ch5.pdf

CHAPTER FIVE PHYSICAL OPTICS 1|P a g e  Q) Explain an activity in which you demonstrate interference of waves. Activity Tools: A ripple tank apparatus Power Supply. Vibrator. pecker with two tapered ends as two- point sources (S1, S2) emitting circul...

CHAPTER FIVE PHYSICAL OPTICS 1|P a g e  Q) Explain an activity in which you demonstrate interference of waves. Activity Tools: A ripple tank apparatus Power Supply. Vibrator. pecker with two tapered ends as two- point sources (S1, S2) emitting circular waves propagating on the surface of water of the same wavelength. Activity Steps: We get the ripple tank ready to work as it touches both sides of the pointer the surface of the water in the container. When the operating the motor, we see the interference pattern on the water surface as a result of superposition of waves generated by vibration of the symmetrical pointer sources (S1, S2), Conclusion: From our observation for the produced waves at the water surface, it appears that there are two kinds of interference which are: 1) Constructive Interference: When the two waves are in same phase and same amplitude at a certain point, the two waves will combine with each other and in this case the resulting wave amplitude will be equal to the twice amplitude of any of the original waves. This is resulted from overlapping two crests or two troughs for two waves resulting strengthening. 2) Destructive Interference: If the interference is a result of combination of two sequence of waves which are out of phase and equal in amplitudes, in which case the crest of a wave overlaps the trough of another wave, in this case the result will be that the effect of one of the waves cancels the effect of the other wave. This means that the amplitude of the resulting wave will be zero. 2|P a g e  Q) Mention the types of interference, and explain each type. 1) Constructive Interference: When the two waves have the same phase and amplitude, where the two waves warn at a certain point to strengthen each other and in this case (the amplitude of the resulting wave equal to twice the amplitude of either of the original two waves). This interference results from the superposition of two peaks or troughs that result in strengthening. 2) Destructive Interference: Generated when the two waves have two opposite phases of equal amplitude and this interference results from the superposition of one-wave peaks with the other wave trough, and this resulted in the effect of one wave cancels out the effect of the other one (that the amplitude of the resulted wave equal zero). Q) Define the concepts of (Coherent Waves, Waves Interference & Light Interference). 1) the coherent waves: are waves of equal frequency, equal in amplitude and the phase difference between them is constant. 2) Waves Interference: It is the superposition of two or more sequences of coherent waves when its propagates in one plane at the same time and in the same media. 3) the light Interference: it is the phenomenon of redistribution of light energy resulting from the superposition of two or more sequences of coherent light waves when its propagates in one plane at the same time and in the same media. Q) What are the conditions for permanent interference? 1. If the two waves are coherent. 2. If they vibrate in the same plane and in the same media and pass through the same point at the same time. 3|P a g e  Q) What is meant by the optical path? What is the mathematical relationship to calculate the optical path difference? The optical path is the displacement done by the light in vacuum by the same period of time which is done in the transparent material media. the optical path difference between these waves by the following relation: ∅λ ∆ℓ = 2𝜋 Where: (∆ℓ): Optical Path Difference, (∅): Phase difference Angle, and (λ): Wavelength Q) What is the magnitude of the phase difference (∅) and the optical path difference (∆𝓵) in the constructive interference? ∅ = [0, 2𝜋, 4𝜋, 6𝜋, … … ] the phase difference angle is equals zero or an even numbers of (𝜋), and we obtain the optical path difference from the relationship: ∅λ ∆ℓ = 2𝜋 And we obtain: ∆ℓ = [0, λ, 2λ, 3λ, … … ] Where, the optical path difference equal zero or integer numbers of the wavelength, then: ∆ℓ = 𝑚λ Where, [m= 0, 1, 2, 3, 4…...] Q) What is the magnitude of the phase difference (∅) and the optical path difference (∆𝓵) in the destructive interference? ∅ = [0, 3𝜋, 5𝜋, 7𝜋, … … ] the phase difference equals to odd numbers from (π rad), and we obtain the optical path difference from the relationship: ∅λ ∆ℓ = 2𝜋 And we obtain: 1 3 5 ∆ℓ = 3 λ, λ, λ, … … 5 2 2 2 Where, the optical path difference equal odd numbers of half wavelength, then: 1 ∆ℓ = 6𝑚 + 8 λ 2 Where, [m= 0, 1, 2, 3,..….] 4|P a g e  Q) What are the conditions to obtain constructive and destructive interference? The condition to obtain constructive interference ∆ℓ = 𝑚λ Where, [m= 0, 1, 2, 3, 4…...] The condition to obtain destructive interference 1 ∆ℓ = 6𝑚 + 8 λ 2 Where, [m= 0, 1, 2, 3...