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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 16: OPTICS
MIRRORS AND LENSES CLASS NOTES
PLANE MIRRORS
The image formed by a plane mirror has the following properties:
1. The image is as far behind the mirror as the object is in front.
2. The image is unmagnified, virtual, and erect.
3. The image has right-left reversal.
IMAGES FORMED BY SPHERICAL MIRRORS
A spherical mirror has the shape of a segment of a sphere. Real images are formed at a point where
reflected light actually passes through the point. Virtual images are formed at a point where light rays
appear to diverge from the point.
RAY DIAGRAMS FOR MIRRORS
The point of intersection of any two of the following rays for mirrors locates the image. There are three
steps to drawing the rays.
RAY DIAGRAMS FOR MIRRORS
The following three rays form the ray diagram for a curved mirror.
1. The first ray is drawn from the top of the object (the object is drawn as an arrow). The ray is parallel
to the optical axis and is reflected back through the focal point, F.
2. The second ray is drawn from the top of the object to the vertex (center) of the mirror and is reflected
with the angle of incidence equal to the angle of reflection.
3. The third ray is drawn from the top of the object through the center of curvature, C, which is reflected
back on itself.
MIRROR DIAGRAM EXAMPLE
RULES FOR MIRROR DIAGRAMS
1. If O > 2F, the image is inverted, smaller, and located between F and 2F.
2. If O = 2F (at C), the image is inverted, the same size as the object, i.e. the distances of both the object
and image to the mirror are equal.
3. If 2F < O < F, the image is inverted, larger than the object, and located > 2F.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
REAL IMAGES
Real images form at the intersection of light rays. They are often projected on screens. Any image
formed on a screen is a real inverted image.
VIRTUAL IMAGES
A virtual image is formed because the reflected rays diverge from the surface of the mirror. The virtual
image is upright, enlarged, and behind the mirror. Virtual images are always upright.
VIRTUAL IMAGE EXAMPLE
THE MIRROR EQUATION
The mirror equation determines the location of an image:
1/si + 1/so = 1/f
si = distance of image to mirror
so = distance of object to mirror
f = focal length of mirror
Important note: si is negative in the case of virtual images.
MIRROR FOCAL POINTS
The focal point of a spherical mirror is located midway between the center of curvature and the vertex
of the mirror:
f=R/2
Example 1. An object is placed at a distance of 40.0 cm from a convex mirror along an axis. If a
virtual image forms at a distance of 50.0 cm from the mirror, on the same side of the object, what is the
focal length of the mirror?
1A.
(1) 1/si + 1/so = 1/f
(2) 1 /(- 50 cm) + 1/(40 cm) = 1/f
(3) 1/(40 cm) – 1/(50 cm) = 1/f
(4) 1/f = 5/(200 cm) – 4/(200 cm) = 1/(200 cm)
(5) f = 200 cm
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
MAGNIFICATION
For spherical mirrors, the magnification can be expressed as the ratio of image distance to object
distance. Magnification can also be expressed as the ratio between image and object height.
MAGNIFICATION EQUATIONS
M = hi / ho
M = - si / s o
Note: The negative sign in front of si means that the image is upside down.
Example 2. If a man’s face is 30.0 cm in front of a concave shaving mirror creating an erect image 1.50
times as large as the object, what is the mirror’s focal length?
2A.
(1) M = - si / so
(2) - si = M(so)
(3) si = - (1.50)(30.0 cm)
(4) si = - 45.0 cm
(5) 1/so - 1/si = 1/f
(6) 1/f = 1/(30 cm) - 1/(45 cm)
(7) 1/f = 3/(90 cm) - 2/(90 cm) = 1/(90 cm)
(8) f = 90 cm
Example 3. A concave spherical mirror has a radius of curvature of 40 cm. An object is placed 60 cm
from the mirror’s vertex.
3a. What is the focal length of this mirror?
A.
(1) f = R / 2
(2) f = 40 cm / 2
(3) f = 20 cm
3b. Sketch a ray diagram.
A.
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MR. SURRETTE
3c.
A.
(1)
(2)
(3)
(4)
(5)
(6)
VAN NUYS HIGH SCHOOL
Calculate the image distance.
1/si + 1/so = 1/f
1/si = 1/f – 1/so
1/si = 1/(20 cm) – 1/(60 cm)
1/si = 3/(60 cm) – 1/(60 cm)
1/si = 2/(60 cm)
si = 30 cm
RAY DIAGRAMS FOR THIN LENSES
The following three rays form the ray diagram for a thin lens.
1. The first ray is drawn parallel to the optic axis. After being refracted by the lens, this ray passes
through (or appears to come from) one of the focal points.
