optics and optical design 2011 problems - atomic physics · 2011-08-28 12:31 this material is ......

Optics

and

Optical design

2011

Problems

Sven-Gran Pettersson

2011-08-28 12:31

This material is taken from several sources. Some problems are from the book

Vglra och Optik by Gran Jnsson and Elisabeth Nilsson. Some are from

the book: Fundamental of Photonics, second edition by Bahaa Saleh and

Malvin Carl Teich. Another source is Exempelsamling i vglra by Lars

Engstrm. Examples and figures that are referenced to in the textbook

Fundamental of Photonics are labelled with FoP.

The exercise marked with * are the most relevant for the course. The other

exercises are meant as a repetition and/or as preparation for the laboratory

exercises.

Chapter 1: Ray Optics

P1-1. Show that the distance p that a ray is displaced parallel to its initial direction when

transmitted through a plane parallel glass plate of thickness d, is given by

1 2

2

sin( )

cos

dp

where 1

is the angle of incidence and 2

is the angle of the transmitted ray. See the following

figure (figure 1-1).

Figure 1-1. A plane parallel plate does not change the direction of a light ray. However, the

ray is translated a distance p that is dependent on the thickness d of the plate.

P1-2. At the bottom of a 2.0 m deep swimming pool there are illuminating lamps. Due to total

reflection, the light from a single lamp can only be seen within a limited region. Calculate the

area of this region. Assume that the refractive index for water is 1.33.

* P1-3. The high index of refraction of diamonds is utilised to obtain repeated total reflections which causes a beautiful scattering of the light. The most common type of diamond polishing

is called brilliant. In figure 1-2 the refraction and reflection of a red light ray is shown as it is

propagating through a diamond. The refractive index for diamond and red light is 2.407.

a) Show that the ray is totally reflected at the point P b) Calculate the angle of incidence at the upper surface of the diamond.

Figure 1-2. Total reflection in a diamond.

P1-4. The yellow reflexes which are attached to the bike wheel have cubic corners of pressed

plastics at the back. See figure 1-3. The bike reflex acts as a total reflector only as long as the

incident light is quite near the normal to the reflex. In that case the light is totally reflected as

shown in the left part of the figure. On right a tilted ray is shown which is not totally reflected.

For which angles of incidence (in relation to the normal of the bike reflex) will this type of

skew ray be totally reflected at the back? Assume that the refractive index of the plastics is

1.50 and that the reflex is surrounded by air.

Figure 1-3. A reflex on a bike is surrounded by air and acts only as a retro reflector when the

light impinging on the reflex is near the normal.

P1-5. Start with the relation

1 1 1

a b f

and show that a plane mirror always creates an image with the magnification 1.0.

P1-6. At the distance of 3.0 cm from a convex mirror there is an object with the height 4.5 cm.

The radius of curvature of the mirror is 12.0 cm. Make a figure where you construct the

position and size of the image.

* P1-7. Proof of the Snells Law. The proof of Snells law is an exercise in the application of Fermats principle. Referring to figure 1-4, we seek to minimize the optical path length

1 2n AB n BC between points A and C. We therefore have the following optimization

problem: Minimize 1 1 1 2 2 2

sec secn d n d with respect to the angles 1

and 2, subject to the

condition 1 1 2 2

tan tan .d d d Show that the solution of this constrained minimization

problem yields Snells law.

Figure 1-4. Construction to prove Snells law.

* P1-8. Lens in Water. Determine the focal length f of a biconvex lens with radii 20 cm and 30 cm and refractive index n = 1. 5. What is the focal length when the lens is immersed in water

(n = 1.33)?

* P1-9. In an otoscope (medical instrument for the examination of the ear), the imaging is schematically performed with a cylindrical glass rod of the type shown in figure 1-5. The rod

has a refractive index of 1.49 and the end surfaces have the radii of curvature 1

R = 15 mm

and 2

R = 35 mm.

a) Calculate the final image of an object placed 10.0 mm to the left of the surface to the left.

b) Is the image virtual or real?

Figur 1-5. A simplified drawing of an otoscope.

P1-10. One ray is drawn in figure 1-6 from the bottom of an object and through a thin lens.

Construct the image of the object and find the focal length of the lens from the figure. The

scale is given in the figure.

Opticalaxis

12 cm

Object

Lens Figure 1-6. An object is placed 12 cm in front of a thin lens. A ray is drawn from the bottom

of the object and through the lens.

P1-11.

a) In figure 1-7 is shown the start of the ray diagram through a lens. Determine the focal

length of the lens by drawing rays and measuring in the figure.

b) The lens that has the refractive index 1.50, has a plane surface on one side. Calculate

the radius of curvature of the other side of the lens.

c) A positive lens with the focal length 4.0 cm is now placed 2.0 cm to the right of the lens

in the figure. Calculate the size of the final image of the object in the figure. Is the image

upright or upside down?

Opticalaxis

10 cm

Figur 1-7. An object is placed 10 cm from a thin lens. A ray is drawn from the bottom of the

object and through the lens.

* P1-12. An object is placed 10 cm to the left of a lens 1

L mounted on an optical bench. The

lens has the focal length + 7.5 cm. Slightly to the right of the lens 1

L , another lens 2

L , is

placed. This lens creates a final image 10 cm to the right of 2

L . This image is 6.0 times larger

than the object and upside down.

a) Determine the distance between the lenses 1

L and 2

L .

b) What is the focal length of the second lens2

L ?

P1-13. A thin biconvex lens with refractive index 1.50 has the radii of curvature 20 cm and 40

cm.

a) What is the focal length of the lens?

b) A candle light is placed 40 cm to the left of the lens. Describe the image.

P1-14. A mountain at a distance of 10 km is imaged by a concave mirror. The size of the

image in the focal plane of the mirror is 1.5 cm. The mirror has a radius of curvature of

40 cm. Calculate the height of the mountain.

P1-15. A slide projector images sharply an image on a screen 3.10 m from the objective. The

focal length of the objective of the projector is 90 mm.

a) Calculate the magnification of the imaging. b) What is the size of the image on the screen if the slide has the size 24 x 36 mm?

P1-16. Construct the continuation of the beams below (figure 1-8) after they have been

refracted at the surfaces of the lenses. Mark also the focal points F1 and F2. Hint. Use a

suitable help ray (a ray that travels through the centre of the lens)!