….] Q) What does the type of interference depend on? Q) What factor determines the type of interference between waves? Q) How do you know the type of interference? Explain this. By optical path difference (∆ℓ). If the optical path difference is zero or integers of wavelengths, then it is constructive interference: ∆ℓ = 𝑚λ Where, [m= 0, 1, 2, 3, 4…...] If the optical path difference is odd numbers of half-wavelengths, then it is interference Destructive: 1 ∆ℓ = 6𝑚 + 8 λ 2 Where, [m= 0, 1, 2, 3...….] Ministerial Exams: Q.1) What is the difference between the light coherent and non- coherent sources? Q.2) When does the permanent interference between two light waves occur? Q.3) What is meant by light coherent waves? Q.4) What happens when two light coherent waves are interfering if the optical path 𝟑 difference between them: ( 𝟐𝛌 and 𝛌)? 𝟐 5|P a g e Q.5) What is the difference between constructive interference and destructive interference in terms of the optical path difference for each of them between two light coherent waves? Q.6) Put the mark (true) or (false) then correct the mistake: Destructive interference occurs if the optical path difference between the two interfering waves equals zero or integers of wavelength. Q.7) What is meant by (light interference)? Q.8) What are the most important properties of electromagnetic waves? (2) YOUNG’S DOUBLE SLIT EXPERIMENT Q) Explain Young’s experiment in proving the light interference phenomenon. We put a barrier with a tiny slit and monochromatic light, then the light was projected on a barrier containing two identical narrow slits called (double slit) these two slits are situated at the same distance from the slit of the first plate. Then he put a screen at a distance of some meters from the two-slit barrier. The result obtained by Young was the appearance of some bright regions and some other dark regions consequently and this was called the fringe. Q) What is the conclusion reached by scientist Young? Light has a wave nature as it can calculate the wavelength of light used in the experiment. Q) What is the practical benefit of Young's experiment? Calculates the wavelength (λ) of monochromatic light. Q) How do you explain the appearance of bright and dark fringes in Young's experiment? The reason for this is that light waves interfere constructively and destructively. 6|P a g e  Q) Derive the general equation of Young's law of interference. Where (𝜽): Diffraction Angle, (𝒚): Distance between bright fringe and central fringe, (𝑳): Distance between screen and two slits, (∆𝓵): Optical path difference, (𝒅): Distance between the two slits, (𝛌): Wavelength, and (𝒎): Fringe value [m= 0, ±1, ±2, … …. ] 7|P a g e  Q) Derive the general form for the fringe spacing (ΔY). Q) What is the fringe spacing (the space between fringes) based on in Young's experiment? 1. The amount of fringe spacing (Δy) increases when the distance between the two slits and the screen (L) is increases. 2. The amount of the fringe spacing (Δy) increases if the distance between the slits (d) decreases. 3. The amount of the fringe spacing (Δy) increases when the monochromatic light wavelength used increases in young’s experiment. According to the relationship: 𝛌𝐋 ∆𝒀 = 𝒅 8|P a g e  In the case of using red light in the two slit young’s experiment, you will observe the distances between the interference fringe larger than what it is if you use the blue light. Why? Q) When using red light instead of violet light in the Young's experiment, what happens? Why? The separation between fringes which is (∆𝑌) will increase because the spacing are directly proportional to the wavelength and according to the relationship: 𝛌𝐋 ∆𝒀 = 𝒅 Since the wavelength of red light is greater than the wavelength of blue or violet light, the fringing interval is therefore greater. Q) What happens to the distance between the interference fringes in the Young’s experiment when the distance between the two slits decreases? The smaller the distance between the two slits the greater the distance between the fringes, so the fringe appears wider. Q) What is the effect of the wavelength change on the distance between the interference fringes? The relationship is directly proportional, so if the wavelength increases, the distance between the fringes increases. Q) If the white light is used in young’s experiment. How would the colour of the central light fringe appear on either side of the central fringe? The central fringe appears in white colour and on both sides white colour spectra will appear continuously. Gradually each spectra of the violet colour changes to the red colour Q) What do you expect to happen if the two sources of the light are incoherent? Would the constructive and destructive interference happen? In fact, both constructive and destructive interference take place consequently with a very high speed, so the eye can not follow such rapid change, because both of the sources are emitted waves with random change phases, and a high-speed making it impossible to obtain a constant phase difference between the interfered waves at any point of the medium. So the eye can see sustainable lighting as a result of vision persistence. 