 Phyllis Fleming Physics Physics 106
 Outline - Geometric Optics Note that I have used:           so = object distance,  si = image distance,           Θi for the incident ray,  Θr for the reflected ray,           Θt  for the transmitted ray.
1. Descriptive Terms for Wave Motion

1. Velocity is the distance moved by the wave in one second. The velocity of a wave depends on the medium in which the wave travels.

2. Wavelength λ is the distance moved by the wave in one complete oscillation of the source. The wavelength depends on the medium in which the wave travels.

3. Frequency f is the number of oscillations of the source per second. Frequency depends on source and does not change when wave goes into a different medium.

4. Period T is the time for one complete oscillation. The period depends on the source and does not change when the wave enters a different medium.

5. Velocity v is the distance moved by the wave in one second.

6. (distance moved/oscillation)(#of oscillation/s)=(distance/s)
( λ)(f) = v

2. Light in a medium

1. The index of refraction n of a medium is defined as the ratio of the speed of light in a vacuum (or air) c to the speed of light in a medium v.
n = c/v

2. In a vacuum (or air), for light of frequency f, wavelength λo,
and velocity c,
λof = c       (Equation 1)

3. In a medium of index of refraction n,  for light of frequency f,
wavelength λn, and velocity v,
λnf = v      (Equation 2)

Dividing Eq. 1 by Eq. 2,
λo/ λn = c/v = n  or  λn= λo/n.
Wavelength is directly proportional to velocity. When the velocity of a wave in a medium decreases the wavelength decreases.

3. Descriptive Terms for Ray Diagrams

1. The Descriptive Terms

1. Angle of incidence Θi is the angle between the incident ray and the normal to the surface.

2. Angle of reflection Θr is the angle between the incident ray and the normal to the surface.

3. Angle of refraction or transmission Θt is the angle between the refracted (transmitted) ray and the normal to the surface.

2. From Snell's law,
ni sin Θi = nt sin Θt
sin Θt = ni sin Θi / nt

1. If the transmitted medium is more dense than the incident medium
(nt > ni),
sin Θt < sin Θi,
Θt < Θi,
and the light is bent toward the normal. See Fig. 1a below. 2. If the transmitted medium is less dense than the incident medium
(nt < ni),
sin Θt > sin Θi,
Θt > Θi,
and the light is bent away from the normal. See Fig. 1b below. 3. When an incident ray hits a surface normally, the angle between the incident ray and the normal is 0o and, by definition, the angle of incidence is 0o. For an angle of incidence of 0o, the angle of reflection is 0o, and the angle of refraction is 0o regardless of the relative values of ni and nt, as shown in Fig. 2 below. 3. Sample problems in 106 Problem Set for Geometric Optics: 1-3, 4, 6, 9, 11.

4. Total Internal Reflection

1. Occurs only when light goes from a denser medium to a less dense medium, that is, ni is greater than nt.  In Fig. 3 below, the smallest angle of incidence from the more dense material is shown with one arrow. The corresponding refracted ray is also shown with one arrow. As you increase the angle of incidence (ray with two arrows), the angle of refraction increases. 2. When you hit the critical angle Θc (ray with three arrows), the refracted ray skims along the interface between the two media.
For an incident angle = Θc, the angle of refraction Θt = 90o.
In general, ni sin Θi = nt sin = Θt.
For Θi = Θc,  n1 sin Θc = nt sin 90o  or  sin Θc = nt/ni.
For any angle greater than Θc,  the light is totally internally reflected as shown by the ray with four arrows in Fig. 3 above.

3. Sample problems in 106 Problem Set for Geometric Optics: 7, 8, 12.

5. Locating Images

1. You need two rays to find the location of an image.

1. For a real image, the two rays actually meet at a point.

2. For a virtual image, you trace the two rays back until you find the point at which they appear to meet.

2. Image for a plane mirror 1. In Fig. 4a, a ray from the top of the object O hits the mirror normally. The angle of incidence is 0o and the angle of reflection is 0o. The incident and reflected ray are shown with one arrow.

2. In Fig. 4b, an incident ray at angle Θi and a reflected wave ray at angle Θr are shown with two arrows.

3. In Fig. 4c, only the two reflected rays are shown. The eye "thinks" light travels in straight lines. It traces the two rays back and finds they appear to meet at the location of the top of the image I. The foot of the object, which rests on the normal to the surface, is automatically positioned below the top of the arrow.

4. For a plane mirror, the image is as far behind the mirror as the image is in front of the mirror. The object distance so = the image distance si. See solution to Problem 4 in 106 Solution Set for Geometric Optics.

3. Apparent depth of an object in a medium 1. In Fig. 5a, two rays are shown leaving the object O. The ray with one arrow hits the surface normally and continues into the other medium with no refraction. The ray at incident angle Θi in the medium of index of refraction ni> nt is refracted at angle Θt in the medium of index of refraction nt.

