Geometrical Optics

We have been emphasizing waves for several chapters, and rightly so at the atomic level. However, for everyday, real world, discussions of light we can treat the light as if it is a beam of particles that travel in a straight line (unless the light encounters a lens or mirror). This approach (non-wave approach) to the study of light is called geometrical optics.

* Why are the interference and diffraction effects of light negligible for objects you encounter in day to day activities?

REFLECTION

In Figure 36-2 where is the surface of the object located?

… what represents the wavefront?

A ray is a line that is perpendicular to the wavefront with a direction in the direction the wave is traveling.

The normal is the line perpendicular to the surface at the point at which the ray strikes the surface.

Where is the normal in Figure 36-2?

Law of Reflection The angle of the incident ray with respect to the normal is equal to the angle of the reflected ray with respect to the normal.

*If a ray is directed towards a surface such that the ray is

drawn on top of the normal (angle = 0 degrees),

what is the angle of reflection?

Make a sketch of this situation on the right side of this page.

If a ray is directed towards a surface such that the angle

between the normal and the ray is 80 degrees, what is the

angle of reflection? Make a sketch of this situation.

REFRACTION

Summary: When a ray of light passes from one optical medium into (not reflected from the surface) another optical medium (e.g. from air into glass), the ray may change direction. Let the two media be labeled 1 and 2. There is a number called the index of refraction, n, that describes the change in the speed of light in the medium. Snell’s Law describes refraction: n1 sin(theta 1) = n2 sin(theta 2) This law will be explored more later in the chapter.

The law of reflection and Snell’s Law give us the tools to discuss image formation by mirrors and lenses, microscopes and telescopes.

page 3 Reflection from Curved Surfaces

Suppose that a ray leaves focus 1 of the ellipse in Figure 2 such that it strikes the ellipse a little to the right of point A. Will this ray pass through focus 2?

page 4 Parabolic Reflection

It is an important point that a bundle of parallel rays coming directly into a parabolic mirror will be brought to a common point called the focus. The optic axis is the line that passes through the focus on its way to the deepest part

of the parabola. Make a sketch of a parabolic mirror

and show the focus and optic axis. Spherical mirrors

fail to do this, especially for rays far from the optic

axis of the mirror.

The focal length of a mirror is the distance from the mirror to the focal point on the optic axis.

page 6 Mirror Images

DEMO flat mirrors

What questions do you have on Figures 5a and 5b?

The image at A is called a virtual image. Why?

You should be able to recall that left and right are reversed when you look in a mirror. What emergency vehicle takes advantage of this fact?

Write a word with the letters reversed left to right. View the word in a mirror. Do the letters now look normal?

What questions do you have on Figure 7? You will get to practice this in lab.

page 7 The Corner Reflector

Imagine someone is riding a bicycle at night. The bicycle rider may use a flashlight to alert people ahead that the bicycle is approaching. What can the bicycle rider do (before starting the bicycle ride) to warn a car that is approaching the bicycle from the side?

What questions do you have on the analysis of the angles shown in Figure 8a?

Could the corner reflector have any use for bicycle riders?

There are four working corner reflectors on the moon. The first was placed on the surface in 1969.

http://www.anthonares.net/2006/03/bouncing-laser-beams-off-of-the-moon.html

How can a corner reflector be used to measure the distance to the moon?

d = r t If an accuracy of 1 cm is desired in the distance to the moon what should the accuracy in the time of flight be?

Measurements of the distance to the moon are made regularly. The data indicates that the size of the moon’s orbit is increasing by 3.8 cm/year. The first person to send me an email with the correct physical reason that the moon’s orbit is increasing in size will earn one bonus point.

page 8 Motion of Light Through a Medium

Light travels through a vacuum at a speed of 3 x 10⁸ m/s. The speed of light in air is only very slightly slower than the speed of light in a vacuum. By the 1850’s it was known that light travels slower in water than in air. Light travels at a "noticeably" slower speed when it is traveling through water, glass, diamond, vegetable oil, etc. However, the frequency of the light wave remains constant as the light passes through different optical media. Comment on the wavelength of the light when it is in water compared to the wavelength when it is in air.

Is there any evidence here that the speed of light is slower in glass than in air?

Why is the curvature of the wave in Figure 9 reversed for the light that passed through the lens?

page 9 Index of Refraction

The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the medium. n = c/v where v is the speed of light in the medium. The values of n for light having a wavelength of 589 nm is given in Table 1. Calculate the speed of light in Zircon.

page 10 Cerenkov Radiation

Nothing can go from a low speed to a speed faster than the speed of light in a vacuum. However, objects are allowed to travel at speeds up to 3 x 10⁸ m/s. An object can travel faster than the speed of light in water, or glass, etc. When we discussed the Doppler Effect we also observed that shock waves form when an object moves faster than the speed of sound (sonic boom). In the case of a charged particle moving faster than the speed of light in a medium this shock wave of light is called Cerenkov Radiation.