+2,5 cm

-2,5 cm

+1,5 cm

-1,5 cm

Figure 1-8. Rays propagating to lenses.

P1-17 The largest telescope on Mount Palomar has a concave mirror with the radius of

curvature of 34 m and the diameter of the mirror is 5.08 m. How big will the image of the

moon be when the moon is full? The distance to the moon is 83.844 10 m and the diameter of

the moon is 63.476 10 m .

* P1-18. Numerical Aperture and Angle of Acceptance of an Optical Fiber. An optical fiber is illuminated by light from a source (e.g., a light-emitting diode, LED). The refractive indices

of the core and cladding of the fiber are 1

n and 2,n respectively, and the refractive index of

air is 1 (figure 1-9). Show that the angle a

of the cone of rays accepted by the

Figure 1-9. Acceptance angle of an optical fiber.

fiber (transmitted through the fiber without undergoing refraction at the cladding) is given by

2 2 1 / 2

1 2N A sin ( )

an n (FoP1.2-15)

The parameter NA sina

is known as the numerical aperture of the fiber. Calculate the

numerical aperture and acceptance angle for a silica glass fiber with 1

n = 1.475 and 2

n =

1.460.

P1-19. Numerical aperture of a Claddless Fiber Determine the numerical aperture and the

acceptance angle of an optical fiber if the refractive index of the core is 1

n = 1.46 and the

cladding is stripped out (replaced with air 2

1).n

* P1-20. Fiber Coupling Spheres. Tiny glass balls are often used as lenses to couple light into and out of optical fibers. The fiber end is located at a distance f from the sphere. For a sphere

of radius a = 1 mm and refractive index n = 1.8, determine f such that a ray parallel to the

optical axis at a distance y = 0.7 mm is focused onto the fiber, as illustrated in figure 1-10.

Figure 1-10. Focusing light into an optical fiber with a spherical glass ball.

* P1-21. The Grin Slab as a Lens. Show that a SELFOC slab (see figure 1-11) of length / 2d and refractive index given by

2 2 2 2

0( ) (1 )n y n y acts as a cylindrical lens (a lens

with focusing power in the y-z plane) of focal length

)sin(

1

0dn

f

(FoP1.3-13)

Show that the principal point (defined in the figure) lies at a distance from the slab edge

0(1 / ) tan( / 2).AH n d Sketch the ray trajectories in the special cases /d and / 2 .

Figure 1-11. The SELFOC slab used as a lens; F is the focal point and H is the principal

point.

* P1-22. For a gradient-index lens, with a diameter of 2.0 mm, the refractive index on the axis is 1.608 and at the edge 1.534. Calculate its numerical aperture and period. What length is

needed in order to image the surface of the first end on the other end? Note that for a gradient-

index lens the profile can be approximated with 2 20

1( ) (1 )

2n y n y .

* P1-23. Numerical Aperture of the Graded-Index Fiber. Consider a graded-index fiber with the index profile given by 2222

0

21 yxnn and a radius a. A ray is incident from air

into the fiber at its center, which then makes an angle 0

with the fiber axis in the medium

(see FoP Figure 1.3-8). Show, in the paraxial approximation, that the numerical aperture is

0

sina

NA n a (FoP 1.3-16)

where a

is the maximum acceptance angle for which the ray trajectory is confined within the

fiber. Compare this to the numerical aperture of a step-index fiber such as the one discussed in

FoP Ex1.2-5. To make the comparison fair, take the refractive indices of the core and

cladding of the step-index fiber to be 1 0

n n and 2 2 2 22 0 0

11 (1 ),

2n n a n a

respectively.

* P1-24. A Set of Parallel Transparent Plates. Consider a set of N parallel planar transparent plates of refractive indices

1 2, , .....,

Nn n n and thicknesses

1 2, , ......,

Nd d d placed in air (n = 1)

normal to the z axis. Show that the ray-transfer matrix is

Note that the order of placing the plates does not affect the overall ray-transfer matrix. What

is the ray-transfer matrix of an inhomogeneous transparent plate of thickness 0

d and

refractive index n(z)?

* P1-25. A Gap Followed by a Thin Lens. Show that the ray-transfer matrix of a distance d of free space followed by a lens of focal length f is

* P1-26. Imaging with a Thin Lens. Derive an expression for the ray-tranfer matrix of a system comprised of free space/thin lens/free space, as shown in Figure 1-12. Show that if the

imaging condition 1 2

(1 / 1 / 1 / )d d f is satisfied, all rays originating from a single point in

the input plane reach the output plane at the single point 2

y , regardless of their angles. Also

show that if 2

,d f all parallel incident rays are focused by the lens onto a single point in the

output plane.

Figure 1-12. Single lens imaging system.

* P1-27. A Periodic Set of Pairs of Different Lenses. Examine the trajectories of paraxial rays through a periodic system comprising a sequence of lens pairs with alternating focal lengths

1f and

2,f as shown in Figure 1-13. Show that the ray trajectory is bounded (stable) if

1 2

0 (1 )(1 ) 12 2

d d

f f (FoP 1.4-35)

Figure 1-13. A periodic set of lenses

* P1-28. The ray transfer matrix for a curved boundary with curvature R and refractive index n1 before the surface and refractive index n2 after the surface is given by:

2 1 1

2 2

1 0A B

M n n nC D

n R n

a) Calculate the ABCD matrix of a thin spherical lens, made up of two closely spaced dielectric interfaces, of radii R1 and R2 enclosing a material of refractive index n2. The

lens is immersed in a medium of refractive index n1.

b) From the ABCD matrix it is easy to find the focal length f of the lens. Give an expression for f.

* P1-29. The ABCD matrix is not only useful for describing the propagation of specific rays but can also be used to describe the propagation of a spherical wave. Consider in fact a spherical

wave originating at point P1 in the following figure and propagating in the positive z-

direction. After traversing an optical element described by a given ABCD matrix, this wave is

generally transformed into a new spherical wave whose centre is point P2.

r1-r'

2r'1 r

2

Optical elementP1 P2

z1 z2

z

Use the figure to obtain value for the two radii of curvature R1 and R2 and show that the

curvature of the output wave R2 is given by

12

1

A R BR

C R D

This means that in a simple way the curvature of the output wave R2 is found from the

curvature of the input wave curvature R1 by using the ABCD matrix for the optical element.

This means that we have here a very important rule, the ABCD-rule for describing the

transformation of spherical waves. In a similar way transformation of Gaussian waves is

described by the same rule.