9|P a g e Ministerial Exams: Q.1) What does the interference type depend on in young’s experience? Q.2) Do the fringes appear in the Young’s experiment if the two light sources are non- coherent? Why? Q.3) What is the purpose of young’s experiment? Q.4) What does the fringe spacing (the distance between two consecutive fringes) depends on, in young’s experiment? Q.5) Why did a bright fringe and a dark fringe occur in young’s experiment? Q.6) How does the magnitude of the fringe spacing change in the Young’s experiment by changing (the distance between the two slits from the screen, the distance between the two slits, the wavelength of the monochrome light used)? Q.7) Explain Young’s experiment to obtain interference in light, demonstrating the practical benefit from conducting the experiment (or how to calculate the wavelength of the light used). (3) INTERFERENCE IN THIN FILMS Q) How do you explain the interference in the thin films and the fringes occurrence in them? The incident light waves on the film some of it is reflected from front surface of the thin film and get phase shift equals (π rad) because each wave reflected from a media with refractive index greater than the refractive index of the media which came from it, will get a shift in phase by (180o), The remainder of the light wave transmitted in the film and suffer refraction inside the film, when it reflected from the back surface of the film, (with thickness t), will not suffer phase shift, it will make extra optical path equals twice the film’s optical thickness (2nt). So, there is interference between the two opposite waves from the front surface and the back surface according to the amount of the phase difference. 10 | P a g e  Q) we sometimes see the floating oil spill on the surface of the water colorful with bright colors. We also see sometimes soap bubble with the colors of the rainbow colors? Why The cause of interference between white light waves reflected from the front and the back of the thin films surface. Q) What does interference in thin films depends on? 1) Film Thickness of the film: The waves reflected from the back surface makes a distance more than that reflected waves from the front surface path which is equal to twice the thickness of the film. 2) Phase shifts: The reflected waves from the front surface get a phase shift as (π rad). Q) What is the magnitude of the phase difference (") between the incident waves and the reflected waves from the front surface of the thin film, and why? (! rad) because each wave reflected from a media with refractive index greater than the refractive index of the media which came from it, will get a shift in phase by (180o), Q) What is the magnitude of the phase difference between the incident wave and the reflected wave from the back surface of the thin film? Why? (Zero) The reason is that each wave is reflected from a medium whose refractive index is equal to the refractive index of the medium from which it emitted does not have a phase shift. 11 | P a g e  Q) What is the magnitude of the optical thickness (nt) of the thin film in the constructive interference? Optical thickness (nt) in constructive interference are odd numbers of quadrants of wavelength. 𝟏 𝟑 𝟓 𝒏𝒕 = [ 𝝀, 𝝀, 𝝀, … … ] 𝟒 𝟒 𝟒 Q) What is the optical thickness (nt) of the thin film in destructive interference (when the film appears dark)? The optical thickness (nt) in destructive interference is even numbers of quarters of the wavelength. 𝟐 𝟒 𝒏𝒕 = [𝟎, 𝝀, 𝝀, … … ] 𝟒 𝟒 Q) What type of interference is in thin films if the optical thickness of the film is (nt) odd numbers of quarters of wavelengths? Q) Does the thin film appear bright or dark if the optical thickness (nt) is odd numbers of quarters of the wavelength? The type of interference is constructive, as the film appears illuminated. Q) What type of interference is in thin films if the optical thickness of the film is (nt) even numbers of quarters of wavelengths? Q) Does the thin film appear bright or dark if the optical thickness (nt) is even numbers of quarters of the wavelength? The type of interference is destructive, as the film appears dark. Ministerial Exams: 𝟏 𝟑 Q.1) What kind of interference in thin films if the thickness of the optical film ( 𝝀, 𝝀)? 𝟐 𝟒 Q.2) What happens to the incident light on a thin film (such as a soap bubble film)? 12 | P a g e Q.3) How much should the optical thickness of the thin film be in order to obtain constructive interference for the monochromatic light incident on the film? Q.4) Brightly colored oil stains floating on the surface of the water. Give reason. Q.5) Explain what happens to the incident light on a thin film (such as a soap bubble film)? Q.6) What does interference in thin films depend on? Q.7) What should be the optical thickness of the thin film in order to obtain (destructive interference)? (4) DIFFRACTION OF LIGHT Q) Explain an experiment (activity) that demonstrates the phenomenon of diffraction of light. Activity Tools: Glass Plate. Pin. Black Paint. Monochromatic Light Source. Activity Steps: 1) Paint the glass plate with black paint. 