2. In Fig. 5b, the eye is shown tracing the rays back to the image I.

3. In Fig. 5c , the angle of incidence for the second ray is drawn small so the approximation of sin Θ ≈ tan Θ can be used.
From Fig. 5c, we see tan Θt = x/so and tan Θ1 = x/si.
From Snell's law, ni sin Θi = nt sin Θt.
For our approximation, ni x/so = nt x/si  or  si = nt so /ni.
For nt = 1,  si = so /ni.

4. Sample problems in 106 Problem Set for Geometric Optics: 4, 10.

6. Mirrors

1. For all mirrors, the relationship between the object distance so,  the image distance si, and the focal length f is
1/ so + 1/ si = 1/f.
When solving this equation, remember to find the reciprocal.  For example, if you solve for 1/si by subtracting 1/so from 1/f, you must then take the reciprocal to find si. The magnification m = - si/ so.  If m is negative, the image is inverted. For all mirrors, we take f = R/2, where R is the radius of curvature of the mirror. Draw rays to and from a straight line, but indicate what type of mirror you are using.

2. Concave Mirrors

1. Focal length is positive. The center of curvature is in front of the mirror.

2. For object distance so > f,  image is real and inverted.

3. For so > 2f,  image is diminished in size.
2f < so < f,  image is enlarged.

In Fig. 6a below, ray 1 (one arrow) goes through the focal point F and is reflected back parallel to the principal axis. Ray 2 goes through the center of Curvature C, hits the mirror normally and is reflected back along it. Ray 3 is parallel to the principal axis and reflected back through the focal point. These three rays meet at the top of the image at I’. The foot of the object O rests on the principal axis and its image I also rests there in a line with the top of the image. Image is real, inverted, diminished in size. 4. For so < f, image is virtual, erect, and magnified. The same incident rays used in Fig. 6a above are used in Fig. 6b below. The eye traces back the reflected rays, which appear to meet at the top of the image at I’. 3. Convex Mirrors

1. Focal length for convex mirror is negative. The center of curvature is at the back of the mirror.

2. Images of a convex mirror are always virtual, erect, and diminished in size.

3. In Fig. 7 below, Incident Ray 1 (one arrow) is drawn to the focal point F behind the mirror. It is reflected back parallel to the principal axis. Incident Ray 2 is drawn to the center of curvature C behind the mirror and reflected back on itself. Incident Ray 3 is drawn parallel to the principal axis and is reflected back as though it had come from the focal point F behind the mirror. The eye traces the rays back to find the head of the image I’ back of the mirror. 4. For Fig. 7,  let f = -3.4 cm  and  so = 5.1 cm.
1/ si = 1/f – 1/so.
1/si = -1/3.4 cm – 1/5.1 cm = 0.490 cm-1.
si = -2.0 cm.
The negative value of si indicates that it is a virtual image.
The magnification is m = - si/so = - (-2.0 cm)/5.1 cm = +0.39.
The image is erect and diminished in size.

4. Sample problems in 106 Problem Set for Geometric Optics: 21, 22.

7. Lenses

1. For all lenses, the relationship between the object distance so,  the image distance si, and the focal length f is
1/so + 1/si = 1/f.
The magnification m = - si/so.  Draw refracted rays from line through the center of the lens, indicating whether the lens is converging or diverging.

2. Converging Lens

1. Focal length is positive.

2. For so > f,  the image is real and inverted.

3. For so > 2f,  the image is diminished in size.
For so = 2f,  m = 1.

4. For 2f > so > f,  the image is enlarged.

5. In Fig. 8a below, Ray 1 (one arrow) is drawn through the focal point F and is refracted parallel to the principle axis on the left. Ray 2 is drawn through the center of the lens and passes through it without being refracted. Ray 3 is drawn parallel to the principal axis and is refracted through the focal point F on the right. The three rays from the top of the object meet at the top of the image I’. The foot of the image I is on the principal axis. The image is real, inverted, and diminished in size. 6. For so < f,  the image is virtual, erect, and magnified.

In Fig. 8b above, the same three rays are drawn, but these rays diverge on the left side. The eye traces the rays and “sees” them as though the top of the image was at I’. For a converging lens and virtual image, the image distance is negative but greater than the object distance. The image forms behind the object.

3. Diverging Lens

1. The focal length is negative. For all object distances, the image is virtual, erect and diminished in size. The image distance is negative and less than the object distance. The image forms in front of the object. 2. In Fig. 9 above ray 1 is drawn so that it would go through the focal point on the right side of the lens. It is diverged so that it comes out parallel to the principal axis. Ray 2 continues through the center of the lens. Ray 3 is parallel to the principal axis and diverges as though it had come from the focal point to the left of the lens. For all values of so, the rays diverge as they pass through the lens and do not form an image on the right. To the eye these refracted rays appear to come from the top of the image at I’.