However in this case the wave is characterized by a complex beam parameter which is given

by:

2

1 1j

q R w

where w is dependent on the beam size. More on this will be discussed in the course Lasers.

* P1-30. Ray-Transfer Matrix of a GRIN Plate. Determine the ray-transfer matrix of a SELFOC plate [i.e. a graded-index material with parabolic refractive index

2 2

0

1( ) (1 )

2n y n y ] of thickness d.

Chapter 2: Wave Optics

P2-1. We can write a plane wave with the z axis taken in the direction of the wave vector k as

( , ) cos 2 arg( )u t A t kz A r

As /c , 1 /T and 2 /k we can rewrite the plane wave as

( , ) cos 2 ( ) arg( )t z

u t A AT

r

If the wave travels in the opposite direction to the z axis the wave is described by

( , ) cos 2 ( ) arg( )t z

u t A AT

r

When writing the wave in this way we describe the wave in any point and at any time. If we

want to represent the wave by a diagram we must choose a particular value of t as when we

take a picture of the wave. We also need to describe the variation of the optical amplitude

with time at a certain point z on the wave to fully determine the wave.

Determine A, T, and arg(A) for the optical wave that is described by the diagrams in figure

2-1. Determine also the direction of propagation, e.g. if the sign is positive or negative in the

wave equation.

Figure 2-1 In the diagram to the left we see the wave ( , )u tr as a function of z when t = 0 s.

To the right we see the optical wave amplitude at the origin as a function of t.

P2-2. An optical wave is described by

( , ) cos 2 ( ) arg( )t z

u t A AT

r

in air to the left and in a medium with refractive index n to the right.

a) Determine by studying the diagrams in figure 2-2 values for A, T, , n and arg(A) and

decide the sign + or . Do the same for both regions.

b) Determine the phase vector that represents the complex amplitude ( )U r according to

figure 2-2, at the positions z =2.0 m and z = 5.0 m at the time t = 0.0 s.

Figure 2-2. In the diagram to the left we see the wave ( , )u tr as a function of z when t = 0 s.

To the right we see the optical wave amplitude at the point z = 5.0 m as a function of t.

P2-3. Laser light is propagating in a transparent material. The electric field of the laser light

varies according to

0

sin( )E E t kx

where 15 -1 7 -1

03.5 kV /m , 2.272 10 s and 1.287 10 mE k

a) What is the wavelength of the laser light in the material?

b) What is the refractive index of the material through which the light is propagating.

c) What is the frequency of the laser light?

* P2-4. Validity of the Fresnel Approximation. Determine the radius of a circle within which a spherical wave of wavelength = 633 nm, originating at a distance 1 m away, may be

approximated by a paraboloidal wave. Determine the maximum angle m

and the Fresnel

number F

N .

* P2-5.Three antennas receive radio waves with the wavelength 21 cm from space. The distance between the antennas is 1.50 m and the radiation arrives at an angle of 15 according to figure

2-3. The antennas are connected with a mixer with equally long wires. (In the mixer the

waves are added). If only one antenna is connected to the mixer a signal with the amplitude

2.5 mV is obtained. What will the amplitude of the voltage be when all the three antennas are

connected?

Voltmeter

Mixer 15

1,5 m

1,5 m

Figure 2-3. A set of antennas receiving radio waves from space.

* P2-6 A system of vertical slits is illuminated by a laser of the wavelength 600 nm. The diffraction pattern is observed on a screen at a distance of 2.4 m. The intensity distribution is

shown in figure 2-4. The horizontal distance of the screen is 160 mm.

Figure 2-4. The diffraction pattern of a slit system as observed on a screen 2.4 m away

a) How many slits were illuminated?

b) What was the distance between the slits?

c) Calculate the width of the slits.

d) Calculate the width of the central maximum.

* P2-7. Two loudspeakers are connected in phase to a frequency generator adjusted to the frequency 680 Hz. See figure 2-5. Assume the speed of sound to be 340 m/s. The amplitude

of the sound is assumed to be U from one speaker only if measured at the point Q.

a) How many sound maxima are observed between the points Q and P?

b) Calculate the length of the phase vector at point R, 0.8 m from the point Q.

2.0 m

A

B

4.0 m

3.5 m

P

Q

R

0.8 m

Figure 2-5. Observation of interference pattern for sound.

* P2-8. A CD is illuminated with light from an Argon laser, see figure 2-6. The laser light consists of 6 visible lines of which the two strongest have the wavelength 488.0 nm and 514.5

nm. On a screen parallel with the laser ray a bright light spot is observed at a distance 373 mm

(see the figure) and a bit from that position also 6 dots, of which the two strongest are at a

distance 216 mm and 222 mm (see the figure). All the dots are positioned at the same height

above the laser table as the incident laser ray. Calculate the distance between the tracks on the

disc. Make the calculations for both wavelengths.

222 mm

373 mm

300 mm

CDLaser

Screen

200 mm

Figure 2-6. Observation of diffraction spots from a CD.

P2-9. The determination of the wavelength of X-ray radiation is performed by using a

reflection grating. If light with the wavelength 643.87 nm from a Cd-lamp is impinging

perpendicularly against the grating, the first order is observed at a deflection angle of 39.408.

To reflect X-ray the light has to be at grazing incidence according to figure 2-7. The zero

order beam (m = 0) is observed at D and the first order is observed at E. Calculate the

wavelength of the X-ray radiation. The distance d between grating and screen is 1.0 m.

Figure 2-7. X-ray diffraction from a grating.

* P2-10. The light from the slit in a spectrometer is collimated with a lens with the focal length 2.00 m. The light meets a reflection grating at the incidence angle 19.00. The returning light

is focused with the same lens and in the slit plane there is a photographic plate. Se figure 2-8.

The grating has 1500 rules/mm.

a) At which angle is the first order reflected from light with the wavelength 466.8 nm?

b) Another spectral line is also observed 45 m from the line in problem a). Calculate the

difference in wavelength between the two lines.

19o

Lightsource Fotographic

filmLens Reflection

grating

Figure 2-8. Reflection of light in a spectrometer.