2) Make a thin slit in the glass plate, using the tip of pin. 3) Look through the slit at the light source. Observation: we will notice bright regins interspersed with some dark regins and the middle region is wide and very bright. The bright fringe gradually decreases in intensity as you move away from the central bright fringe. Conclusion: The appearance of bright and other dark regins on both sides of the slit Indicate that the light deviating from its path. 13 | P a g e  Q) What are the conditions for obtaining dark or bright fringes? The condition for obtaining a dark fringe is: 𝓵𝒔𝒊𝒏𝜽 = 𝒎𝝀 The condition for obtaining a bright fringe is: 𝟏 𝓵𝒔𝒊𝒏𝜽 = 6𝒎 + 8 𝝀 𝟐 Where (ℓ): Slit width Ministerial Exams: Q.1) In the phenomenon of light diffraction, what is the condition for obtaining a dark fringe and a bright fringe in the single-slit experiment? Q.2) Explain an activity showing diffraction of light waves. (5) DIFFRACTION GRATING Q) Define a diffraction grating, and how is it manufactured? What is its use? Diffraction Grating: It is a glass plate consisting of a large number of parallel grooves with equal width. It is manufactured: can be made by printing grooves on a glass plate in a ruling very accurate machine. The widths between the grooves will be transparent as its doing separate slits and the groove its dark region.Ranging Number of the slits (line) in each centimeter will be between: (1000 – 10000 line/cm). Its usefulness: 1) Study of spectra. 2) Light analysis. 3) Measurement of the wavelength of light. 14 | P a g e  Q) What is the grating constant (d) and what is the mathematical relationship with which it is given? The grating constant (d) has a very small value, and it represents the distance between two successive grooves. 𝑾 𝒅= 𝑵 Where (d): Grating Constant, (W): Grating Width, (N): Gratings Number Q) What is the magnitude of the optical path difference between the two adjacent rays in the diffraction grating? Optical path difference (∆𝓵) is equal to ∆𝓵 = 𝒅 𝒔𝒊𝒏𝜽 Q) When is the fringe generated in the diffraction experiment by the grating illuminated? If the optical path difference is equal to one wavelength (λ) or integers number of the wavelength (mλ) according to the following relationship: 𝒅 𝒔𝒊𝒏𝜽 = 𝒎𝝀 Where m = [0, 1, 2, 3, ……] Q) What is the practical benefit of using a spectrometer? It is used to calculate the wavelength of monochromatic light according to the relationship: 𝒅 𝒔𝒊𝒏𝜽 = 𝒎𝝀 Where m = [0, 1, 2, 3, ……] 15 | P a g e Ministerial Exams: Q.1) How does the diffraction angle of a bright fringe with known value change by decreasing of the grating constant? Explain this. Q.2) What is the purpose (what is the use) of the diffraction grating? (6) POLARIZATION OF LIGHT Q) Explain an experiment (activity) that demonstrates the phenomenon of mechanical wave polarization. Activity Tools: A rope which is fixed at one of the ends to a wall A Barrier with a slit. Activity Steps: 1) Pass the free end of the rope through the slit of the barrier, such that the slit is vertical upwards, and the rope line is perpendicular to the slit. 2) Tighten the rope and then hit it so that it creating transverse wave which passes through the slit 3) Now we make the slit horizontal and tighten the rope and then hit it. We observe that the transverse waves cannot pass through the slit, Conclusion: Because the waves whose plane of oscillation is parallel to the slit are those, that pass through the slit and whose plane of oscillation is vertical cannot pass through the slit. 16 | P a g e  Q) Explain an experiment (activity) about the polarization of light waves. Activity Tools: Two slices of tourmaline. source of light. Activity Steps: 1) Take a slice of tourmaline and place it on the way of the light source. 2) Rotate the slice about its central and perpendicular axial. Note: The light intensity of the tourmaline slide does not change. 3) Put two slices of tourmaline, fix one of the slices and start rotating the other one slowly about the light rays. We notice the change in the intensity of the light. Conclusion: Unpolarized light is transverse waves, and its electric field vibrates in all directions. The crystallization of the tourmaline will have the molecules arranged as long chain which does not allow light waves to pass through unless the level of its electric field vibration perpendicular to the line of the chain. Whereas it observes the rest of the waves. This operation is called Polarization and the light waves are called Polarized light waves. The slice which performs this operation the Polarizer and the second slice is called the Analyzer. Q) Explain an experiment (activity) that demonstrates the effect of a polarized material on the intensity of light transmitted through it. Activity Tools: A monochromatic light source Two slices of tourmaline. photocell. Activity Steps: 1) We put the light source in front of the polarizer plate. Then we put the second plate analyzer behind it. We notice the decrease in the intensity of the light passing through the two plates. 2) We start rotating the analyzer plate until the intensity of the light completely disappears. Conclusion: The unpolarized light passing through the polarizer plate has been plane polarized and its intensity decreased, when it passes through the analyzer plate, its intensity has decreased more. When the analyzer plate is rotated at a certain position, we observe that the intensity of the light completely disappears by looking through it. This proves that the polarized light has been obstructed by the analyzer completely. 