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Optical Theory Simplified: 9 Fundamentals to Becoming an Optical Genius APPLICATION NOTES

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LEARNING – UNDERSTANDING – INTRODUCING – APPLYING Optical Theory Simplified: 9 Fundamentals To Becoming An Optical Genius APPLICATION NOTES Optics Application Examples Introduction To Prisms Understanding Optical Specifications Optical Cage System Design Examples All About Aspheric Lenses Optical Glass Optical Filters An Introduction to Optical Coatings Why Use An Achromatic Lens www.edmundoptics.com OPTICS APPLICATION EXAMPLES APPLICATION 1: DETECTOR SYSTEMS Every optical system requires some sort of preliminary design. system will help establish an initial plan. The following ques- Getting started with the design is often the most intimidating tions will illustrate the process of designing a simple detector step, but identifying several important specifications of the or emitter system. GOAL: WHERE WILL THE LIGHT GO? Although simple lenses are often used in imaging applications, such as a plano-convex (PCX) lens or double-convex (DCX) in many cases their goal is to project light from one point to lens, can be used. another within a system. Nearly all emitters, detectors, lasers, and fiber optics require a lens for this type of light manipula- Figure 1 shows a PCX lens, along with several important speci- tion. Before determining which type of system to design, an fications: Diameter of the lens (D1) and Focal Length (f). Figure important question to answer is “Where will the light go?” If 1 also illustrates how the diameter of the detector limits the the goal of the design is to get all incident light to fill a detector, Field of View (FOV) of the system, as shown by the approxi- with as few aberrations as possible, then a simple singlet lens, mation for Full Field of View (FFOV): (1.1) D Figure 1: 1 D θ PCX Lens as FOV Limit FF0V = 2 β in Detector Application ƒ Detector (D ) 2 or, by the exact equation: (1.2) / Field of View (β) -1 D FF0V = 2 tan ( 2 ) f 2ƒ For detectors used in scanning systems, the important mea- sure is the Instantaneous Field of View (IFOV), which is the angle subtended by the detector at any instant during scan- ning. (1.3) Pixel Size IF0V = ƒ Continue www.edmundoptics.com GOAL: WHERE WILL THE LIGHT GO? (CONT.) D Figure 2: 1 Figure 3: θ Instantaneous FOV β PCX Lens as FOV Limit in Emitter Emitter (D ) FFOV IFOV Detector (D2) 2 Application / Field of View (β) (radians) f Considered in reverse, Figure 1 can also represent an emitting setup will be the premise of the application example. system (Figure 3), with the lens used to collimate the light. This LIGHT TRANSMISSION: HOW MUCH LIGHT EXISTS INITIALLY? Knowing where the light will go is only the first step in design- Correspondingly, this relationship can be mathematically ex- ing a light-projecting system; it is just as important to know pressed according to Equation 1.5. It is important to note that how much light is transmitted from the object, or the source. the larger the Diameter, the smaller the F/#; this allows more The efficiency is based on how much light is received by the light to enter the system. To create the most efficient system, detector, thereby answering the question “How much light it is best to match the emitted cone of light from the source to exists initially?” The Numerical Aperture (NA) and F-number the acceptance cone of the lens, as this avoids over or under (F/#) of a lens measure the amount of light it can collect filling the lens area. based on f, D, index of refraction (n), and Acceptance Angle (θ). Figure 4 illustrates the relationship between F/# and NA. Figure 4: (1.4) DCX Lens Showing F/# and ƒ NA F/# = ( D ) F D (1.5) 1 NA = ( 2(F/#) ) f (1.6) NA =n sinθ ≈nθ OPTICAL THROUGHPUT: HOW MUCH LIGHT GETS THROUGH THE SYSTEM? When using a lens as a tool to transfer light from an emitter affects TP even when the ratio between Diameter and Focal to a detector, it is important to consider Throughput (TP), a Length (specified by F/#) remains constant. quantitative measurement of transmitted light energy. In other words, answering the question “How much light gets through Figure 5 shows the basic definition of throughput (TP) as ex- the system?” dictates the geometry of the lens used and the pressed in Equation 1.7, where A is the Area of the object, configuration of the system. Because emitters and detectors (light source), Ω is the Solid Angle, and z is the Object Distance are areas of light and not point sources, the diameter of a lens (with their conjugates in image space as A’, Ω’, and z’). (1.7) Figure 5: Ω’ 2 DCX Lens Illustrating Through- TP = n AΩ putz Ω A A’ z z’ Need Help? Our team is ready to assist you! www.edmundoptics.com/contactContinue www.edmundoptics.com OPTICAL THROUGHPUT: HOW MUCH LIGHT GETS THROUGH THE SYSTEM? Solid angle is defined as Ω = A/r2, with the area of the lens within the system due to lens aperture limitations. However, surface and the radius (r) being the distance from the lens to some systems benefit from intentional vignetting, as it can the object (z) or image plane (z’), for Ω or Ω’, respectively. eliminate stray light that would negatively affect the quality of the image. It is important to note that properly aligning the The amount of light reaching the detector can be reduced by system reduces stray light and unintentional vignetting. vignetting, which is the result of light being physically blocked ABERRATIONS: HOW DOES THE IMAGE LOOK? Determining how much light passes through the system is im- performance with cost is an important decision for any de- portant, but aberrations within the system also play a major signer. Several basic aberrations, such as coma (variation in role. Answering “How does the image look?” can lead to im- magnification or image size with aperture), spherical (light rays proving the system’s design in order to reduce aberrations and focusing in front of or behind paraxial focus), and astigmatism improve image quality. Aberrations are errors inherent with (having one focus point for horizontal rays and another for ver- any optical system, regardless of fabrication or alignment. tical) can be reduced by a large F/#, as shown in the following Since every optical system contains aberrations, balancing relations. (1.8) (1.10) 1 1 Spherical ∝ (F/#)3 Astigmatism ∝ (F/#) (1.9) 1 Coma ∝ (F/#)2 APPLICATION EXAMPLE: DETECTOR SYSTEM As an example, consider a system in which light is emitted from ●Calculated Parameters a 1/4” diameter fiber optic light guide, as shown in Figure 3. F- Number (F/#) ●Initial Parameters (1.11) NA of Light Guide = 0.55 1 Diameter of Source (Emitter) = 6.35mm NA = 2(F/#) Index of Refraction of Air = 1 1 0.55 = 2(F/#) F/# = 0.9 A PCX lens of F/1, meaning the F/# is 1, would be ideal to place in front of the light guide in order to collimate as much light as possible. According to Equation 1.4, if the F/# is 1, then the diameter and focal length of a lens are equal. In other words, if we consider a lens with a diameter of 12mm, then the focal length is also 12mm. Continue www.edmundoptics.com APPLICATION EXAMPLE: DETECTOR SYSTEM Full Field of View (FFOV) (1.12) 6.35mm D2 Dsource FFOV = = = = 12mm = 0.529 radians ƒ Flens Throughput (TP) (1.13) (1.15) 2 2 6.35mm 2 2 Arealens 113.097mm π ( ) = 31.669mm 2 = 0.7854 steradians Asource = π r = 2 Ω = Radius = (12mm)2 (1.14) 2 π ( 12mm )2 = 113.097mm2 Alens = π r = 2 Steradians correspond to a 2-dimensional angle in 3-dimen- In order to calculate Throughput (TP) of this system, we need sional space, as the angle from the edge to edge of the lens to first calculate the Area of the Source (Equation 1.11), the is in two dimensions. A higher value in steradians is given by Area of the Lens (Equation 1.12) and the Solid Angle (Equa- a shorter distance from emitter to lens, or a larger diameter tion 1.13). As a rule of thumb for collimating light from a di- of the lens. The largest value a solid angle can have is 4π, or vergent source (i.e. the light guide in this example), place the about 12.57, as this would be equivalent to the solid angle of lens a distance equal to one focal length away from the source. all space. (1.16) 2 2 2 TP = n AΩ = (1)(31.669mm )(0.7854steradians) = 24.873mm steradians Since the system is in free space, where n is approximated as 1, n2 does not factor into the final calculation. APPLICATION 2: SELECTING THE RIGHT LENS High image quality is synonymous with low aberrations. As a Figures 6a - 6e compare a variety of lens systems for a relay result, designers often utilize two or more lens elements in or- lens, or 1:1 imaging, application. In this specific example, out- der to obtain higher image quality compared to a single lens lined in the following series of comparisons, it is easy to see solution. Many factors contribute to selecting the right lens for how image quality is affected by the inherent geometry and an application: type of source, space constraints, cost, etc. optical properties of the lenses chosen. Figure 6a: Figure 6b: DCX Lens Relay System: PCX Lens Relay System: 25mm EFL x 20mm En- 50mm EFL x 20mm En- trance Pupil Diameter trance Pupil Diameter (Left is Color and Right (Left is Color and Right is Monochromatic) is Monochromatic) Connect With Us! www.facebook.com/edmundoptics Tech Tuesday, Geeky Friday and More!Continue www.edmundoptics.com APPLICATION 2: SELECTING THE RIGHT LENS (CONT.) Figure 6c: Figure 6d: Achromatic Lens Relay Aspherized Achromatic System: 50mm EFL x Lens Relay System: 20mm Entrance Pupil 50mm EFL x 50mm En- Diameter (Left is Color trance Pupil Diameter and Right is Monochro- (Left is Color and Right matic) is Monochromatic) Figure 6e: Aspheric Lens Relay System: 50mm EFL x 40mm Entrance Pupil Diameter (Left is Color and Right is Monochro- matic) APPLICATION EXAMPLE: SINGLE ELEMENT LENS SYSTEM A double-convex (DCX) lens is regarded as the best single ele- For applications that only need one lens, with the object or ment for 1:1 imaging because of its symmetrical shape, as both source at infinity, a better shape factor can be found, to reduce sides of the lens have equal power, instead of one side bending whichever aberration is most detrimental to the system.
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