P2-11. Bragg Reflection. Light is reflected at an angle from M parallel reflecting planes

separated by a distance d as shown in figure 2-9. Assume that only a small fraction of the light

is reflected from each plane, so that the amplitudes of the M reflected waves are

approximately equal. Show that the reflected waves have a phase difference (2 sin )k d

and that the angle at which the intensity of the total reflected light is maximum satisfies

sin2d

(2.5-11)

This angle is known as the Bragg angle. Such reflections are encountered when x-ray waves

are reflected from atomic planes in crystalline structures. It also occurs when light is reflected

from a periodic structure created by an acoustic wave.

Figure 2-9. Reflection of a plane wave from M planes separated from each other by a

distance d. The reflected waves interfere constructively and yield maximum intensity when

the angle is the Bragg angle.

* P2-12. In a salt crystal (according to figure 2-10) the deviation (angle between the transmitted and Bragg reflected light) of two X-ray wavelengths is 26.30 and 29.50. The distance

between close crystal planes in the NaCl crystal is 282 pm. Calculate the two X-ray

wavelengths if you assume that it is the first order Bragg reflection that causes the deviation.

Figure 2-10. Sodium chloride is crystallized in a cubic grating. At the corners of the small

cubes a sodium or a chloride atom is situated. In the figure the size of the atoms is too small in

relation to the distance between them.

* P2-13. Yellow light from a Sodium light source consists of two frequencies. In air the wavelengths are 588.9953 nm and 589.5923 nm. How long shall the distance between the

mirrors in a Fabry-Perot interferometer be so that one misses to see that the light consists of

two frequencies?

Hint! Assume that one ring with order number m from one wavelength coincides with the ring

with order number m+1 from the other wavelength. You can also calculate the frequency

difference between the two wavelengths and use the relation / 2f c d .

Figure 2-11. Interference pattern observed from two point sources.

* P2-14. Interference of Two Spherical Waves. Two spherical waves of equal intensity 0

I

originates at the points (a,0,0) and (-a,0,0) interfere in the plane z = d as illustrated in figure 2-

11. The system is similar to that used by Thomas Young in his celebrated double-slit

experiment in which he demonstrated interference. Use the paraboloidal approximation for

the spherical waves to show that the detected intensity is

0

2( , , ) 2 1 cos .

xI x y d I

(FoP 2.5-8)

where 2 /a d is approximately the angle subtended by the centers of the two waves at the

observation plane. The intensity pattern is periodic with period / .

* P2-15. Two coherent equally intense laser beams with the wavelength 532 nm intersect with an angle according to the figure 2-12. In the region of overlap an interference pattern is

obtained. This pattern consists of parallel planes with either high or low intensity. Calculate

the distance between two nearby planes of high intensity.

This is the principle of making holographic gratings. The method of producing a reflecting

grating is the following. A plane glass surface is covered with photoresist. After exposure and

development (where the exposed resist is removed) the surface is covered with a thin

reflective layer.

a

Figure 2-12. Inteference pattern observed where two coherent beams overlap.

* P2-16. Fringe visibility. The visibility of an interference pattern such as that described by (FoP 2.5-4) and plotted in FoP, Figure 2.5-1, is defined as the ratio:

max min max min

( ) /( )V I I I I

where max

I and m in

I are the maximum and minimum values of I. Derive an expression for V

as a function of the ratio 1 2

/r I I of the two interfering waves and determine the ratio 1 2

/I I

for which the visibility is maximum.

* P2-17. A schematic of the Michelson interferometer is shown in figure 2-13a. The light (1) from an extended light source (a transparent screen illuminated by a light source) is split by a

beam splitter BS into two beams (2) and (3). The beams are reflected at the mirrors M1 and

M2 and returned to the beam splitter. In the beam splitter two beams are created, one that goes

back to the light source and one (4) that is directed to the observer. If the light source is a

white lamp with a broad spectrum it is a necessity to use a compensator plate C to observe

fringes. If the interferometer is correctly adjusted a circular ring system is observed localized

at infinity. The fringes are due to the interference of two images of the light source as shown

in figure 2-13b. An off axis point P will thus be imaged as two points 1

P and 2

P separated by

a distance 2d where d is the difference in distance that is due to a different position of the

mirrors. In the following we assume that we use a HeNe light with the wave length 633 nm.

a) Derive a relation between the optical path difference p as a function of the angle

and distance d.

b) We adjust the interferometer so that the path difference between the two paths is zero. After that we move one mirror a distance of 4.0 mm and adjust slightly to observe a

dark area in the center. At which angle do we observe the nearby dark ring?

c) A gas cell is now placed in one of the arms of the interferometer. The length of the gas cell is 100 mm. We start with vacuum and we fill the gas cell with gas. After we have

stopped filling with gas, we find that 155 central interference fringes have passed.

Calculate the refractive index of the gas.

S

M1

M2

BS

C

(1)

(4)

(3)

(2)

p2d

d

S P

P

P

1

2

M1

M2

S1

S2

Figure 2-13a. Schematic view of a Figure 2-13b An equivalent drawing for the rays

Michelson interferometer. in the Michelson interferometer in figure 2-13a.

Chapter 4: Fourier Optics P4-1. Calculate the Fourier transform of the function rect(2x)rect(y/3)

The rectangular function rect(x) is given by

1 / 21

( ) when 1 / 20

xrect x

x

* P4-2. Assume that ( ( , )) ( , )x y

g x y G e.g. the function G is the fourier transform of g.

a) Calculate ( ( 0.4, 2.3))g x y

b) For which values of x

and y

is the following true:

x y

( ( 0.4, 2.3)) ( , )g x y G

* P4-3. Assume that the rectangle in figure 3-1a below has a length of c (bottom side). It is placed in front of a lens with a focal length f. Calculate the distance to the first horizontal

minimum in the Fourier transform of the object by:

a) using the two figures (Figure 3-1b) below.

b) using ordinary diffraction relations e.g. building up the image from by adding waves.

Figure 3-1a. A rectangular opening Figure 3-1b. A function and its Fourier transform.

* P4-4. A transparency with the transmission function t(x,y) is placed close to a lens with a

focal length of 20 cm. How far from the optical axis is it possible to observe the vertical

spatial frequency 30 cycles/mm, if the wavelength of the light is 633 nm.

* P4-5. Using the figure 3-2 below, derive the following formula:

w

fw

0,

which gives the beam waist 2w0 when a Gaussian beams of width 2w is focused by a lens with

the focal length f. The beam waist is the diameter of the beam defined where the electric field

amplitude has dropped to 1/e of the top value.

Figure 3-2. A Gaussian function and its Fourier transform.