If you watch a color video of the water surrounding the core of a nuclear reactor you will notice that the light has a blue glow. This is due to the fast subatomic particles released in the reactor creating Cerenkov Radiation in the water.

page 11 Snell’s Law

Imagine that you are part of a high school marching band. The band members are arranged by rows and are marching across the concrete parking lot. Everyone is in step and traveling at the same speed. For some unknown reason the band director is not paying attention to the band. The band is moving at an angle with respect to the straight edge of the parking lot. The band marches off the end of the

parking lot into a muddy field. Describe the speed of the

band members in the mud compared to the speed of the band

members who are still on the concrete. The band members

are maintaining a steady beat with their steps (their frequency

of stepping remains constant). What happens to the distance

between the rows (the wavelength)? Make a sketch of this

on the right side of the page. Does the band formation take

on a new direction in the mud field?

Examine Figure 10. Which section of the figure has the

smallest wave velocity?

The wavelength in the new medium is l _n = l / n where n is the index of refraction.

page 12 We will skip the derivation of Snell’s Law.

n1 sin(theta 1) = n2 sin(theta 2)

A ray of light that has a wavelength of 589 nm is moving in air. It strikes a piece of crown glass at an angle of 63 degrees with respect to the normal. Calculate the angle of the ray inside the glass.

A ray of light that has a wavelength of 589 nm is moving in crown glass. It strikes the surface at an angle of 35.89 degrees with respect to the normal and then is refracted out into the air. Calculate the angle of the ray in air just outside the glass.

page 13 Internal Reflection

A ray of light that has a wavelength of 589 nm is moving in crown glass. It strikes the surface at an angle of 63 degrees with respect to the normal. Use Snell’s Law to calculate the angle of the ray in air just outside the glass.

In Figure 14 the vertical ray was added to the photo to show the location of the normal. The first ray on the right went through the water and is shown as a broader ray on a card above the water. Why does it appear that the two rays on the left were reflected back down into the water instead of refracted out into the air?

Calculate the angle of the ray in water that would be refracted with an angle of 90 degrees from the normal in air. This angle is called the critical angle.

Aside from the scattering that occurs at the boundary between the two media, 100% of the light is reflected when the angle of the ray in the more optically dense medium is larger than the critical angle.

page 14 Fiber Optics

Other than wireless communication, what is used to carry telephone and data communications (i.e. Internet) from one location to another across this country?

Fiber optic cable is made from very clear glass. The fibers are flexible (to a degree) and can be bundled together and wrapped in a protective layer. As shown in Figure 16b, one fiber optic strand can carry much more information and separate channels of communication than one copper wire.

Page 15 Medical Imaging

The flexible nature of optical fibers allow for the imaging of various regions inside the human body. Do you think this is just a matter of guiding the bundle of fibers as shown in Figure 17 to the site of interest or is something else necessary to make the image shown in Figure 18?

Prisms

You may have wondered why the indices of refraction given on page 9 were noted to be values for light having a wavelength of 589 nanometers. The answer to your question is that the index of refraction changes as the wavelength changes. Some typical values for the index of refraction of glass at selected wavelengths is shown in Figure 20. Why is blue light "bent" more (refracted more) than red light by the prism?

Is this the same behavior we saw for the diffraction grating?

Page 16 Rainbows

*What two effects are involved in producing a rainbow?

Think back to the last time you saw a rainbow. Where was the sun in the sky in relation to the rainbow?

What questions do you have on the Figures in the text?

Page 17 The Green Flash we will skip this section

I am very much opposed to the author’s suggestion that you use binoculars to view the sunset at "just the last couple of seconds." Binoculars gather light to make the image of an object brighter. If you view the sun through binoculars you run a great risk of damaging your retina from a deposit of too much light. Don’t do this activity. Instead, go to this web site and click on the various links to view pictures of the green flash. http://mintaka.sdsu.edu/GF/pictures.html

Page 18 Halos and Sun Dogs We will not discuss this section.

Lenses

The formation of images by lenses is an everyday and important effect. We will discuss the simple models of lenses called the thin lens. A thin lens has a thickness much smaller than its diameter. We will use simple ray tracing techniques and a simple formula to locate the images produced by thin lenses.

