91. (36.1) Mirrors.
a)
Mirror - a highly reflective surface, which creates
images.
b) Plane mirror - the reflective surface is flat. If the rays
are emitted from a point (real object) or converge to a point
(virtual object), after they are reflected by the mirror, the
rays or their extensions meet at a point to form an image.
image
object
s
s’
ray diagram
The position of the object s (with respect to the mirror)
is related to the position of the image s' by the mirror
equation.
1+ 1 = 1
s s' ∞
92. (36.2, 36.5) Spherical mirror
121
a) If the rays are emitted from a point (real object) or
converge to a point (virtual object), after they are reflected
by the mirror, the rays or their extensions meet
approximately at a point to form an image. The failure to
meet exactly at a point is called spherical aberration.
O
(object) parallel ray
(object) chief ray
(object)
normal ray
(object)
focal ray
F
principal
axis
C
(image) parallel ray
I
(image)
chief ray
(image)
focal ray
(image)
normal ray
s
s’
0
The position of the object s is related to the position of
the image s' by the mirror equation.
1+ 1 = 1
s s' f
where f is called the focal length of the mirror and
represents the positions of the focal points from the optical
element.
122
b) For a spherical mirror, the absolute value of the focal
length is approximately two times smaller than the
curvature radius.
f≈R
2
For a concave mirror the focal length is positive and for a
convex mirror the focal length is negative.
The inverse of the focal length is called the focusing
power of the mirror. The SI unit of focusing power is
1 diopter (dioptre).
Proof.
R
O
h
h’
C
I
s’
s
0
from which
h s
=
h' s'
h s−R
=
h' R − s'
;
2 1 1
= +
R s' s
123
93. (36.4) Magnification of an optical element or
instrument.
The magnification is defined as the ratio of the image
height to the object height. If the orientation of the image is
the same as that of the object, a positive value is assigned to
the magnification. For an inverted image the magnification
is negative.
M≡
h'
h
94. (36.2) Magnification in a mirror
The magnification of a mirror is related to the object
position and the image position.
M=−
124
s'
s
95. (36.3, 36.4) Lens - a piece of transparent refracting
substance, with two opposite regular surfaces, used in
optical systems to converge or diverge light rays to form an
image.
96. (36.4, 36.5) Thin spherical lens - the surfaces are parts
of spheres and the thickness of the lens is small relative to
the diameter. If the rays are emitted from a point (real
object) or converge to a point (virtual object), after they are
refracted by the lens, the rays or their extensions meet
approximately at a point to form an image.
O
parallel ray
chief ray
focal ray
Fi
focal ray
s
s
object real side
image virtual side
s’
Fo
principal
axis
chief ray
object real side
image virtual side
parallel ray
0
I
The failure to meet exactly at a point is called
aberration (spherical and chromatic). Spherical aberration
result from the shape of the lens. Chromatic aberration
result from dispersive property of the lens material.
125
s’
97. (36.4) Thin lens equation
a) The position (with respect to the lens) of the object s is
related to the position of the image s' by the thin-lens
equation.
1 1 1
+ =
s s' f
where f is called the focal length of the lens and represents
the positions of the focal points from the optical element.
b) The focal length of a thin-lens is determined by the
curvatures of the two surfaces, the index of refraction of
the lens material, and the index of refraction of the
surrounding medium (lens maker's equation)
 1
1  n
1 
=  − 1
−