P4-6. Calculate a) the auto-correlation and b) the self-convolution of the following

asymmetric function:

( ) 0 w hen 0 and 2

1 w hen 0 1

2 w hen 1 2

f u u u

u

u

The auto-correlation is the function

( ) ( ) ( )ak

h x f u f u x du

and the convolution (swedish: faltningen) is normally given between two different

functions f and g by the following relation

( ) ( ) ( ) ( )f

h f u g u f u g x u du

When g = f this type of convolution is called self-convolution.

P4-7. The normalized auto-correlation function of the function f(x) is often denoted (x),

defined as:

*

*

( ) ( + )

( )

( ) ( )

f u f u x du

x

f u f u du

Suppose the function f(x) is defined by:

0 >

1 -

( ) = 0 w hen

* P4-9. The optical processor shown in figure 3-4a is used for spatial filtering. A cross grating according to figure 3-4b is the object. The result of a vertical and horizontal spatial filtering of

the object is shown in figure 3-4c. Assume now that we insert a narrow slit at a 45 angle in

the transform plane of figure 3-4a below. How will the spatially filtered image of the object in

figure 3-4c change?

Object plane

Transform plane

Image plane

y

x

x y

y

x

Plane wave

f1

f 1f

2f

2

L 1

L 2

The Fourier transformof the letter E

Figure 3-4a An optical processor Figure 3-4b A cross grating and its Fourier

transform.

Figure 3-4c. Vertical and horizontal spatial filtering.

P4-10. Assume that a slide with the amplitude transmission function given as

0g( ) cos 2x x is placed in an imaging system with the cut-off frequency

01.2 . Is it

better with coherent illumination than with incoherent?

Hint:

02 2

0 2

12cos 2

1 2

n

j n x

n

x e

n

Draw the spectrum and the transfer functions.

Hint! For a coherent system the system is linear for amplitudes. The transfer function is

given by the function ( , )x y

H as shown in FoP figure 4.4-10. For an incoherent system

the transfer function is *H H which has the double cut-off frequency 0

2.4 .

Note also that the transfer function operates on the intensity of the image instead of the

amplitude.

P4-11. The first minimum in the Airy pattern is given by equation 4.3-8:

1.22D

The second minimum is given by:

2.23D

The following minima can be described with:

kD

where k = m + 0.25 with integers m 3.

To determine the diameter of a circular hole it is illuminated by red light from a He-Ne laser

( = 632.8 nm). The diffraction pattern is studied on a screen at a distance of 5.00 m from the

hole. The diameter of the fifth dark ring as measured from the bright central spot is 62 mm.

Calculate the diameter of the hole.

* P4-12. Figure 3-5a below shows a bubble chamber photograph and figure 3-5b is a filtered version of the same image. Figure 3-5c illustrates the Fourier Transform of figure 3-5a.

a) Show in a figure what the Fourier transform of figure 3-5b would look like. Indicate the

main difference compared to figure 3-5c.

b) Describe in a figure what the spatial filter should look like in order to obtain the filtered

image in figure 3-5b.

c) With the lens L4 removed, the Fourier transform of the original image (figure 3-5a) is

observed on a screen (figure 3-5d). Calculate the distance (on the screen) from the optical

axis to the first vertical minimum in the Fourier transform of the horizontal lines (average

width of the lines is 0.24 mm ) if the image was placed 9 cm after L2 as shown in figure 3-

5d. The wavelength of the light is 633 nm. Note that the scale of the Fourier transform is

determined by the distance from the object to the Fourier plane.

d) If the lens L4 is introduced in the set-up, an enlarged image of the object is obtained at

the screen. Calculate the focal length of the lens L4 that will make this image sharp.

Figure 3-5a Figure 3-5b Figure 3-5c

L1 L2: f = 38 cm

9 cm 13 cm

156 cm

Screen24 cm

L3:f= 12 cm

L4

Figure 3-5d. Optical set-up for observation of Fourier transforms and spatial filtering.

P4-13. A hologram can, in some cases, have a resolution of 5000 lines/mm. How many values

of photographic density must be known to completely describe a hologram with an area of 1

mm2?

* P4-14. A photographic plate C used to create a hologram (figure 3-6) is illuminated with light from two coherent light sources A and B. The wavelength of the light is 0.6328 m .

a) What will be the distance between nearby fringes in the point P?

b) When the plate is developed a hologram is formed. This means that in P you have

locally a grating with the grating constant given from a). This hologram is mounted in the

same position as before development and illuminated with laser light coming only from

the source in A. Calculate the direction for which you observe light on the other side of

the plate coming from the point P.

A

BP

C

1000 mm

500 mm

100 mm

A

BP

C

1000 mm

500 mm

Figure 3-6 a) Two point sources (A and B) illuminates a photographic plate C.

c) The illumination of the plate (hologram) after development.

Chapter 5: Electromagnetic Optics

* P5-1. An Electromagnetic Wave. An electromagnetic wave in free space has an electric-field vector

0( / )f t z c x where x is a unit vector in the x direction, and

2 2

0( ) exp( / ) exp( 2 ),f t t j t where is a constant. Describe the physical nature of this

wave and determine an expression for the magnetic-field vector.

* P5-2. Dielectric Media. Identify the media described by the following equations, regarding linearity, dispersion, isotropy, and homogeneity. Assume that all media are spatially

nondispersive.

a) 0

P E a E

b) 2

0P aP E

c) 2

1 2 02

P Pa a P E

t t

d) 2 20 1 2 expP a a x y E

P5-3. A 3V-torch is driven by a current of 0.25 A. About one per cent of the power is

converted to light. Assume that the light is monochromatic with the wavelength 550 nm. How

many photons are emitted each second?

* P5-4. An isotropic and approximately monochromatic point source transmits light in air with a power of 100 W.

a) Calculate the intensity at the distance 1.0 m. b) Calculate the amplitude of the E- and the B-field.

* P5-5. In a microwave oven the frequency is always 2.45 GHz (this is a frequency that is absorbed by the water molecules in the food). The transmitter in an ordinary microwave oven

has a power of 750 W. Estimate the maximum electric field inside the microwave oven

assuming that the effect is distributed on an area of 20.10 m .

P5-6. A cell phone that transmits at the frequency 900 MHz releases a power of 2.0 W.

Assume that the antenna of the phone distribute the microwaves evenly in all directions.