2. The second ray is drawn through the center of the lens. This ray continues in a straight line.
3. The third ray is drawn through the focal point F, and emerges from the lens parallel to the optic axis.
CONVERGING LENS EXAMPLES
DIVERGING LENS EXAMPLE
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LENSES PLACED IN SERIES
When two or more thin lenses are placed together, their combined focal point is:
1/ fs = 1/f1 + 1/ f2 …
1/ fs =  (1/fi)
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 16: OPTICS
WAVE OPTICS CLASS NOTES
WAVE THEORY OF LIGHT
The wave theory of light analyzes phenomena that cannot be explained with ray optics.
INTERFERENCE
In order to observe sustained interference in light waves, the following conditions must be met:
1. The sources must maintain a constant phase (coherent).
2. The sources must be a single wavelength (monochromatic).
3. The superposition principle must apply.
YOUNG’S DOUBLE-SLIT APPARATUS
The two slits S1 and S2 serve as coherent monochromatic sources. The path difference:
(r2 – r1) = dsin
YOUNG’S DOUBLE-SLIT APPARATUS
YOUNG’S DOUBLE-SLIT APPARATUS
The maximum bright fringe m occurs in the center of the screen where m = 0. The bright fringes on
either side of m= 0 are m = +1 and – 1. The equation to find the central bright fringe (bright central line
on the screen) is:
xm =mL / d
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
YOUNG’S DOUBLE-SLIT APPARATUS
To determine the position of all bright fringes, use the equation:
dsin = m
(m = 0, + 1, + 2, etc. The fringes get fainter as the value of m increases.)
Example 1. A Young’s double slit apparatus has a slit separation of 4.00 x 10-5 m on which a
monochromatic light beam is directed. The resultant bright fringes on a screen 1.20 m away are
separated by 2.15 x 10-2 m. What is the wavelength of the beam?
1A.
(1) xm =mL / d
(2)  = dxm / mL
(3)  = [(4.00 x 10-5 m)(2.15 x 10-2 m)] / [(1)(1.20 m)]
(4)  = 7.17 x 10-7 m
(5)  = 717 nm
Example 2. Light in the form of plane waves of a single wavelength are incident on two parallel slits.
A viewing screen is a distance D from the slits. A point P on the screen is a distance r1 from one slit and
r2 from the other.
Example 2. Diagram
2a. If the difference in the distances (r2 – r1) is 1.5 wavelengths (3/2) what would be observed at P?
A. A minimum. Destructive interference occurs when the extra distance is an odd integer multiple
of the wavelength divided by 2.
2b. If the difference in the distances (r2 – r1) is 2.0 wavelengths (2) what would be observed at P?
A. A maximum. Constructive interference occurs when the extra distance is a whole number multiple
of wavelength.
CHANGE OF PHASE DUE TO REFLECTION
An electromagnetic wave undergoes a phase change of 180o upon reflection from a medium that is
optically more dense than the one it was traveling (n2 > n1).
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VAN NUYS HIGH SCHOOL
INTERFERENCE IN THIN FILMS
Interference in thin films (like soap bubbles) will be constructive if the waves are out of phase by a
multiple of ½ . Destructive interference will occur when the phase difference is ¼ , ¾ , etc. This is
expressed:
2nt = (m + ½)
THIN FILM INTERFERENCE
Example 3. Constructive interference occurs when light of wavelength 500 nm shines on soap bubble
film (n = 1.46). What is the minimum thickness of the film?
3A.
(1) 2nt = (m + ½)
(2) 2nt = (m + ½)
(3) 2nt = ½ 
(4) 4nt = 
(5) t =  / 4n
(6) t = (500 x 10-9 m) / (4)(1.46)
(7) t = 8.56 x 10-8 m
Example 4. A possible means for making an airplane radar-invisible is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the
polymer is n = 1.50, how thick would the coating be?
4A.
(1) 2nt = (m + ½)
(2) 2nt = (m + ½)
(3) 2nt = ½ 
(4) 4nt = 
(5) t =  / 4n
(6) t = (3.00 x 10-2 m) / (4)(1.50)
(7) t = 5.00 mm
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
POLARIZATION
Light can be polarized because light waves are transverse (the electric and magnetic fields that make
up light are perpendicular to each other). Compressed longitudinal waves like sound cannot be
polarized.
POLARIZATION
Polarized light has vibrations confined to a single plane that is perpendicular to the direction of motion.
POLARIZED LIGHT
BREWSTER’S LAW
Brewster’s law gives the value of the polarizing angle for a surface of index of refraction n:
n = tan p
BREWSTER’S ANGLE
Example 5. A beam of unpolarized light in air strikes a flat piece of glass at an angle of incidence of
57.33o to the normal. If the reflected beam is completely polarized, what is the index of refraction of the
glass?
5A.
(1) n = tan p
(2) n = tan 57.33o
(3) n = 1.56
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