Optics Lecture Notes PDF

Optics By: Dr. SABAH IBRAHIM ABBAS Al-Karakh university of Sciences References: 1- Fundamental of optics by [ Jenkins and white ] 2- Introduction to modern optics by [ Grant R. Fowles]  Nature and propagation of light:- What is the light? Light has both wave Theory (Huygens’s) and particle...

Optics By: Dr. SABAH IBRAHIM ABBAS Al-Karakh university of Sciences References: 1- Fundamental of optics by [ Jenkins and white ] 2- Introduction to modern optics by [ Grant R. Fowles]  Nature and propagation of light:- What is the light? Light has both wave Theory (Huygens’s) and particle theory (Newton’s) Wave-like: propagation, interference, Diffraction. Particle-like: emission, absorption, photoelectric effect, Compton Effect Raman effect Particle-like energy exchange (explained by quantum mechanics) only discrete quanta of energy can be absorbed or emitted; therefore it must exist in only discrete energy units. In a particle theory of light travels in straight-line paths called light rays represent the paths of particles. Branch of optics dealing with ray model-geometric optics , wave behavior – physical optics. The speed of light in a vacuum is expressed as m/s. Light travels in a vacuum at a constant speed, and this speed is considered a universal constant. In a vacuum, light will travel in a straight line at fixed speed, carrying energy from one place to another. Two key properties of light interacting with a medium are: 1. It can be deflected upon passing from one medium to another (refraction). 2. It can be bounced off a surface (reflection). Dual Nature of Light:- Scientists have observed that light energy can behave like a wave as it moves through space, or it can behave like a discrete particle with a discrete amount of energy (quantum) that can be absorbed and emitted. Concept of a photon:- The particle-like nature of light is modeled with photons. A photon has no mass and no charge. It is a carrier of electromagnetic energy and interacts with other discrete particles (e.g., electrons, atoms, and molecules). A beam of light is modeled as a stream of photons, each carrying a well-defined energy that is dependent upon the wavelength of the light. Energy (E): measured in J/mole, J/photon, or electron volt (eV, 1eV=1.6 * 10-19 J per photon or per mole of photons). Energy of the wave can be calculated directly from the wavelength or frequency. The energy (E) of a given photon can be calculated by: E = hc/λ Where E is in joules; h = Planck’s constant = ; c = Speed of light = ; λ = Wavelength of the light in meters Example (1) Photons in a pale blue light have a wavelength of 500 nm. The symbol nm is defined as a nanometer = m. What is the energy of this photon? Solution: The energy of photon is: The energy of photon in (e V) is calculated Wave Model :- The particle-like model of light describes large-scale effects such as light passing through lenses or bouncing off mirrors. However, a wavelike model must be used to describe fine-scale effects such as interference and diffraction that occur when light passes through small openings or by sharp edges. The propagation of light or electromagnetic energy through space can be described in terms of a traveling wave motion. The wave moves energy- without moving mass-from one place to another at a speed independent of its intensity or wavelength. This wave nature of light is the basis of physical optics and describes the interaction of light with media. Characteristics of light waves:- To understand light waves, it is important to understand basic wave motion itself. Water waves are sequences of crests (high points) and troughs (low points) that “move” along the surface of the water. When ocean waves roll in toward the beach, the line of crests and troughs is seen as profiles parallel to the beach. An electromagnetic wave is made of an electric field and a magnetic field that alternately get weaker and stronger. The directions of the fields are at right angles to the direction the wave is moving, just as the motion of the water is up and down while a water wave moves horizontally. is a one-dimensional representation of the electric field. One-dimensional representation of the electromagnetic wave The maximum value of the wave displacement is called the amplitude (A) of the wave. Wavelength (): The distance from one peak to the next, Measured in nanometers (nm), micrometers (m), or angstroms (  , 1  = 0.1 nm). Wavenumber (𝜐): The wavenumber is how many waves fit in the distance of 1 cm. and measured in inverse centimeters (cm-1). Also the inverse of the wavelength (1/λ) is the wave number (𝜐), which is The wave propagates at a wave speed (v). This wave speed in a vacuum is equal to c, and is less than c in a medium. The time to complete one cycle is called the cycle time or period (τ) and can be calculated Period or cycle time is Frequency ( ). It is measured as the number of waves that pass a given point in one second. The unit for frequency is cycles per second, also called hertz (Hz). Example 2 Blue light in a vacuum, we can calculate the cycle time and frequency. From a previous example, we know that the wavelength of blue light is 500 nm and the velocity of light in a vacuum is c. Sol: Calculate the frequency using Equation Example 3: Consider a 1550 nm wave. What are the frequency, wavenumber and energy (in a vacuum) ? Frequency: ⁄ Wavenumber:  1   1nm   1m  -1   *  9  *   = 6451.61cm  1550nm   10 m   100cm  Energy:   1 -19 E  hf  6.626 x10 34 Js * 1.94 x1014  = 1.285x10 J/photon  s  The Electromagnetic Spectrum:- The electromagnetic (EM) spectrum is the range of all types of EM radiation. Radio: Radio waves are emitted and received by antennas, which consist of conductors such as metal rod resonators. In artificial generation of radio waves, an electronic device called a transmitter generates an AC electric current which is applied to an antenna. The oscillating electrons in the antenna generate oscillating electric and magnetic fields that radiate away from the antenna as radio waves. Radio waves are also emitted by stars and gases in space. Microwave: Microwave radiation will cook your popcorn in just a few minutes, but is also used by astronomers to learn about the structure of nearby galaxies. Infrared: Night vision goggles pick up the infrared light emitted by our skin and objects with heat. In space, infrared light helps us map the dust between stars. Visible: Our eyes detect visible light, light bulbs, and stars all emit visible light. Ultraviolet: Ultraviolet radiation is emitted by the Sun and is the reason skin tans and burns. "Hot" objects in space emit UV radiation as well is the longest wavelength radiation whose photons are energetic enough to ionize atoms, separating electrons from them, and thus causing chemical reactions. Short wavelength UV and the shorter wavelength radiation above it (X-rays and gamma rays) are called ionizing radiation, and exposure to them can damage living tissue, making them a health hazard. At the middle range of UV, UV rays cannot ionize but can break chemical bonds, making molecules unusually reactive. Sunburn, for example, is caused by the disruptive effects of middle range UV radiation on skin cells, which is the main cause of skin cancer. UV rays in the middle range can irreparably damage the complex DNA molecules in the cells producing thymine dimers making it a very potent mutagen. X-ray: After UV come X-rays, which, like the upper ranges of UV are also ionizing. However, due to their higher energies, X-rays can also interact with matter by means of the Compton effect. Hard X-rays have shorter wavelengths than soft X-rays and as they can pass through many substances with little absorption, they can be used to 'see through' objects with 'thicknesses' less than that equivalent to a few meters of water. One notable use is diagnostic X-ray imaging in medicine (a process known as radiography). Gamma ray: Doctors use gamma-ray imaging to see inside your body. The biggest gamma-ray generator of all is the Universe. Gamma rays are used experimentally by physicists for their penetrating ability and are produced by a number of radioisotopes. They are used for irradiation of foods and seeds for sterilization, and in medicine they are occasionally used in radiation cancer therapy. More commonly, gamma rays are used for diagnostic imaging in nuclear medicine, an example being PET scans. The wavelength of gamma rays can be measured with high accuracy through the effects of Compton scattering.  Spectrum of Electromagnetic Radiation Region Wavelength Wavelength Frequency(Hz)(THz= Energy (nm) (m)(nm= [e V] Radio > > Wavelength of the visible region: Violet 380 nm to 435nm Blue 435 nm to 500nm Cyen 500 nm to 520 nm Green 520 nm to 565 nm Yellow 565 nm to 590 nm Orange 590 nm to 625nm Red 625 nm to 740 nm Refractive Index:- Refractive index is measured of the bending of a ray of light when passing from one medium into another. If is the angle of incidence of a ray in vacuum (angle between the incoming ray and the perpendicular to the surface of a medium, called the normal) and is the angle of refraction (angle between the ray in the medium and the normal), the refractive index n is defined the ratio of the speed of light in a vacuum c to the speed of light through the material v. i r nair = 1 nglass = 1.5 t Optical Path Length :- Suppose that the light has a distance QR in vacuum at the t time, in the same time (t ) the light has a distance PS in the medium. Speed of light in the vacuum Q ∆ P R n Speed of light S in medium And Let ∆ ( ) ∆ ∆ If a light ray travel through a series optical materials with thickness and refractive indices , the total optical path length is just of the sum of separate values ∆ l 𝑙 𝑙 n 𝑛 𝑛 ∆ Fermat’s Principle:- Fermat’s principle states that light will take path with the shortest travel time to go from one point to another, as shown in figure. Illustration of Fermat’s principle, light will travel in a straight line from one point to another since a straight