Calculate the amplitude of the electric field and the magnetic density at the ear 5.0 cm from

the antenna.

P5-7. A plane, harmonic wave is propagation trough glass. The electric field is given by:

150

cos 100.65

z

xE E t

c

Calculate

a) the frequency b) the wavelength in vacuum c) the wavelength in glass d) the propagation velocity in glass

e) the refractive index

* P5-8. Laser light propagates in a transparent nonmagnetic material (r

1.000 ). The electric

field of the laser light varies according to

0

sin( )E E t kx

where 0

3.5 kV/m,E 15 -1

2.272 10 s and 7 -1

1.287 10 m .k

a) What is the wavelength of the laser light inside the material? b) What is the refractive index in the material where the wave propagates? c) Calculate the amplitude of the magnetic density in the material. d) Calculate the intensity of the laser light.

* P5-9. a) Make an estimation of the dispersion coefficient D

of quarts at the wavelength 852 nm

by using the following table:

Wavelength/nm Refractive index 707 1.45515 852 1.45247 997 1.45043

b) What would the broadening of a pulse at 852 nm with the spectral width 80 nm be at a

distance of 12 km? Hint: Make a numerical calculation of the second derivative of the refractive index in terms of

the wavelength.

Chapter 6: Polarization and Crystal Optics

* P6-1. Cascaded Wave Retarders. Show that two cascaded quarter-wave retarders with parallel fast axes are equivalent to a half-wave retarder. What is the result if the fast axes are

orthogonal.

P6-2. Jones Matrix of a Polarizer. Show that the Jones matrix of a linear polarizer with a

transmission axis making an angle with the x axis is 2

2

cos sin cos

sin cos sinT

(FoP 6.1-25)

Hint! Derive (FoP 6.1-25) using (FoP 6.1-18), (FoP 6.1-22) and (FoP 6.1-24)

* P6-3. Three polarizers are placed after each other. The first is illuminated by unpolarized light with the intensity

0.I The transmission direction for the second and the third polarizer is

rotated 45 and 90 in relation to the first respectively. See figure 6-1.

Figure 6-1. Unpolarized light with the intensity

0I is passing through three polarizers with

different transmission directions.

a) Give the intensity between polarizer 1 and 2 in relation to 0.I

b) Give the intensity between polarizer 2 and 3 in relation to 0.I

c) Give the intensity after polarizer 3 in relation to 0.I

* P6-4. Two polarizers can be used as a continuously variable grey filter. What is the angle be between the transmission axes, so that 5.0 % of incoming light is transmitted? We assume that

we can neglect reflections.

P6-5. Give the propagation direction and polarization state for the following waves:

a) tkxiezyiE 32Re

b) tkyiezixE Re

c) tkzii eeyxE 6/3Re

* P6-6. Give the polarization state of the following wave: a)

0 0 cos( ) cos( / 2 )E x E kz t y E kz t

b) 0 0

cos( ) cos( / 2 )E x E t kz y E t kz

c) 0 0

cos( ) cos( )E x E t kz y E t kz

d) 0 0

cos( ) cos( / 4 )E x E t kz y E t kz

e) ( )

Re 2 3i t kx

E iy z e

* P6-7. A plane, linearly polarized light wave, with intensity 0,I is transmitted through a system

of perfect linear polarizers (we assume that all light is transmitted in the transmission

direction but in the perpendicular direction all light is absorbed).

Give for the following systems of polarizers and transmission directions the total transmitted

intensity: (angles are measured in the same direction and relatively to the polarization

direction of the incident light).

a) one at 90 angle

b) two at the angles 45 and 90 .

c) three at the angles 30 , 60 and 90 .

d) N polarizers an the angles 90 / , 2 90 / , 3 90 / , 4 90 / , ......90N N N N

e) from d) we let N

P6-8. Brewster window. At what angle is a TM-polarized beam of light transmitted through a

glass plate of refractive index n = 1.5 placed in air (n = 1) without suffering reflection losses

at either surface? Such plates, known as Brewster windows (figure 6-2), are used in lasers, as

described in FoP Sec. 15.D.

Figure 6-2. The Brewster window transmits TM-polarized light with no reflection loss.

* P6-9. Reflectance of Glass. A plane wave is incident from air (n =1) onto a glass plate (n = 1.5) at an angle of incidence of 45. Determine the power reflectances of the TE and TM

waves. What is the average reflectance for unpolarized light (light carrying TE and TM waves

of equal intensities)?

P6-10. Left elliptically polarized light impinges from air (n = 1) to a glass surface (n = 1.56).

The ratio of the long and short axis of the ellipse is 3:2 and the long axis is in the plane of

incidence. Calculate the angle of incidence for which the reflected light is right circularly

polarized.

P6-11. Light impinges on a glass plate with the refractive index 1.5.

a) Calculate the Brewster angle. b) Calculate the angle between the reflected beam and the transmitted beam.

* P6-12. When white light is reflected on a glass plate one can obtain plane polarized light if the plate is oriented at Brewster angle.

a) Calculate the efficiency of the polarizer e.g. the intensity ratio between reflected plane polarized light and incident unpolarized light. The refractive index of the glass is 1.54.

b) Higher efficiency can be obtained if transmitted light is used instead. In this case the light will not be perfectly polarized. It is then better to use several glass plates

mounted in Brewster angle after each other. Calculate the number of plates needed so

that the polarization degree# is higher than 99%. As in a) the refractive index of the

glass is 1.54

# Polarisation degree is given by:

deg

perp par

perp par

I IP

I I

P6-13. Start with Fresnel equations (FoP 6.2-8 and FoP 6.2-9) and show that when light

impinges along the normal we have the following formula:

2

1 2

1 2

n n

n n

Utilize that, for small angles, we can approximate sin and tan with the angle .

* P6-14. a) Calculate the thickness of the thinnest possible quarter wave plate of crystalline quartz

with 0

n = 1.5497, e

n =1.5590 for the vacuum wavelength 486 nm. Such a plate can be

made thicker if it consists of two plates sandwiched together that counteract.

b) A plane monochromatic, linearly polarized wave is incident on two thin quarter wave retarders according to a). They are placed after each other so that their optical axes

have the angle respectively relative to the incident linear polarization direction.

Give for the following combinations of and the final polarization state after the

two retarders.

In the case when the resulted light is elliptically polarized, give the ratio between the short

axis and the long axis in the ellipse describing the rotation of the E-vector. It is not

necessary to give the rotation direction.

