lens.

Short Answer

11. In both microscopes and telescopes, at least two converging lenses are used: one for the objective and one for the eyepiece. These lenses must be positioned in such a way that the final image is virtual and very much enlarged. In terms of the focal points of the two lenses, how must the lenses be positioned?

Short Answer, continued

11. In both microscopes and telescopes, at least two converging lenses are used: one for the objective and one for the eyepiece. These lenses must be positioned in such a way that the final image is virtual and very much enlarged. In terms of the focal points of the two lenses, how must the lenses be positioned?

Answer: The focal point of the objective must lie within the focal point of the eyepiece.


12. A beam of light passes from the fused quartz of a bottle (n = 1.46) into the ethyl alcohol (n = 1.36) that is contained inside the bottle. If the beam of the light inside the quartz makes an angle of 25.0° with respect to the normal of both substances, at what angle to the normal will the light enter the alcohol?


12. A beam of light passes from the fused quartz of a bottle (n = 1.46) into the ethyl alcohol (n = 1.36) that is contained inside the bottle. If the beam of the light inside the quartz makes an angle of 25.0° with respect to the normal of both substances, at what angle to the normal will the light enter the alcohol?

Answer: 27.0°


13. A layer of glycerine (n = 1.47) covers a zircon slab (n = 1.92). At what angle to the normal must a beam of light pass through the zircon toward the glycerine so that the light undergoes total internal reflection?


13. A layer of glycerine (n = 1.47) covers a zircon slab (n = 1.92). At what angle to the normal must a beam of light pass through the zircon toward the glycerine so that the light undergoes total internal reflection?

Answer: 50.0º

Extended Response

14. Explain how light passing through raindrops is reflected and dispersed so that a rainbow is produced. Include in your explanation why the lower band of the rainbow is violet and the outer band is red.


14. Explain how light passing through raindrops is reflected and dispersed so that a rainbow is produced. Include in your explanation why the lower band of the rainbow is violet and the outer band is red.

Answer: There are three effects—a refraction, a reflection, and then a final refraction. The light of each wavelength in the visible spectrum is refracted by a different amount: the red light undergoes the least amount of refraction, and the violet light undergoes the most. (Answer continued on next slide.)


14. Answer (continued): At the far side of the raindrop, the light is internally reflected and undergoes refraction again when it leaves the front side of the raindrop. Because of the internal reflection, the final dispersion of the light is such that the violet light makes an angle of 40° with the incident ray, and the red light makes an angle of 42° with the incident ray. For an observer, the upper edge of the rainbow has the color of the light that bends farthest from the incident light, so the outer band of the rainbow is red. Similarly, the lower edge has the color of the light that bends least from the incident light, so the inner band is violet. The net effect is that the ray that is refracted the most ends up closest to the incident light, that is, the smallest angular displacement.


A collector wishes to observe a coin in detail and so places it 5.00 cm in front of a converging lens. An image forms 7.50 cm in front of the lens, as shown in the figure below.






Answer: 15 cm




16. What is the magnifi-cation of the coin’s image?




16. What is the magnifi-cation of the coin’s image?

Answer: 1.5





17. If the coin has a diameter of 2.8 cm, what is the diameter of the coin’s image?




17. If the coin has a diameter of 2.8 cm, what is the diameter of the coin’s image?Answer: 4.2 cm




18. Is the coin’s image virtual or real? upright or inverted?



18. Is the coin’s image virtual or real? upright or inverted?

Answer: virtual; upright


Interference

Combining Light Waves

• Interference takes place only between waves with the same wavelength. A light source that has a single wavelength is called monochromatic.

• In constructive interference, component waves combine to form a resultant wave with the same wavelength but with an amplitude that is greater than the either of the individual component waves.

• In the case of destructive interference, the resultant amplitude is less than the amplitude of the larger component wave.

Interference Between Transverse Waves

Interference

Interference

Combining Light Waves, continued

• Waves must have a constant phase difference for interference to be observed.

• Coherence is the correlation between the phases of two or more waves.– Sources of light for which the phase difference is

constant are said to be coherent.– Sources of light for which the phase difference is

not constant are said to be incoherent.

Incoherent and Coherent Light

Lasers

Interference

Demonstrating Interference

• Interference can be demonstrated by passing light through two narrow parallel slits.

• If monochromatic light is used, the light from the two slits produces a series of bright and dark parallel bands, or fringes, on a viewing screen.

Conditions for Interference of Light Waves

Interference

Interference

Demonstrating Interference, continued

• The location of interference fringes can be predicted.

• The path difference is the difference in the distance traveled by two beams when they are scattered in the same direction from different points.