line will be the quickest path. A ray of light is incident on the boundary separating between two different medium, part of the ray is reflected back into the first medium and the remainder is refracted and enters the second medium. Incident Reflected ray 𝜃𝑖 𝑟 ray nair = 1 nglass = 1.5 Refracted ray 𝜃𝑡 A light ray incident upon a reflective surface will be reflected at an angle equal to the incident angle. Both angles are typically measured with respect to the normal to the surface. This law of reflection can be derived from Fermat’s principle. Angle of incidence =Angle of reflection  The incident ray , the normal and the reflected ray lies in the same plane which is perpendicular to the interface that separating of two media by using the fermat’s principle prove the law of reflection. Light follows the path of least time: the path length from A to B Angle of incidence =Angle of reflection Suppose the path length from A to B is equal to (L) as shown in figure. A B a 𝜃𝑖 𝜃𝑟 b o x d-x d √ √ ( ) Since the speed of light in vacuum is constant, the minimum time path is simply the minimum distance path. This may be found by setting the derivative of t with respect to x equal to zero. √ √ ( ) [√ √ ( ) ]  ( )( ) [ ] √ √ ( ) But ( ) is not equal to zero ( ) =0 √ √ ( ) ( ) This reduce to √ √ ( ) For low reflection angle Fermat’s principle: light follows the path of least time. Snell’s law can be derived by setting the derivative of the time equal to zero. A Fast Medium smaller refractive index 𝑛 a 𝜃 o Slow Medium higher refractive 𝜃 b index 𝑛 B x d √ √ ( ) ( ) √ √ ( ) Critical Angle and Total Internal Reflection The critical angle Is the angle of incidence of a light ray which travels from high optical dense medium to the lower one which results in it being refracted at 90 degree to the normal. As an example of the critical angle suppose that a light traveling through water towards the boundary with a less dense material such as air. When the angle of incidence in water reaches a certain critical value, the refracted ray lies along the boundary, having an angle of refraction of (90-degrees). This angle of incidence is known as the critical angle; it is the largest angle of incidence for which refraction can still occur. Derivation of the Critical Angle formula Let's consider two different media and the critical angle is that gives a value of (90-degrees). If this information is substituted into Snell's Law equation, a general equation for predicting the critical angle can be derived. 𝑛 𝑛 =? Using Snell’s law  ( ) ( ) ( ) is the critical angle formula ( ) Total internal reflection Is the phenomenon which occurs when a propagated wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal of the surface. When the refractive index is lower on the other side of the boundary and the incident angle is greater than the critical angle, the wave cannot pass through and is entirely reflected. (1) If θ ≤ θc, the ray will split; some of the ray will reflect off the boundary, and some will refract as it passes through. This is not total internal reflection. (2) If θ > θc, the entire ray reflects from the boundary. None passes through. This is called total internal reflection as shown in figure. 𝜃𝑖 < 𝜃𝑐 𝜃𝑖 𝜃𝑐 𝜃𝑖 𝜃𝑐 𝑛 𝑛 𝑛 Summarize the subject with important notes. (1)A ray traveling from a higher index to a lower index bends away from the normal. (2) At an angle called the critical angle, the refracted angle is 90°; the ray travels along the surface (3) For larger incidence angles, Snell’s law says the sine of the refracted angle is greater than one, this is impossible (4)For incident angles larger than the critical angle, the light is completely reflected, there is no refracted ray. Explains this phenomenon according to Snell’s law if , that is if a ray of light enters from rarer medium to a denser medium, it is deviated towards the normal and if < , that is if the ray of light enters from denser to a rarer medium it is deviated away from the normal. Example1: A light pulse passes through a glass rod 3m long and refractive index 1.5 with another light pulse that passes through the same distance in the air which of the two pulses arrive first and what is difference time between them? Sol: A light pulse in the air is first arrival Time of light pulse ⁄ Time of light pulse in the medium Example2: Refractive indices of water and glass are 1.33 and 1.54 respectively; find the relative refractive index for (a) glass with respect to water, (b) water with respect to glass. Sol: Find the velocity of light in the water, and the velocity of light in the glass. = ( ) The velocity of light in the glass = ( ) The relative refractive index of glass with respect to water = = The relative refractive index of water with respect to glass = Example3: Refractive index of glass with respect to air 1.5 and the refractive index of diamond with respect to air 2.4, in which medium would the light bend more? Solution: Refractive index glass w.r.t air = 1.5 ( ) Refractive index of diamond w.r.t air = 