Angles I II III IV

45 45 0 0

45 0 45 20

P6-15. Between two crossed polarizers (figure 6-3) there is a glass plate with thickness d. This

plate can be slightly birefringent if a mechanical force is applied at 45 relative to the two

linear polarizers. Give how the transmitted irradiance I after the second polarizer depends on

the incident intensity o

I on the plate, the thickness d, the vacuum wavelength 0

and

0en n n the difference between the extraordinary and the ordinary refractive index for

the plate.

Figure 6-3. Adjusting the transmitted beam by a birefringent plate.

* P6-16. We have the following set-up (figure 6-5) for the wavelength 0.53 m.

PolarizerOptically

activecrystal

Vertical polarization

Verticalpolarization

45

Opticalaxisd d

MirrorFaraday rotator

Figure 6-4. A Faraday rotator combined with an optically active crystal.

After a linear polarizer there is a Faraday rotator that turns the polarization direction 45. This

rotator is followed by an optically active crystal that turns the polarization direction back to

the original direction. Answer the following questions:

a) A certain glass type with the Verdet constant 5.28 1

T cm

is used in the Faraday

rotator. It is possible to use a magnetic field of 1.0 T. How long glass rod has to be used?

b) As optically active crystal, a quartz crystal is used. The refractive index is -

n =1.544204

and +

n =1.544271. How thick must the crystal be?

c) After the optically active crystal there is a mirror that reflects the light back the same

path as it came from. Give the polarization state after the optically active crystal and after

the Faraday rotator. Give also the part (in %) of the reflex that passes back through the

first linear polarizer.

* P6-17. A Wollaston prism according to the figure 6-5 is used. The surrounding medium is air and the optical axes are marked.

30

30

Figure 6-5. A Wollaston prism.

The two parts of the prism are in contact, e.g. we can neglect any influence of air or glue at

the border surface. The material is calcite with 0

n =1.6584 and e

n =1.4864. Unpolarized light

is impinging from left. Draw the rays through the material with an accurate mark of the

polarization states. Calculate also the angle between the resulting rays.

* P6-18 A 0.900 mm thick plane parallel plate of crystalline quartz has the end surfaces parallel with the optical axes. It is placed between two polarizers which have their transmission

directions parallel so that the optical axis of the quartz plate makes an angle of 45 with the

transmission axes of the polarizers. A number of wavelengths are missing e.g. the two

adjacent wavelengths 443 nm and 467 nm. Calculate 0

.e

n n n

* P6-19. The following figure (figure 6-6) shows the cross section of a Rochon prism with the

optical axes marked. The prism is of calcite with 0

n =1.6584 and e

n =1.4864.

Draw the rays through the prism for an incident ray of unpolarized light that impinges

perpendicular to the surface on the left. Mark the polarization state of the beam and calculate

also the angle between the two emergent beams.

30

30

Figure 6-6. A Rochon prism.

Answers to the problems:

P1-2: 16 2m P1-3: b) 20

P1-4: 4.8

P1-6:

b = -2.0 cm, 3.0 cmb

y

F

P1-8: f = 24 cm in air and f = 94 cm if immersed in water.

P1-9: a) 181 mm to the left of the first surface b) virtual

P1-10: f = 4 cm

P1-11: f = -10 cm, 2

R = 5 cm, y(final image) -2.6 cm, upside down

P1-12 : a) 25 cm b) -10 cm

P1-13 : a) 27 cm b) 80 cm to the right of the lens. The image is upside down and enlarged 2x.

P1-14 : 750 m

P1-15 : a) -33 b) 80 cm x 1.2 m

P1-16 : +2,5 cm

-2,5 cm

+1,5 cm

-1,5 cm

Focal plane

P1-17 : 15 cm

P1-18 : NA = 0.21, a

= 12

P1-19 : NA = 1

P1-20 : 0.03 mm

P1-21 : See FoP Figure 1.3-5

P1-22 : NA = 0.49, period = 20,7 mm, length = 10,4 mm

P1-23 . Graded-index fiber: NA = 0.2104, Step-index fiber: NA = 0.2098

P1-24:

0

0

1( )

0 1

d dz

n zM

P1-26:

2 1

1 2

1

1 (1 )

11

d dd d

f fM

d

f f

P1-28: a) 1

11

01

122

12

RRRn

nnDC

BAM b)

212

21111

RRn

nn

f

P1-29: 1 21 2' '

1 2

;r r

R Rr r

P1-30:

sincos

sin cos

dd

M

d d

P2-1: A 1.5 mV/m, T = 20ns, = 6 m, arg (A) = 32

,+ ,since the wave travels to the left.

P2-2: a) A 3.5 mV/m, T = 20ns, = 6 m, arg(A) = 3

, ,since the wave travels to the

right. In the right region we have A 2.8 mV, T = 20 ns, = 4.2 m, n = 1.43, arg(A)

=1.23

b) 4 / 33.5j

u e

mV/m

P2-3: a) 488 nm b) n = 1,70 c) 143.6 10 Hz

P2-4: a

P4-7: ( )xx x

x x x 13

2 21

3

when - 1 < < 1, ( ) = 0 when

P4-8: a) a-f, c-d, e-i, b is the correlation of h and g

b) c is a sum of two object: a triangle and a pattern of small openings. d is the sum of

the FT of a triangle and the FT of the openings

c) First take the FT of each function. Multiply the result and take IFT of the new

result. This is the convolution of the original functions. This seams complicated but

is done quickly by a FFT(Fast Fourier Transform) algorithm.

P4-9: The lines will lean 45 and the distance will change to 1 2/ times what it was before.

P4-10:The incoherent system is the best.

P4-11: 0.54 mm

P4-12

From lines in figure 3-5a.