• The path difference equals dsin.

Interference


• The number assigned to interference fringes with respect to the central bright fringe is called the order number. The order number is represented by the symbol m.

• The central bright fringe at q = 0 (m = 0) is called the zeroth-order maximum, or the central maximum.

• The first maximum on either side of the central maximum (m = 1) is called the first-order maximum.

Interference


• Equation for constructive interferenced sin = ±m m = 0, 1, 2, 3, …

The path difference between two waves = an integer multiple of the wavelength

• Equation for destructive interferenced sin = ±(m + 1/2) m = 0, 1, 2, 3, …

The path difference between two waves = an odd number of half wavelength

Interference

Sample Problem

Interference

The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is measured on a viewing screen at an angle of 2.15º from the central maximum. Determine the wavelength of the light.

Interference


Interference1. DefineGiven: d = 3.0 10–5 m

m = 2= 2.15º

Unknown: = ?

Diagram:

Interference


Interference2. Plan

Choose an equation or situation: Use the equation for constructive interference.

d sin = m


d sinm

Interference


Interference3. Calculate


–5

–7 2

2

3.0 10 m sin2.15º

2

5.6 10 m 5.6 10 nm

5.6 10 nm

Interference


Interference4. Evaluate

This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellow-green color.

Diffraction

The Bending of Light Waves

• Diffraction is a change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge.

• Light waves form a diffraction pattern by passing around an obstacle or bending through a slit and interfering with each other.

• Wavelets (as in Huygens’ principle) in a wave front interfere with each other.

Destructive Interference in Single-Slit Diffraction

Diffraction

Diffraction

The Bending of Light Waves, continued

• In a diffraction pattern, the central maximum is twice as wide as the secondary maxima.

• Light diffracted by an obstacle also produces a pattern.

Diffraction

Diffraction Gratings

• A diffraction grating uses diffraction and interference to disperse light into its component colors.

• The position of a maximum depends on the separation of the slits in the grating, d, the order of the maximum m,, and the wavelength of the light, .

d sin = ±m m = 0, 1, 2, 3, …

Constructive Interference by a Diffraction Grating

Diffraction

Diffraction

Sample Problem


Monochromatic light from a helium-neon laser ( = 632.8 nm) shines at a right angle to the surface of a diffraction grating that contains 150 500 lines/m. Find the angles at which one would observe the first-order and second-order maxima.

Diffraction



1. Define

Given: = 632.8 nm = 6.328 10–7 m

m = 1 and 2

Unknown: = ? 2 = ?

d 1

150 500lines

m

1

150 500m

Diffraction



1. Define, continued

Diagram:

Diffraction



2. Plan

Choose an equation or situation: Use the equation for a diffraction grating.

d sin = ±mRearrange the equation to isolate the unknown:

–1sinm

d

Diffraction



3. Calculate


For the first-order maximum, m = 1:

–7–1 –1

1

1

6.328 10 msin sin

1m

150 500

5.465º

d

Diffraction



3. Calculate, continued

For m = 2:

–12

–7

–12

2

2sin

2 6.328 10 msin

1m

150 500

10.98º

d

Diffraction



4. Evaluate

The second-order maximum is spread slightly more than twice as far from the center as the first-order maximum. This diffraction grating does not have high dispersion, and it can produce spectral lines up to the tenth-order maxima (where sin = 0.9524).

Diffraction

Diffraction and Instrument Resolution

• The ability of an optical system to distinguish between closely spaced objects is limited by the wave nature of light.

• Resolving power is the ability of an optical instrument to form separate images of two objects that are close together.

• Resolution depends on wavelength and aperture width. For a circular aperture of diameter D:

1.22

D

Resolution of Two Light Sources

Diffraction

Lasers

Lasers and Coherence

• A laser is a device that produces coherent light at a single wavelength.

• The word laser is an acronym of “light amplification by stimulated emission of radiation.”

• Lasers transform other forms of energy into coherent light.

Lasers

Applications of Lasers

• Lasers are used to measure distances with great precision.

• Compact disc and DVD players use lasers to read digital data on these discs.

• Lasers have many applications in medicine.– Eye surgery– Tumor removal– Scar removal

Components of a Compact Disc Player

Lasers

Multiple Choice

1. In the equations for interference, what does the term d represent?

A. the distance from the midpoint between the two slits to the viewing screen

B. the distance between the two slits through which a light wave passes

C. the distance between two bright interference fringes

D. the distance between two dark interference fringes


1. In the equations for interference, what does the term d represent?

A. the distance from the midpoint between the two slits to the viewing screen

B. the distance between the two slits through which a light wave passes

C. the distance between two bright interference fringes

D. the distance between two dark interference fringes


2. Which of the following must be true for two waves with identical amplitudes and wavelengths to undergo complete destructive interference?