f  n0
R
R
 1
2
126
98. (36.8) Angular magnification
The angular magnification (magnifying power) of an
instrument is defined as the ratio of the "angular size" of the
final image and the "angular size" of the observed object.
θ'
m≡
θ
(Angular size is defined as the measure of the angle in
which the object is confined.)
Example. The angular magnification of a magnifying
glass (used properly) depends on the refractive power of
the glass, the location of the observed object and the
distance to the near point from the eye.
From geometrical
consideration
h ;
θ≈
N
N
θ
h
N
h'
s =h
θ' ≈ =
N
N
s
h⋅
N
Therefore
h’
θ' N
m= ≈
θ s
θ’
s
127
99. Combinations of optical elements.
The combination of optical elements (an optical
system) is an arrangement such that light "passes" the
elements in a certain order.
Example 1. (36.9) The compound microscope.
If the object was observed directly
it would be confined in an angle eye piece
determined by the size of the
object
h
θ≈ .
L
L
The objective produces the
first image almost at the focal point
of the ocular. The size of the
image is determined by the length objective
of the microscope and the focal
length of the objective.
The final image is confined to
an angle dependent on the focal
length of the ocular
h'
θ' ≈
fe
From the definition, the angular magnification of the
microscope is
θ' h ' L
Lh L
L2
m= = ⋅ =−
⋅ =−
θ fe h
f o fe h
f of e
128
Example 2. (36.10) Keplerian telescope
h
θ
θ’
h”
h’
The objective produces the first image almost at its focal
point. Therefore
h'
θ≈ .
fo
Since this image is almost at the focal point of the ocular
h'
θ' ≈ −
fe
Hence, the angular magnification of a telescope is
m = − fo .
fe
129
Example 3. (36.4) Two thin lenses in contact
Fi
Fo
F2i
F1o
If the position of the object is s, then the first lens
produces an image at location satisfying the thin lens
equation
1+ 1 = 1.
s s' f1
This image becomes an object for the second lens. If the
lenses are very close to each other, the object (for the
second lens) has the position opposite to the position of the
image produced by the first lens. Therefore using again the
lens equation we can determine the position of the final
image
1 +1= 1
−s' s" f2
Combining both equations lead to the relationship
between the position of the object and final image
1+ 1 = 1 + 1
s s" f1 f2
The system of two close lenses behaves like a single
lens with a focusing power equal to the sum of the powers
of the two lenses separately.
1 1 1
= +
f f1 f 2
130
100. The wave nature of radiation
The wave nature of any type of radiation is manifested
by the following two processes:
a) Interference - a phenomenon that results from the mutual
effect of two or more waves passing through a given region
at the same time, producing reinforcement of oscillations at
some points and neutralization at other points. In order to
obtain interference, the sources of the waves must maintain
a constant phase difference (coherent).
b) Diffraction - bending of radiation around obstacles or the
edges of openings.
101. (35.6) Huygen’s Principle
All points on a given wave front become sources of
spherical secondary waves, called wavelets, which
propagate through a medium with a speed characteristic on
waves in that medium.
131
Example 1. Reflection and refraction
A’
D
θ’1
θ1
A θ2
C
B
The triangles ADC and AA’C are congruent (they share the
hypotenuse and since reflected light travels with the same speed as the
incident light A’C = AD, and they both have one right angle).
Therefore
sin θ1 =
from which θ1 = θ'1 .
A' C AD
=
= sin θ'1
AC AC
For the two media
AB v 2 n1 ;
=
=
A' C v1 n 2
By substitution
from which
AB = AC ⋅ sin θ2 ;
sin θ2 n1
=
sin θ1 n 2
n 2 sin θ2 = n 2 sin θ1
132
A' C = AC ⋅ sin θ1
Example 2. Young’s double slit experiment
d
θ
θ
δ
The path difference δ determines the phase relationship between the
two wavelets of monochromatic light.
a) constructive interference
For constructive interference along a chosen direction, the
phase difference must be an even multiple of π, which requires
d sin θ = δ = mλ
m = 0, ±1, ±2, …
b) destructive interference
For destructive interference along a chosen direction, the phase
difference must be an odd multiple of π, which requires
1
d sin θ = δ =  m + λ
2

133
m = 0, ±1, ±2, …
102. (37.3, 37.4) Phasor addition of monochromatic waves.
Since each monochromatic waves are sinusoidal
functions they can be represented by a complex wave
functions. (The complex oscillating function representing a
physical quantity is defined in such a way that its imaginary
part is equal to the quantity.)
Example. Intensity distribution in the double slit interference pattern
E1 = E 0ei (kr1 −ωt )
E 2 = E 0ei (kr2 − ωt)
E
ϕ
E2
where k (r2-r1) = ϕ corresponds to the
phase difference between the two
wavelets.
E1
P
r1
The net electric field
r2
E = E1 + E 2 =
(
(1 + e
d
)
ikd sin θ
)e
θ
O
= E 0eikr1 1 + eik (r2 − r1 ) eiωt =
= E 0eikr1
y
L
i ωt
Hence the intensity at point P is
(
)(
)
(
)
I = cε 0E 20 1 + eikd sin θ 1 + e −ikd sin θ = cε 0 E 20 2 + eikd sin θ + e−ikd sin θ =
2
ikd sin θ
 ikd sin θ

−
kd sin θ
2
= cε 0E 0 e 2 + e 2  = 2cε 0 E 20 cos
=


2


 πd sin θ 
2
2  πd 
= 2cε 0 E 20 cos2 
y
 ≈ 2cε 0 E 0 cos 
 λ 
 2L 
134
Example. Multiple-slit diffraction pattern
For n-slit grating the electric field at a point is a superposition
of n waves for which the complex electric fields are related by the
same phase difference
E1 = E 0 ;
E1 = E 0eiϕ ;
E1 = E 0ei 2 ϕ ; ...
where phase difference depends of the separation between the slits
and the wavelength of the wavelets
ϕ=
d sin θ
λ
In result the interference patters results in primary maxima when the
phase of consecutive wavelets differs by even multiple of π.
Depending on the number of slits, a precise number of secondary
maxima occurs in the pattern
ϕ=0
ϕ=π
ϕ = 2π/3
ϕ=
π/3
I
I
N=2
N=4
dsinθ
primary
maxima
dsinθ
secondary
maxima
135
103. (37.5) Phase in reflection
An electromagnetic wave undergoes a phase change of
π upon reflection from a medium that has higher index of
refraction than the one in which the wave is traveling.
There is no change if the second medium has a lower index
of refraction. (These rule can be deducted from Maxwell’s
equations.)
n1
n2 > n1
incident wave
n1
reflected wave
reflected wave
Example. Lloyd’s Mirror
mirror
Following the argument used in the double slit experiment the conditions
for the maxima and the minima are:
a) constructive interference:
b) destructive interference:
1

d sin θ = δ =  m + λ
2

d sin θ = m λ
136
104. (37.6) Interference in thin films
Light reflected from the
1 2
two surfaces of a thin film is
coherent and therefore subject
to interference effects. At
film
normal
incidence
the
constructive interference in
reflection (and destructive
4
interference in transmission)
3
occurs when the wavelengths
λ n and the thickness of the film t are related by
1