2.4 ( ) Dividing (2) by (1) ( ) Medium one 𝜃𝑖 ( ) ( ) ( ) Medium two 𝜃𝑟𝑑 𝜃𝑟𝑔 Angle of refraction for glass > Angle of refraction for diamond Angle of refraction is increased and bending of light decreased So, light bends more in diamond. Velocity of light in glass > velocity of light in diamond and diamond is optically denser than glass. Example 1:- A step-index fiber 0.0025 inch in diameter has a core index of 1.53 and a cladding index of 1.39. See drawing. Such clad fibers are used frequently in applications involving communication, sensing, and medical  imaging. What is the maximum acceptance angle for a cone of light rays incident on the fiber face such that the refracted ray in the core of the fiber is incident on the cladding at the critical angle? Cladding 𝑛 𝑛𝑎𝑖𝑟 𝜃𝐴 𝜃𝑐 𝜃𝑟 Fiber core n=1.53 𝜃𝐴 Solution: First find the critical angle in the core, at the core-cladding interface. Then, from geometry, identify and use Snell’s law to find at the core-cladding interface ( ) From right-triangle geometry, θr = 90 − 65.3 = 24.7° From Snell’s law, at the fiber face, ( ) ( ) Thus, the maximum acceptance angle is 39.7° and the acceptance cone is twice that, or 2 = 79.4°. The acceptance cone indicates that any light ray incident on the fiber face within the acceptance angle will undergo total internal reflection at the core-cladding face and remain trapped in the fiber as it propagates along the fiber. Example1: A light pulse passes through a glass rod 3m long and refractive index 1.5 with another light pulse that passes through the same distance in the air which of the two pulses arrive first and what is difference time between them? Sol: A light pulse in the air is first arrival Time of light pulse ⁄ Time of light pulse in the medium Example2: Refractive indices of water and glass are 1.33 and 1.54 respectively; find the relative refractive index for (a) glass with respect to water, (b) water with respect to glass. Sol: Find the velocity of light in the water, and the velocity of light in the glass. = The velocity of light in the glass = The relative refractive index of glass with respect to water = = The relative refractive index of water with respect to glass = Example 1:- A step-index fiber 0.0025 inch in diameter has a core index of 1.53 and a cladding index of 1.39. See drawing. Such clad fibers are used frequently in applications involving communication, sensing, and medical imaging. What is the maximum acceptance angle for a cone of light rays  incident on the fiber face such that the refracted ray in the core of the fiber is incident on the cladding at the critical angle? Cladding 𝑛 𝑛𝑎𝑖𝑟 𝜃𝐴 𝜃𝑐 𝜃𝑟 Fiber core n=1.53 𝜃𝐴 Solution: First find the critical angle in the core, at the core-cladding interface. Then, from geometry, identify and use Snell’s law to find at the core-cladding interface ( ) From right-triangle geometry, θr = 90 − 65.3 = 24.7° From Snell’s law, at the fiber face, ( ) ( ) Thus, the maximum acceptance angle is 39.7° and the acceptance cone is twice that, or 2 = 79.4°. The acceptance cone indicates that any light ray incident on the fiber face within the acceptance angle will undergo total internal reflection at the core-cladding face and remain trapped in the fiber as it propagates along the fiber.  Refraction by prism:- Prism:- Is a transparent optical object with flat , polished surfaces that refract light. at least two of the flat surfaces must have an angle between them. Bending of light:- Ligth changes its speed when it moves from one medium to another. this speed change causes the light to be refracted and to enter the new medium at a different angle. The degree of bending of the light’s path depends on the angle that the incident ray of the light makes with the surface , and on the ratio between the refractive index of the two media. We'll express the emergent angle and the deviation angle δ as functions of the incident angle i1and after this we'll find the condition that minimizes the deviation angle δ. The deviation angle δ equal:  ( ) ‫االشتقاق لالطالع‬ THE REFRACTIVE INDEX OF THE PRISM IS CALUCALTED BY ( ) For small angle less than :- ( ) ( ) Angle of Deviation 𝛿 𝛿𝑚 Angle of Incident 𝑖 𝑖 𝑖 Figure2.16: A graph of the minimum angle of deviation A beam passing through an object like a prism or water drop is deflected twice: once entering, and again when exiting. The sum of these two deflections is called the deviation angle. The deviation angle in a prism depends upon: Refractive index of the prism: The refractive index depends on the material and the wavelength of the light. The larger the refractive index, the larger the deviation angle. Angle of the prism: The larger the prism angle, the larger the deviation angle. Angle of incidence: The deviation angle depends on the angle that the beam enters the object, called angle of incidence. The deviation angle first decreases with increasing incidence angle, and then it increases. There is an angle of incidence at which the sum of the two deflections is a minimum. The deviation angle at this point is called the "minimum deviation" angle, or "angle of minimum deviation".At the minimum deviation angle, the incidence and exit angles of the ray are identical. One of the factors that cause a rainbow is the bunching of light rays at the minimum deviation angle that is close to the rainbow angle.  