Page 19 Spherical Lens Surface

Spherical lenses, just as was the case for spherical mirrors, suffer from spherical aberration. One easy solution for this problem is to use an aperture that only allows light to enter the lens near the optic axis. This is shown in Figure 26b.

Page 20 Focal Length of a Spherical Surface

What questions do you have on the derivation of equation 13?

Calculate the focal length for crown glass that has a radius of curvature of 14 cm. The glass is in air.

Calculate the focal length for crown glass that has a radius of curvature of 14 cm. The glass is in water.

Page 21 Aberrations

What is one cure for spherical aberration? (Be a little more specific than the cure mentioned above.)

*Define chromatic aberration.

What is an achromat?

We will not discuss astigmatism.

Page 22 Why was the Hubble space telescope out of focus when it was placed in orbit?

What was the cure for this condition?

Why don’t reflecting telescopes suffer from chromatic aberration?

Page 23 Thin Lenses

You should make a drawing of a lens (Figure 29) and verify that two parallel rays are brought to a focus by the process of refraction.

We will do some calculations with the lens maker’s equation.

Page 24 The Lens Equation

For thin lenses we can use a simpler equation than the lens maker’s equation.

What questions do you have on the derivation of equation 16?

f = focal length i = image distance o = object distance

(1/f) = (1/o) + (1/i)

Ray Tracing

Types of lenses: A converging lens is thicker in the middle than at the edge. It will bend incoming rays that are parallel to a focus. A diverging lens is thinner in the middle than at the edge. It will bend incoming rays that are parallel into a diverging pattern of rays.

In order to locate the position of the image formed by a lens you need to draw at least 2 rays through the lens. There are three rays that are easy to draw for a converging lens.

We will practice ray tracing and calculations that use the lens equation.

For my drawings o is positive to the left of the lens and i is positive to the right of the lens.

Page 26 Negative Image Distance

When a converging lens is used and the object is closer to the lens than the focal length a virtual image is formed. This image has a negative number for the focal length. Suppose that f = 10 cm and o = 7 cm. Calculate the image location, i. Make a ray tracing that locates the image.

Negative Focal Length and Diverging Lenses

There are three rays that are easy to draw for a diverging lens.

A diverging lens has a negative focal length value for use in the lens equation. By tradition the lens equation has only one form. By assigning negative values to the focal length the same equation can be used for image calculations with diverging lenses.

Suppose that f = -10 cm and o = 15 cm. Calculate the image location, i. Make a ray tracing that locates the image.

*Does a bi-convex lens have a negative or positive focal length?

page 28 Multiple Lens Systems

Suppose that you have two converging lenses. The light from an object to the left of the lens passes through one lens and an image is formed 15 cm to the right of the lens. Describe the new image location (no calculation) when the second converging lens is placed 5 cm to the right of the first lens.

Suppose that an object is 40 cm to the left of a lens that has a focal length of 25 cm. A second lens having a focal length of 15 cm is placed 5 cm to the right of the first lens. Make a ray tracing that locates the image with only the first lens in place. Draw in the second lens and perform the ray tracing that locates the final image.

Perform the calculations necessary to compare to your ray tracing. The image formed by the first lens becomes the object for the second lens. If this image is to the right of the second lens you need to use a negative value for the object distance to the second lens.

page 29 Two Lenses Together We will skip this section.

page 30 Magnification

You noticed that the image size is not usually equal to the object size. The ratio of these sizes is the magnification value m = -i/o . If the magnification calculation yields a negative for the final value then the image is inverted with respect to the orientation of the object. If m is a positive value than the image is upright, in the same orientation as the object.

When two lenses are used the overall magnification is the multiplication of the magnification numbers for each lens.

Calculate the magnification values for the calculations done earlier for lenses.

page 31 The Human Eye (optics only, no A&P)

We will consider just a few parts of the eye: cornea, lens, retina and pupil.

The cornea and lens form an image of the objects we see. Where should the focused image be located in the eye?

We will consider the cornea and lens to be a single lens for purposes of our calculations. Consider the lens equation (1/f) = (1/o) + (1/i) . Which of these variables is fixed for a human eye?

How is it possible that objects at different distances from the eye (o) all have the same image distance (i)?

Suppose that a certain eye has a cornea/lens focal length of 1.8 cm. Calculate the image distance for these cases: o = 30 cm o = 200 cm o = 200,000 cm

The ability of the eye to focus on objects at different distances is called accommodation. The ciliary muscle changes the shape of the lens and thus changes its focal length. The smallest object distance at which a focused image can be formed is called the near point. Pick up this page and hold it far from your eye. If the letters are in focus then move the page slowly toward your eye until the letters become blurry. Make a rough measurement of this distance. This is your near point. 25 cm is often used as an average value of the near point of a young adult (you!). Babies have a near point of a few cm. Senior citizens often have near points (without the aid of glasses) of more than 100 cm.