2 t =  m +  λ n m = 0,1, …
2

and the destructive interference reflection
constructive in transmission) occurs when
2 t = mλ n m = 0,1, …
air
t
air
(and
proof.
For constructive interference the phase difference between
the two reflected beams must be an even multiple of π. Since the
beams reflected by the top surface changes its phase on reflection
by π, the film must result in a phase change equal to an odd
multiple π of the beam reflected from the second interface. It
requires that the double thickness of the film is a half-multiple of
the wavelength.
For destructive interference the phase difference between the
two reflected beams must be an odd multiple of π. It requires that
the double thickness of the film is a whole-multiple of the
wavelength. (Note that in transmission there is no change in phase
at the interface from which conclusions about interference effects
in transmission can be concluded.)
137
105. (37.7) Michelson Interferometer.
image
fixed
mirror
light
source
moving
mirror
beam
splitter
telescope
A single beam of light is split into two equal in
intensity beams by the beam splitter. The path difference
results in an interference pattern. As the moving mirror
travels along its axis, the interference pattern moves
across the field of view.
In
a
different
configuration,
Michelson
interferometer is widely used in Fourier Transform
Infrared Spectroscopy (FTIR).
138
106. (38.6) Polarization
Light may be polarized by selective absorption, reflection,
birefringence, or scattering.
a)
polarization by selective absorption
Polaroid is a material that selectively transmits light
depending on its linear polarization. Only that part of light
with polarization (of the electric field) parallel to the
transmission axis passes through the polaroid.
No matter how the transmission axis is oriented, half
of the unpolarized light is transmitted through a polaroid. If
polarized light passes a polarizer, the law of Malus relates
the intensities of light on both sides of the polaroid.
S1 = S0 cos2 θ
unpolarized
light
polarized light
θ
polarized light
trasmission
axis
139
Polarization by reflection
When light is incident on a surface at an angle θB
(called the Brewster angle), such that the reflected and
refracted rays form a right angle, the reflected light is
linearly polarized. The value of the angle is given by
Brewster's law, relating the angle of incidence with the
refraction indices of the two media.
b)
tanθ B =
unpolarized
light
θB θB
n2
n1
linearly
polarized light
θ2
partially
polarized light
140
c)
Polarization by birefringence (double refraction)
There are substances that exhibit double-refracting
properties. The substance has one index of refraction, no,
which does not depend on the direction of light propagation
(isotropic index of refraction). Light with one linear
orientation (related to the crystal) is refracted with this
index of refraction (an ordinary ray). The second index of
refraction, ne, depends on the direction of the light
propagation (an anisotropic index of refraction). Light with
polarization perpendicular to the polarization of the
ordinary ray, is refracted with the second index of
refraction (an extraordinary ray).
extraordinary ray
unpolarized
light
linearly
polarized light
ordinary ray
linearly
polarized light
141
d)
(Rayleigh) scattering
When light interacts with matter it can be absorbed
and reemitted. Depending on the change in the direction, a
component of the initial electric field appears in the
scattered light resulting in its linear polarization.
unpolarized
light
polarized light
142
107. (38.1, 38.2) Franhofer Diffraction.
Interference of light pasing through a narrow single slit
results in a Franhofer diffraction pattern. Both contractive
and destructive interference is observed. The intensity
distribution depends of the width of the slit, wavelength of
the radiation and the diffraction angle:
1.0
2
0.8
relative intensity [I/Im a x ]
I = I max
θ 
 
 sin  πa sin  
λ 
⋅ 
θ 

π
a
sin

λ 

0.6
0.4
0.2
0.0
-1.0
(Bessel function)
proof:
-0.5
0.0
0.5
1.0
diffraction angle [θ/λ]
Diffration pattern for a 10 µm single silt.
Im
R
a dy
R
θ
E
ϕ
dϕ =
2 π sin θ
dy ;
λ
ϕ=
2 πa sin θ
;
λ
Rϕ = E max
 sin (ϕ / 2) 
 sin(π sin θ / λ ) 
⋅
 = E 2max ⋅ 

ϕ
/
2
π
sin
θ
/
λ




2
I ∝ E = 2R (1 − cos ϕ) =
2
2
E 2max
Re
143
2
108. Resolution (Rayleigh’s criterion)
When the location of the central maximum of one image
coincides with the the location of the first minimum of the
second image, the images are resolved.
For a circular aperture:
θ min = 1.22 ⋅
144
λ
D