FT of figure 3-5a FT of figure 3-b

Opening for the zero sp. frequency

a) b)

c) 9 mm d) f = 6.8 cm

P4-13: 810

P4-14: a) 1.52 m b) 68, 76 , 31.05 , 5.71 , 18.47 and -47.14

P5-1: This is a pulse of width 2 . The pulse is moving in the positive z-direction and the light

is linearly polarized along the x-axis. 0

0

( )z

H f t yc

P5-2:

Relation 1 2 3 4

a) 0

P E a E Yes No No Yes

b) 2

0P aP E No No Yes Yes

c) 2

1 2 02

P Pa a P E

t t

Yes Yes Yes Yes

d) 2 20 1 2 expP a a x y E Yes No Yes No

1) Linearity 2) Dispersion 3) Isotropy 4) Homogeneity

P5-3: 162 10 photons/s

P5-4: a) 27.96 W/m b) 77.4 V/m and 7 2

2.58 10 Vs/m

P5-5: 2.4 kV/m

P5-6: 0.22 kV/m and 0.73 T

P5-7: a) 145 10 Hz b) 0.6 m c) 0.39 m d) 8

1.95 10 m/s e) 1.538

P5-8: a) 488.2 nm b) n = 1.698 c) 20 T d) 228 kW/m

P5-9: a) 58.46 10 s

b) 81 ns

P6-1: If orthogonal then no retardation is obtained.

P6-3: 0 0 0

/ 2, / 4 and / 8I I I

P6-4: 72

P6-5: a) in the positive x-direction, left elliptical polarisation

b) in the positive y-direction, left circular polarisation

c) in the positive z-direction, right elliptical polarization

P6-6: a) right circularly polarized b) left circularly polarized c) linearly polarized d) right

elliptically polarized e) right elliptically polarized

P6-7: a) 0 b) 0

0.25 I c) 0

0.42 I d)

2

0cos

2

N

IN

e)0

I

P6-8: 56.3

P6-9: TE: 0.092 TM: 0.0085 For unpolarized light 0.05

P6-10: 29.7

P6-11: a) 56.3 b) 90 This value is independent of n

P6-12: a) 8.27 % b) at least 15 plates

P6-14: a) 613 10 m

b)

Optical axis Optical axis

I 45 Circular 45 Linear

II 45 Circular 0 Linear

III 0 Linear 45 Circular

IV 90 Linear 20 Elliptic 20

For the elliptic polarization: Long axis sin 20

tan 20 0.36Short axis cos 20

P6-15: If 0

2n d

then

2

0sin / 2I I

P6-16: a) 85 mm b) 2.0 mm c) linear at 45 and at 90, I = 0

P6-17: Se FoP Figure 6.6-3, the angle between the beams is 11.4

P6-18: 0.0096

P6-19: See FoP Figure 6.6-3, the angle between the beams is 5.8

www.lth.se [TYPO3 4.2.16]

http://www.lth.se/typo3/backend.php[2011-08-28 14:48:51]

Atomic Physics

modern_trends

About

Research

Applied molecular spectroscopy..

Attosecond Physics

Attosecond Physics Research

Group Members

Open Positions

Master Projects

Publications

Support and Collaborations

Internal Pages

FLASH2011

Biophotonics

Quantum Information

Ultrafast X-ray science

Ultra-high intensity laser phy...

Attosecond Physics

Biophotonics

Education

Photonics

The Photonics Programme

Elective Courses

FAFF01 Optics and Optical De

Formal course plan

News

News 2010

Literature

Schedule

Exercises

FRED

Laboratory exercises

Examination

FAF080 Atomic and Molecular

FAFN01 Lasers

Formal course plan

News

Literature

Schedule

Exercises

FRED

Laboratory exercises

Examination

Kolumner

Skvg: ...urses/FAFN01 Lasers/Exercises/ [pid: 51683]

Sidinnehll

NORMAL

Swedish (Standard) Text

Indexera: YesProblems

HG

Swedish (Standard)

Indexera: YesSolutionsChapter 3Chapter 10Chapter 10bChapter 13Chapter 14Chapter 15

Visa dolda innehllselement

SK

Skord: Denna sida

Visa element:

Sida

Visa

Lista

Funktioner

Mailformplus Admin

Visa filer

Anvndarinstllningar

Om moduler

Om

Manual

Webb

Fil

Anvndarverktyg

Hjlp

L'Huillier, Anne [fysi-alh]

http://www.typo3.com/http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?noReset=1http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/alt_db_navframe.php?&currentSubScript=sysext%2Fcms%2Flayout%2Fdb_layout.php?AcrobatWebCapTID2#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_new_content_el.php?id=51683&sys_language_uid=0&returnUrl=%2Ftypo3%2Fsysext%2Fcms%2Flayout%2Fdb_layout.php%3Fid%3D51683http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/move_el.php?table=pages&uid=51683&returnUrl=%2Ftypo3%2Fsysext%2Fcms%2Flayout%2Fdb_layout.php%3Fid%3D51683http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/db_list.php?id=51683&returnUrl=%2Ftypo3%2Fsysext%2Fcms%2Flayout%2Fdb_layout.php%3Fid%3D51683http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683&clear_cache=1http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/tce_db.php?&data[tt_content][118087][hidden]=1&redirect=%2Ftypo3%2Fsysext%2Fcms%2Flayout%2Fdb_layout.php%3Fid%3D51683&vC=e00ea98ee0&prErr=1&uPT=1http://www.lth.se/typo3/tce_db.php?&cmd[tt_content][118087][delete]=1&redirect=%2Ftypo3%2Fsysext%2Fcms%2Flayout%2Fdb_layout.php%3Fid%3D51683&vC=e00ea98ee0&prErr=1&uPT=1http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/tce_db.php?&data[tt_content][118089][hidden]=1&redirect=%2Ftypo3%2Fsysext%2Fcms%2Flayout%2Fdb_layout.php%3Fid%3D51683&vC=e00ea98ee0&prErr=1&uPT=1http://www.lth.se/typo3/tce_db.php?&cmd[tt_content][118089][delete]=1&redirect=%2Ftypo3%2Fsysext%2Fcms%2Flayout%2Fdb_layout.php%3Fid%3D51683&vC=e00ea98ee0&prErr=1&uPT=1http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#http://www.lth.se/typo3/sysext/cms/layout/db_layout.php?id=51683?AcrobatWebCapTID4#

www.lth.sewww.lth.se [TYPO3 4.2.16]

hwP0Fjcm9iYXRXZWJDYXBUSUQyAA==: treeFilter:

Y4Mz9BY3JvYmF0V2ViQ2FwVElENAA=: form0: SET[function]: [1]input0: input0_(1): input0_(1)_(2): input0_(1)_(2)_(3): input1: input3: search_field: search_levels: [0]search: showLimit: SET[tt_content_showHidden]: on

guc2UvdHlwbzMvYmFja2VuZC5waHAA: form1: input0:

optics and optical design 2011 problems - atomic physics · 2011-08-28 12:31 this material is ......

Documents