17 | P a g e  Q) Define (Polarization Phenomenon, Polarized Light, and Non-polarized light). 1) Polarization Phenomenon: It is the phenomenon in which the light electric field oscillation is limited to one plane only and perpendicular to the light propagation line. 2) Polarized light: It is the light in which the electric field of the electromagnetic waves oscillates in one direction perpendicular to the wave propagation line. 3) Non-polarized light: It is the light in which the electric field of electromagnetic waves oscillates in random directions and in parallel and perpendicular levels to the wave propagation line. Q) Which phenomena indicate that? 1) Light is wave in nature. 2) Light is a transverse wave. 1) Interference and diffraction. 2) Polarization. Q) What do the phenomena of light diffraction and light interference indicate? Denotes the wave nature of light. Q) What are the methods of polarization with light? 1) Polarization by selective absorption, as in a tourmaline crystal (polar materials). 2) Polarization of light by reflection, obtained from reflective surfaces such as flat mirrors and water surfaces. 18 | P a g e  Q) How does polarization occur by reflection when a non-polarized light falls on a reflective surface such as a mirror or water surface? when light incident on a reflected surfaces such as plane mirrors or the surface of water in a lake or glass, then the reflected light will be partially polarized and, in a plane, parallel to the reflected surface plane While the refracted light in the second media will be at the plane of incident rays. The degree of polarization depends on the incidence angle. If the incidence angle is zero, then there will be no polarization. At the same time the polarization will be increase by increase of the incidence angle until it reaches the total linear polarization at a certain angle called Brewster Angle The refracted ray will be partially polarized and the angle between the reflected ray and the refracted ray equals (90°). The scientist Brewster found a relationship between the angle of polarization (𝜃5 ) and the coefficient refraction of the media, as follows: 𝒕𝒂𝒏𝜽𝑷 = 𝒏 Q) What is the Brewster Angle? It is the angle of incidence of non-polarized light, which makes the reflected light totally polarized and the refracted light partially polarized, and the angle between the refracted and the refracted is right angle (90°) and by which the coefficient refraction of the media can be obtained by the following relationship: 𝒕𝒂𝒏𝜽𝑷 = 𝒏 Q) What does the degree of polarization by the reflection method depend on? It depends on the angle of incidence and the polarization increases with increasing angle of incidence. 19 | P a g e  Q) In polarization by reflection under any conditions: 1) Polarization does not occur. 2) Equatorial total polarization occurs. 1) If the angle of incidence (θ) is equal to zero, the polarization does not occur. 2) At an angle of incidence called Brewster angle (θP) a total equatorial polarization takes place for the reflected ray, and the refracted ray at this angle is partially polarized, and in this case, the angle is right angle (90°) between the reflected ray and the refracted ray. Q) What is the effect of the light incidence angle on the reflective surface in the degree of polarization? If the light incidence angle is equal to zero, polarization does not occur. The degree of polarization increases with the increase in the incidence angle until it reaches a total equatorial polarization at a certain angle called the Brewster angle (θP). Q) What are the optical active materials? it is materials have the ability to rotate polarization plane of the polarized light when it passes through these materials with an angle called the angle of optical rotation. (such as quartz crystal, turpentine liquid, sugar solution in water). Q) What does the optical rotation angle depend on? 1) Material type. 2) Material thickness. 3) Solution concentration. 4) Light wavelength. 20 | P a g e Ministerial Exams: Q.1) Why the sunlight and the ordinary lamps are not polarized? Q.2) What does the light polarization degree by the reflection method depends on? Q.3) What is meant by (polarized light)? Q.4) What does the optical rotation angle depend on in optically active materials? Q.5) If the light is polarized by reflection under any conditions: 1) Polarization does not occurs. 2) Total equatorial polarization occurs. Q.6) What are the methods of polarization in the light? Q.7) How to obtain a linear (equatorial or total) light beam from a non-polarized light beam? Q.8) What is the effect of increasing the light incidence angle on the reflective surface on the degree of polarization? Q.9) How does the polarization of light by reflection can occurs? With drawing. Q.10) What is meant by the optical active substances? Q.11) Explain an activity in which you demonstrate the polarization of the waves. Q.12) Explain with an activity the effect of polarized matter on the intensity of polarized light transmitted through it. Q.13) Explain with an activity the polarization of light waves with inference. 