Dispersion:- In optics, dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency. Media having such a property are termed dispersive media. The most familiar example of dispersion is probably a rainbow, in which dispersion causes the spatial separation of a white light into components of different wavelengths (different colors). However, dispersion also has an effect in many other circumstances: for example, it causes pulses to spread in optical fibers, degrading signals over long distances; also, a cancellation between dispersion refractive index n decreases with increasing wavelength λ. In this case, the medium is said to have normal dispersion. Whereas, if the index increases with increasing wavelength the medium has anomalous dispersion. In optics, one important and familiar consequence of dispersion is the change in the angle of refraction of different colors of light.   Color Dispersion :- It is well known to those who have studied elementary physics that refraction causes a separation of white light into its component colors. Thus, as is shown in Fig, the incident ray of white light gives rise to refracted rays of different colors (really a continuous spectrum) each of which has a different value of. The value of n' must therefore vary with color. White light 𝜃 n=1 Red = C Yellow=D 𝑛 Blue = F 𝜃 F C D The refractive indices of these colors are The dispersive power is defined is a measure of the ability of a substance to disperse light, equal to the quotient of the difference in refractive indices of the substance for two representative wavelengths divided by the difference of the refractive index for an intermediate wavelength, and is defined by the equation: Or in prism it’s defined as the ratio of angular dispersion to the deviation produced by mean color. It is possible to observe the following points from the figure:  The principal axis in each diagram is straight line passes through the center of curvature C.  The point A where the axis crosses the surface is called the vertex.  In diagram (a) rays are shown diverging from a point source F on the axis in the first medium and refracted into a beam everywhere parallel to the axis in the second medium.  Diagram (d) shows a beam converging in the first medium toward the point F and then refracted into a parallel beam in the second medium. F in each of these two cases is called the primary focal point, and the distance f is called the primary focal length.  In diagram (b) a parallel incident beam is refracted and brought to a focus at the point F'.  In diagram (C) a parallel incident beam is refracted to diverge as if it came from the point F'. F' in each case is called the secondary focal point and the distance f ' is called the secondary focal length.  We now state that the primary focal point F is an axial point having the property that any ray coming from it or proceeding toward it travels parallel to the axis after refraction.  the similar statement that the secondary focal point F' is an axial point having the property that any incident ray traveling parallel to the axis will, after refraction, proceed toward, or appear to come from, F'.  A plane perpendicular to the axis and passing through either focal point is called a focal plane. GRAPHICAL CONSTRUCTIONS. THE PARALLEL-RAY METHOD They apply only to images formed by paraxial rays. For such rays the refraction occurs at or very near the vertex of the spherical.  Q y 𝐹′ 𝑀′ C M F y’ 𝑄′ f f‘ S S’ Convention of signs for distances The following set of sign conventions will be adhered to throughout the following chapters on geometrical optics, and it would be well to have them firmly in mind: 1. All figures are drawn with the light traveling from left to right. 2. All object distances (s) are considered positive when they are measured to the left of the vertex and negative when they are measured to the right. 3. All image distances (s') are positive when they are measured to the right of the vertex and negative when to the left. 4. Both focal lengths are positive for a converging system and negative for a diverging system. 5. Object and image dimensions are positive when measured upward from the axis and negative when measured downward. 6. All convex surfaces are taken as having a positive radius, and all concave surfaces are taken as having a negative radius. Derivation of the Gaussian formula The Gaussian formula is general method that may be employed to determine the position and size of the image. A method involving oblique rays used for derivation the Gaussian formula, an oblique ray from an axial object point M is shown incident on the surface at an angle and refracted at an angle ′. The refracted ray crosses the