4. Eye Defects and Corrections

It is customary to speak about the dioptric power D of a lens, which is the reciprocal of the focal length. You must first give the focal length of the lens in meters and then take the reciprocal of this.  1 m-1 = 1 D.

1. Nearsightedness

1. Light rays originating from a long distance away (so = infinity), focus behind the retina. The largest distance the person can see is at her far point. For the contact lens or glasses that correct nearsightedness the image distance si = - far point and the object distance = so = ∞. With this correction, she will see the object at infinity as though it had come from her far point.

Since 1/f = 1/so + 1/si = 1/∞ - 1/far point = 0 - 1/far point
= -1/far point  and  f = - far point.

To correct nearsightedness, you must use a diverging lens.

2. Notice the image distance at the far point is smaller than the object distance at infinity so we could have guessed she needed a diverging lens.

2. Farsightedness

1. The ideal nearness of an object that can be seen by the eye is taken to be 25.4 cm. A person who is farsighted needs to have the object further away than 25.4 cm. The distance for which a farsighted person can see comfortably is called her near point. In correcting for farsightedness, si = - near point. The image distance is negative because it is a virtual image and so = 25.4 cm.  Now 1/f = 1/so + 1/si = 1/25.4 cm - 1/near point.

Since the near point distance is greater than 25.4 cm,
1/near point < 1/25.4 cm  and  f > 0,  so that the corrective lens will be a converging lens.

2. Notice the image distance is greater than the object distance so you would expect to use a converging lens.

Caution: Remember to express the dioptric power in m-1.

5. Sample problems in 106 Problem Set for Geometric Optics: 13-15, 17, 18, 26, 27.

8. Optical Instruments

1. Magnifying glass 1. A magnifying glass uses a converging lens.

2. Magnification is defined as Θau, where Θu is the angle the unaided eye makes with the object of height h (Fig. 10a above), and Θa is the angle the eye makes with the object of height h when assisted by a lens (Fig. 10b above). The lens allows you to move the object closer to the eye than 25.4 cm.

From Fig. 10a, you see so is approximately equal to the focal length f of the lens. Also from Fig. 10a, we find tan Θu = h/25.4 cm and tan Θa = h/f. For small angles, the angle in radians is approximately equal to the tangent of the angle.

m = Θau ≈ tan Θau = (h/f)/(h/25.4 cm) = 25.4 cm/f = magnification of magnifying glass.

2. Microscope 1. A microscope consists of two lenses. The one closest to your eye (lens 2 in Fig. 11) is called the eyepiece and the other lens is called the objective (lens 1 in Fig. 11). In Fig. 11, you locate the image of the objective with ray 1 through the center of the lens and ray 2 through the focal point of lens 1 which refracts parallel to the principal axis. When ray 2 goes through the eyepiece, it is refracted through the focal point of the eyepiece. You can then draw ray 3 because every ray that comes from the top of the object must arrive at the top of the image. You draw ray 3 so it will pass through the center of the eyepiece.

2. Light enters through the objective. You want the image of the objective

1. to be a real image, and

2. to be formed at the focal point of the eyepiece.

3. The image of the objective becomes the object for the eyepiece. Since the object of the eyepiece is at its focal point, the image of the eyepiece will be at infinity. These rays are the most restful on your eyes because your eye is set for this distance and does not need accommodation.

4. The image of the objective is found from 1/fo = 1/so1 + 1/si1, where fo is the focal length for the objective lens.

5. Magnification

1. The magnification of the objective mo = - si1/so1

2. Another formula for mo can be found in the following way taking the "length" L of the tube as si1 – fo.
First solve for si1 from 1/fo = 1/so1 + 1/si1.
1/so1 = 1/fo -1/si1 = (si1 - fo)/ si1fo   or
so1 = si1fo /(si1 - fo) .
Then mo = - si1/so1 = - si1(si1 - fo)/si1fo = - L/fo.

3. The magnification of the eyepiece is the magnification of the magnifying glass.
mE = 24.5 cm/fE
4. Total M = mo mE = - si1/so1(24.5 cm/fE) = - L/fo(24.5 cm/fE).

3. Telescope 1. Parallel rays from a distant object, but not parallel to the principal axis, are brought to focus in the focal plane. Note top of image of height h in Fig. 12 above is in the focal plane but not at the focal point.

2. This image is formed at the focal distance fE of the eyepiece. I have not shown the origin of the two rays that go from the top of the image through the center of the eyepiece and another parallel to the principal axis of the eyepiece emerging through the focal point of the eyepiece. The final image is at infinity.

3. The magnification of a telescope is defined as the
ratio of angle ΘE to Θo,
as shown in Fig. 12.  For small angles the ratio of these angles equals the ratio of their tangents.  From Fig. 12, we see that
m = tan ΘE/tan Θo = (h/fE)/ (h/fo) = fo/fE.

4. Sample problems in 106 Problem Set for Geometric Optics: 16, 19, 20, 24.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming August 8, 2002 April 25, 2003