21 | P a g e  (7) SCATTERING OF LIGHT Q) What is the phenomenon of scattering in light? It is the phenomenon of diffraction of light falling on particles whose diameters are close to the wavelength of the incident light, where the intensity of the scattering of light is inversely proportional to the fourth exponent of the wavelength. 𝟏 Scattering Intensity ∝ 𝝀𝟒 Q) What is the reason for the blue colour of the sky? Because of the scattering of the blue color, because its wavelength is short, the intensity of its scattering is great. 𝟏 Scattering Intensity ∝ 𝝀𝟒 Q) What is the cause of the red and orange colors of the horizon at sunrise and sunset? This is due to the lack of scattering of red and orange light because they are of long wavelengths. Ministerial Exams: Q.1) What is meant by the (scattering phenomenon)? Q.2) What is the reason for seeing the sky blue on the surface of the earth without stars during the day? Q.3) What is the reason for the appearance of the sun disc red during sunrise and sunset? Q.4) Why are short light waves scattered more than long light waves? Q.5) Why the sky appears pale blue when the sun is above the horizon during the day? 22 | P a g e  Chapter Five Laws YOUNG’S DOUBLE SLIT EXPERIMENT Constructive Interference Condition: ∆𝓵 = 𝒅 𝒔𝒊𝒏𝜽 = 𝒎𝝀 Destructive Interference Condition: 𝟏 ∆𝓵 = 𝒅 𝒔𝒊𝒏𝜽 = 6𝒎 + 8 𝝀 𝟐 Constructive and Destructive Interference: ∆𝓵 = 𝓵𝟐 − 𝓵𝟏 If the Interference Constructive: 𝒚𝒅 𝝀= 𝒎𝑳 If the Interference Destructive: 𝒚𝒅 𝝀= 𝟏 T𝒎 + U 𝑳 𝟐 Constructive and Destructive Interference: 𝝀𝑳 ∆𝒚 = 𝒅 23 | P a g e  INTERFERENCE GRATING If the Interference Constructive: ∆𝓵 = 𝒅 𝒔𝒊𝒏𝜽 = 𝒎𝝀 If the Interference Destructive: 𝟏 ∆𝓵 = 𝒅 𝒔𝒊𝒏𝜽 = 6𝒎 + 8 𝝀 𝟐 𝑾 𝒅= 𝑵 Where (d): Grating constant, (W): Interference grating width, (N): Grooves number SINGLE SLIT INTERFERENCE EXPERIMENT If the Interference Constructive: 𝟏 𝓵 𝒔𝒊𝒏𝜽 = 6𝒎 + 8 𝝀 𝟐 If the Interference Destructive: 𝓵 𝒔𝒊𝒏𝜽 = 𝒎𝓵 Where (ℓ): Slit width 24 | P a g e  POLARIZATION OF LIGHT 𝟏 𝒏= 𝒔𝒊𝒏𝜽𝒆 𝒕𝒂𝒏𝜽𝒑 = 𝒏 Where (𝒏): Refraction index, (𝜽𝑷 ): Polarization (Brewster) angle, (𝒏𝒆 ): Critical angle Chapter Five Notes 1) If the question is asked to know the type of interference, it must be obtained (∅) from a law of: 𝟐𝝅∆𝓵 ∅= 𝝀 If it is an even number of (𝝅), then the interference is constructive, and if it is an odd number of (𝝅), then the interference is destructive. 2) In problems of the diffraction grating (pay attention), the value of the grating constant (d) is often in units of (cm) so it must be converted into units of meters (m). 25 | P a g e 26 | P a g e  Chapter Five Questions Q.1. Choose the best statement from the followings: 1. In light diffraction from one slit, the condition that the first bright fringe (non central), is that the width of the slit is equal: a. 𝝀 𝝀 b. 𝟐𝒔𝒊𝒏𝜽 𝟑𝝀 c. 𝟐𝒔𝒊𝒏𝜽 𝝀 d. 𝟐 2. The colors of the soap bubbles occurs because of: a. Interference. b Diffraction. c. Polarization. d. Scattering. 3. The reason of bright and dark fringes in Young’s double slit experiment is: a. Diffraction and interference of light waves together b. Diffraction of light only c. Interference of light only d. Using two sources of light which are not coherent 4. When a green light is used in diffraction grating, the central fringe appears: a. Yellow. b. Red. c. Green. d. White. 5. Light diffraction angle increases with: a. Decrease of wavelength for the used light. b. Increase of wavelength for the used light. c. Constant wavelength for the used light. d. All of the above. 6. If the optical path difference between two optical waves which are coherent and overlapped equals odd numbers of half wavelength, then the following will happen: a. Constructive Interference. b. Scattering. c. Polarization. d. Destructive Interference. 27 | P a g e 7. To get the permanent interference in light waves, their sources must be: a. Coherent. b. Incoherent. c. Two laser sources. d. All above the cases. 8. In Young’s double slit experiment the first bright fringe occurs on both sides of bright central fringe which has been formed on the screen, when the difference of optical path equals: a) 1/2λ b) λ c) 2λ d) 3λ 9. The pattern of interference is generated when the following happens: a. Reflection. b. Refraction. c. Diffraction. d. Polarization. 10. The thin oil films and the soap bubbles appear shiny colored as a result of reflection and: a. Refraction. b. Interference. c. Diffraction. d. Polarization 11. The property of the generated spectrum by diffraction grating will be a. Bright lines very clear. b. Expanded bright lines. c. Non-existence of bright lines. d. Non-existence of dark lines. 12. The unpolarized light ray which electric fields oscillates: a. In one plane. b. In all direction. c. which cannot transmitted through polarized plate. d. In specific direction. 13. The longitudinal waves cannot show: a. Refraction. b. Reflection. c. Diffraction. d. Polarization. 28 | P a g e 14) The sky is blue because. a) Particle of air is blue b) The lens of eye is blue c) The light scattering would be ideal for the short wavelength d) The light scattering would be more ideal for the long wavelength. 