Nearsightedness and Farsightedness

*Describe the distance of in-focus objects from the eye for a person who is nearsighted.

*Describe the distance of in-focus objects from the eye for a person who is farsighted.

Nearsightedness results when the eye

is too long or the cornea/lens system

has a focal length that is too short.

Farsightedness results when the eye

is too short or the cornea/lens system

has a focal length that is too long.

What type of lens, diverging or converging, should be used to allow a nearsighted person to see mountains in the distance?

As individuals age the lens in the eye becomes less flexible. Then the ciliary muscle cannot change the shape of the lens enough to create a focused image for the wide range of object distances that humans generally want to view. What type of glasses provide some help with this condition?

page 33 The Camera

How does the eye and a camera control the amount of light entering the system?

How does the lens of a camera differ from the lens of the eye?

How is a focused image produced in a camera for various object distances?

Compare the detection of light by the camera and by the retina.

f number = focal length of the lens / diameter of the lens

Calculate the f number for the human eye.

page 34 Depth of Field

When the aperture is smaller a wider range of object distances will be in focus.

page 36 Eye Glasses and a Home Lab Experiment

The power of a lens is measured in diopters. The diopter value is found by 1/f where f is measured in meters. This is the number that is written on prescriptions for eyeglasses.

page 37 The Eyepiece (Simple Magnifier)

Take a converging lens and hold it next to this page. Slowly lift the lens away from the page and observe the size of the image. Stop when the image becomes inverted. Before the image became inverted were you viewing a real image or a virtual image?

Calculate the magnification for the situation of f = 8 cm o = 6 cm.

page 38 The Magnifier

We used the lens formula to calculate the magnification for a converging lens used as a magnifier. For the typical near point value of 25 cm a quick way to calculate the magnification is m = 25cm / f . This is for a relaxed eye. If the eye is straining a little then the magnification calculation is m = (25cm / f ) + 1 .

Use the relaxed eye formula for magnification and calculate the magnification for f = 8 cm.

Material not in the text: MIRRORS

The study of image formation by mirrors has many common features to the study of image formation by lenses. We will use the same formula for our calculations 1/f = 1/o + 1/i and similar rules for ray tracing the location of the image.

Converging Mirrors

A concave mirror will converge incoming parallel rays to a focus. It forms images in a similar way to a convex lens. It has a positive focal length, but the image distance is labeled positive when the image is to the left of the mirror (object is to the right again). It can form real or virtual images.

Suppose a concave mirror has a focal length of 20 cm. Let the object distance be 55 cm. Locate the image by calculation and by ray tracing.

Diverging Mirrors

A convex mirror is similar to the diverging (concave) lens. It only forms virtual images. It has a negative value for its focal length. Suppose a convex mirror has a focal length of -20 cm. Let the object distance be 55 cm. Locate the image by calculation and by ray tracing.

page 40 Telescopes

A telescope has at least two optical elements. The first element (the objective) gathers light and forms an image. The second element (eyepiece) is used to view the image. A refracting telescope uses a lens for the objective. A reflecting telescope uses a mirror for the objective.

As is shown in Figure 56b, the objective forms an image between the objective and the eyepiece. The focal length of the objective for real telescopes is quite a bit larger than the focal length shown in the drawing. The overall magnification is found by dividing the focal length of the objective by the focal length of the eyepiece. m = f_o / f_e . How is the magnification of a telescope usually changed?

Why is a large diameter refracting telescope hard to produce and use?

The primary task of a telescope is to gather light. What property of the objective determines how much light is gathered?

Calculate the ratio of light collected by a 12 inch telescope to the light collected by the unaided human eye.

What aberration is not present in reflecting telescopes that is present in refracting telescopes?

pages 43 – 49 This is interesting reading but we will skip these pages.

page 50 Microscopes

We will study the compound microscope that consists of two converging lenses. The objective is the lens near the object. The objective’s main purpose is to produce a magnified image. Should the focal length of the objective be short or long?

As was the case for the telescope, the image produced by the objective will be viewed and further magnified by the eyepiece. What questions do you have on the derivation of equation 49? m = (L/f_o) (25cm/ f_e )

Calculate the magnification for a compound microscope that has a 5 cm focal length objective and a 3cm focal length eyepiece.

How is the magnification of a microscope changed?

page 51 Scanning Tunneling Microscope Interesting reading but we will skip this section.

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