F. The waves must be in phase at all times.

G. The waves must be 90º out of phase at all times.

H. The waves must be 180º out of phase at all times.

J. The waves must be 270º out of phase at all times.


3. Which equation correctly describes the condition for observing the third dark fringe in an interference pattern?

A. dsin = /2

B. dsin = 3/2

C. dsin = 5/2

D. dsin = 3


4. Why is the diffraction of sound easier to observe than the diffraction of visible light?

F. Sound waves are easier to detect than visible light waves.

G. Sound waves have longer wavelengths than visible light waves and so bend more around barriers.

H. Sound waves are longitudinal waves, which diffract more than transverse waves.

J. Sound waves have greater amplitude than visible light waves.


5. Monochromatic infrared waves with a wavelength of 750 nm pass through two narrow slits. If the slits are 25 µm apart, at what angle will the fourth order bright fringe appear on a viewing screen?

A. 4.3º

B. 6.0º

C. 6.9º

D. 7.8º


6. Monochromatic light with a wavelength of 640 nm passes through a diffraction grating that has 5.0 104 lines/m. A bright line on a screen appears at an angle of 11.1º from the central bright fringe.What is the order of this bright line?

F. m = 2

G. m = 4

H. m = 6

J. m = 8


7. For observing the same object, how many times better is the resolution of the telescope shown on the left in the figure below than that of the telescope shown on the right?

A. 4

B. 2

C. 1/2

D. 1/4


8. What steps should you employ to design a telescope with a high degree of resolution?

F. Widen the aperture, or design the telescope to detect light of short wavelength.

G. Narrow the aperture, or design the telescope to detect light of short wavelength.

H. Widen the aperture, or design the telescope to detect light of long wavelength.

J. Narrow the aperture, or design the telescope to detect light of long wavelength.


9. What is the property of a laser called that causes coherent light to be emitted?

A. population inversion

B. light amplification

C. monochromaticity

D. stimulated emission


10. Which of the following is not an essential component of a laser?

F. a partially transparent mirror

G. a fully reflecting mirror

H. a converging lens

J. an active medium

Short Response

11. Why is laser light useful for the purposes of making astronomical measurements and surveying?


11. Why is laser light useful for the purposes of making astronomical measurements and surveying?

Answer: The beam does not spread out much or lose intensity over long distances.


12. A diffraction grating used in a spectrometer causes the third-order maximum of blue light with a wavelength of 490 nm to form at an angle of 6.33º from the central maximum (m = 0). What is the separation between the lines of the grating?


12. A diffraction grating used in a spectrometer causes the third-order maximum of blue light with a wavelength of 490 nm to form at an angle of 6.33º from the central maximum (m = 0). What is the separation between the lines of the grating?

Answer: 7.5 104 lines/m = 750 lines/cm


13. Telescopes that orbit Earth provide better images of distant objects because orbiting telescopes are more able to operate near their theoretical resolution than telescopes on Earth. The orbiting telescopes needed to provide high resolution in the visible part of the spectrum are much larger than the orbiting telescopes that provide similar images in the ultraviolet and X-ray portion of the spectrum. Explain why the sizes must vary.


13. (See previous slide for question.)

Answer: The resolving power of a telescope depends on the ratio of the wavelength to the diameter of the aperture. Telescopes using longer wavelength radiation (visible light) must be larger than those using shorter wavelengths (ultraviolet, X ray) to achieve the same resolving power.

Extended Response

14. Radio signals often reflect from objects and recombine at a distance. Suppose you are moving in a direction perpendicular to a radio signal source and its reflected signal. How would interference between these two signals sound on a radio receiver?


14. (See previous slide for question.)

Answer: The interference pattern for radio signals would “appear” on a radio receiver as an alternating increase in signal intensity followed by a loss of intensity (heard as static or “white noise”).


Base your answers to questions 15–17 on the information below. In each problem, show all of your work.

A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe.

15. What is the wave-length of the light?




15. What is the wave-length of the light?

Answer: 589 nm




16. At what angle would the third-order (m = 3) bright fringe appear?




16. At what angle would the third-order (m = 3) bright fringe appear?

Answer: 6.77º




17. At what angle would the third-order (m = 3) dark fringe appear?




17. At what angle would the third-order (m = 3) dark fringe appear?

Answer: 7.90º

Characteristics of Light Electromagnetic Waves An electromagnetic wave is a wave that consists of...

Documents

Transcript of Characteristics of Light Electromagnetic Waves An electromagnetic wave is a wave that consists of...