15) Young’s double slits are lighted green, wavelength (5 x 10-7m), the distance between the slits is (1mm) and the distance of the screen from the slits is (2m). The distance between the centers of two bright fringe on the screen equals a) 0.1mm b) 0.25mm c) 0.4mm d) 1mm Q.2) Is it possible for the light from source which are incoherent to interfere? Is there any difference between coherent and incoherent sources? In fact, both constructive and destructive interference take place consequently with a very high speed, so the eye can not follow such rapid change, because both of the sources are emitted waves with random change phases, and a high-speed making it impossible to obtain a constant phase difference between the interfered waves at any point of the medium. So, the eye can see sustainable lighting as a result of vision persistence.. This is the main difference between coherent and non- coherent sources. Q.3) Two sources of light situated and they are side to side. The light waves are projected on a screen. Why dose not the interference pattern appear from the superposition of light waves from them on the screen? The light emitted from the two optical sources consists of several waves of different wavelengths, with variable random phases, meaning there is no coherence between the two sources, so the light from the two sources does not lead to a constant phase difference over time, so it is impossible to see the interference pattern. Q.4) When Young’s experiment is carried out under the surface of the water, how would interference pattern be affected? The wavelength of light in water is shorter than that in air, according to the following relationship: 𝝀 𝝀𝒏 = 𝒏 Since the bright and dark beams are in proportion to the wavelength (λ), the widths between the interference fringes will be decrease. 29 | P a g e Q.5) What is the condition applied in the difference in optical path between two interfered coherent waves in case of: a) Constructive interference. b) Destructive interference. a) ∆𝓵 = 𝒎𝝀 If the optical path difference is equal to zero or integer numbers of wavelengths: ∆𝓵 = [𝟎, 𝝀, 𝟐𝝀, 𝟑𝝀, ….. ] 𝟏 b) ∆𝓵 = T𝒎 + 𝝀U If the optical path difference is equal to odd numbers of half- 𝟐 wavelengths: 1 3 5 ∆ℓ = 3 λ, λ, λ, … … 5 2 2 2 Where (m) = (0, 1, 2, 3, ……..). Q.6) An astronaut on surface of moon see sky as black and see star clear during the daylight. At the same time a man on the surface of earth sees the sky blue and cannot see stars. Explain why? During the day, and from the surface of the moon, the astronaut sees the sky black and is able to see the stars clearly due to the absence of the atmosphere and the particles that cause scattering of sunlight. Whereas during the day, on the surface of the earth, the sky is blue and without stars, due to the occurrence of the phenomenon of scattering (color dispersion) due to the presence of the atmosphere. Q.7) What happens to central bright fringe in one slit diffraction if the width of the slit is reduced? The width of the central bright fringe increases and be with low intensity according to the relationship of: 𝓵 𝒔𝒊𝒏𝜽 = 𝒎𝝀 𝟏 𝓵∝ 𝒔𝒊𝒏𝜽 Ministerial Exams: Q.1) If you conduct the Young’s experiment underwater surface, how does this affect the interference model? 30 | P a g e Q.2) Can light emitted from non-coherent light sources interfere. Q.3) What change occurs in the fringe spacing (the distance between fringes) in Young's experiment when the distance between the two slits decreases? Q.4) What condition is met in the difference in the length of the optical path between two coherent waves? In the case of: 1) Constructive interference 2) Destructive interference? Q.5) If white light is used in Young's experiment, how does the color of the central bright fringe appear? and how do the rest of the bright fringes appear on either side of the central bright fringe? Q.6) What happens in the width of the central bright region of the diffraction pattern from a single slit when we make the width of the slit even narrower? Q.7) Do the fringes appear in the Young’s experiment if the two optical sources are non- coherent? Why? Q.8) Choose the correct answer: the thin oil films and the water soap bubble film appear brightly colored due to reflection and (refraction - interference - diffraction - polarization). Q.9) Two light sources placed side by side together projected the light waves from them onto a screen, why does the interference pattern not appear from the superposition of the light waves emitted by them on the screen. Q.10) Choose the correct answer: Longitudinal waves cannot be shown (refraction - polarization - diffraction - polarization). Q.11) Choose the correct answer: The color of soap bubbles is due to the phenomenon of (interference - diffraction - polarization - scattering). Q.12) Choose the correct answer: In the diffraction of light from one slit, the condition of the first non-central bright fringe, where the width of the slit is equal to (λ, λ/2, λ/2sinθ, 3λ/2sinθ). Q.13) The thin oil films and soap bubble film appear brightly colored as a result of: (refraction and diffraction - reflection and diffraction - diffraction and interference - reflection and interference). 31 | P a g e Q.14) Choose the correct answer: the light diffraction angle increases with (decreasing the used light wavelength, increasing the used light wavelength, the constant used light wavelength). Q.15) During the day, and on the surface of the moon, the astronaut sees the sky is black and he can see the stars clearly, while during the day, and on the surface of the earth, he sees the sky blue and without stars. What is the explanation for that? Q.16) Choose the correct answer: The reason for the appearance of a bright fringe and a dark fringe in Young’s experiment is: (diffraction of light waves only - the use of two non-coherent light sources - the interference of light waves only - the diffraction and interference of light waves together). Q.17) Can constructive interference and destructive interference be obtained if the two optical sources are non-coherent? Q.18) In the Young’s experiment, the first bright fringe is obtained on either sides of the central bright fringe formed on the screen when the optical path difference is equal to (1/2λ, 3λ, 2λ, λ). Q.19) Mark (true) or (false) and then correct the error: The reason for the appearance of a bright fringe and a dark fringe in the Young’s experiment is the interference of only light waves. Q.20) Mark (true) or (false) then correct the error: the light diffraction angle increases with the increase in the wavelength of the used light. 32 | P a g e 33 | P a g e █ Example (1) In the adjacent figure, two coherent sources (S1 & S2) emit waves of wavelength (λ=0.1m), their waves interfere simultaneously at point (P) ❶ What type of interference is produced at this point when one of the two waves crosses an optical path of (3.2m) and the other has an optical path of (3m)? Interference type is constructive because the phase difference is an even number of π █ Example (2) If the distance between the two slits of Young’s experiment is (0.2mm) and the screen distance from them is (1m), and the distance between the third bright fringe and the central fringe is (9.49mm). ❶ Calculate the wavelength of light used in this experiment. 34 | P a g e █ Example (3) In the adjacent figure, red light of wavelength (λ=664 nm) was used in Young’s experiment and the distance between the two slits was (d=1.2X10-2 m) and the distance of the screen from the two slits (L=2.75m). ❶ Find the distance (y) on the screen between the third bright fringe and the center of the central fringe. █ Example (4) Monochromatic light of helium-neon laser of wavelength (λ=632.8nm) falls perpendicular to a diffrac- tion grating each one centimeter of which contains (6000 line). ❶ Find the diffraction angles (θ) for the first and second luminous orders. (Note that: sin 49=0.7592 & sin 21.3=0.37968) 35 | P a g e █ Q.1) A screen was placed at a distance of (4.5m) from a two-slit barrier and the two slits were illuminated with a monochromatic light of wavelength in air is (λ=490nm), then the distance between the center of the central bright fringe and the center of the bright fringe (m=1) is equal to (4.5 cm). ❶ What is the distance between the two slits? █ Q.2) White light, its spectrum compounds are distributed by a diffraction grating. If the groove has 2000Line/cm. ❶ What is the measure of the first order diffraction angle for red light of wavelength (λ=640nm)? 36 | P a g e █ Q.3) A beam of light falls on a reflecting surface at different angles of incidence, the reflected ray became fully polarized when the angle of incidence was (48°) ❶ Calculate the refractive index of the medium. █ Q.4) If the critical angle of the light rays of the blue agate material surrounded by air is (34.4°). ❶ Calculate the angle of polarization for the light rays of this material, given that: (sin 34.4°=0.565, tan 60.5°=1.77 37 | P a g e █ 2015 Preliminary - 2017 Round (2) Practical, Outside Country Q.1) If the critical angle of the light rays of a blue agate surrounded by air is (34.4). Calculate the angle of polarization for the light rays of this material, knowing that (sin 34.4°=0.565, tan 60.5o=1). █ 2015 Round (3) Q.2) If the distance between the two slits of Young’s experiment was (0.22mm) and the distance between the screen and them was equal to (1.1m), and the distance between the fourth bright fringe and the central fringe equals (10mm). Calculate the wavelength of the light used. 38 | P a g e █ 2019 Round (1) Biology, Outside Country Q.3) In the adjacent figure, two coherent sources (S1 & S2) emit waves of wavelength (λ=0.1m). The waves emitted from them both interfere at the point (P) at the same time. What kind of interference is resulted at this point when one of the two waves crosses an optical path of (3.2m) and the other crosses an optical path of (2.95m)? █ 2019 Round (1) Biology Q.4) Monochromatic light falls perpendicular to a diffraction grating of which each one centimeter contains (10000Line/cm), so if the diffraction angle of the first bright order is (30°). Find the magnitude of the wavelength of the light used. 39 | P a g e