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PHYS 202 Notes, Week 10 Greg Christian March 28 & 30, 2016 Last updated: 03/31/2016 at 12:28:48 This week we learn about optical instruments, interference, and thin films. Optical Instruments Important points Cameras A camera is a device that is able to capture images and store them either on film or in a digital medium. Figure 1 shows a diagram of how a camera operates. The light coming off the object to be photographed passes through a lens, creating a real and inverted image on the film (for a traditional camera) or CCD array (for a digital camera). The image is then un-flipped by developing the film or with digital processing software, to create what you see when looking at photos. In designing a camera, there are a couple of things which can be altered to affect the image that forms. One is the intensity of light reaching the film or CCD array, which can be adjusted by changing the size of the aperture surrounding the lens. The size of this aperture is usually given in terms of something called the f -number, Focal length f f − number = = , Aperture diameter D (1) where f is the focal length and D is the aperture diameter. • Cameras create a inverted, real image on film or CCD. • Eyes use biological fluids to create a lens that focuses light onto the retina. • Magnifiers change the size of images using lenses; so do microscopes (2 lenses) and telescopes. Important equations • Angular magnification of a magnifier M = θ θ0 = (25 cm) s = (1/ f − 1/s0 ) × 25 cm • Microscope M = m1 M2 = (25 cm) s10 f1 f2 Figure 1: Operation of a camera. As you probably know, cameras can take images of objects at many different distances. This is accomplished by moving the lens in or out. This in turn adjusts the image distance. The goal is to place the lens where the image distance matches the recording medium. Then the phys 202 notes, week 10 camera is “in focus”. Most modern cameras do this automatically, but older and/or high-end cameras still rely on manual focusing by the photographer. Eyes Figure 2: Operation of an eye. Figure 2 shows the operation of the human eye. Basically, the eye works by forming a lens made up of biological material, focusing incoming light onto the retina where rods and cones sense the image and transmit it to the brain. The parts of the eye forming the lens are (in order): • The aqueous humor, index of refraction n = 1.336. • The crystalline lens, index of refraction n = 1.437. • The vitreous humor, index of refraction n = 1.336. In order to form a sharp image, the image point of the eye needs to be exactly on the retina surface. Given that the retina is in a fixed location, the image distance, s0 is always fixed. In order to accommodate different object distances, s, the eye changes its effective focal length. It does that by contracting or relaxing a muscle called the ciliary muscle, which has the effect of changing the radius of curvature, and thus the focal length of the eye. Contracting the ciliary muscle serves to decrease the radius of curvature, and thus focal lngth, of the eye. This allows up-close objects to be in focus. Then the ciliary muscle is relaxed, the focal length increases, creating focus on far-away objects. This process is called accomodation. As people age, the ciliary muscle loses its ability to work, decreasing the range in which the focal length can be changed. This is why eventually nearly everyone starts to have difficulty with seeing up-close 2 phys 202 notes, week 10 objects without reading glasses. The term near point describes the closest distance upon which the eye can focus. For the average person this decreases from about 7 cm at age 10 to 200 cm at age 60. The eye also needs to accommodate different intensities of light. It does this by dilating or contracting the pupils, to let more or less light in. Vision problems Many people have eyes which are either too long or too short. This causes light to be focused either in front of or behind the retina. The first case is called myopia, or nearsightedness, because it causes images of far-away objects to appear out of focus. The second is called hyperopia, or farsightedness, because it causes close-up images to be out of focus. Drawings of these are shown in Figure 3. Each of these vision problems can be fixed using lenses. In the case of hyperiopia, we can use a converging lens to focus the image onto the retina (Figure 4), thus creating a clear image. In the case of myopia, we use a diverging lens to achieve the same result (Figure 5). Figure 3: Myopia and hyperopia. Figure 4: Hyperopia correction. Magnifiers Another optical instrument is a magnifying glass, or simply a magnifier. This instrument uses a converging lens to form a virtual image that is larger and farther away than the original. Magnifiers are typically analyzed in terms of the angular size, or the total angle subtended by the object, as shown in the left of Figure 6. When dealing with magnifiers, we are concerned with the angular magnification, or the change in angular size of the object. This is equal 3 phys 202 notes, week 10 Figure 5: Myopia correction. to θ0 . (2) θ For a magnifier with focal length f , we can calculate the angular magnification as M= (25 cm) s 1 1 = − 0 × 25 cm, f s M= (3) (4) where • s is the object distance; and • We assume that the near point of the eye is 25 cm. Figure 6: Operation of a magnifier. Microscopes To look at very small objects, a simple magnifier isn’t enough. Here we need a microscope. The operation of a microscope is shown in 4 phys 202 notes, week 10 5 Figure 7. A microscope consists of two lenses: 1. The objective lens, which has focal length f 1 . 2. The eyepiece, which has focal length f 2 . It works in the following way: 1. The objective lens creates a real, inverted image inside the focal point F2 of the eyepiece. 2. The eyepiece uses the image I as its object and creates an enlarged, virtual image I 0 which is still inverted. The most important quantity for a microscope is the overall angular magnification. This is given by the following equation, M = m1 M2 = (25 cm) s10 , f1 f2 (5) (6) Figure 7: Operation of a microscope. where • m1 is the lateral magnification of the objective lens (focal length f 1 ). • M2 is the angular magnification of the eyepiece (focal length f 2 ). Telescopes A telescope allows viewing of objects that are large, but very far away. One type of telescope is called the astronomical telescope, and it has a very similar design to a microscope, namely a combination of objective lens and eyepiece. An example is shown in Figure 8. Figure 8: Operation of a astronomical telescope. Because the object to be viewed is so far away, the first image is formed nearly at the second focal point of the objective lens ( f 1 ). This image is then the object for the eyepiece. Ideally, we want this to form a final image at infinity, which means that the first image needs to be located at the focal point of the eyepiece ( f 2 ). This means that the total length of the telescope will be equal to the sum of the focal lengths, f1 + f2 . phys 202 notes, week 10 6 The overall angular magnification for a telescope can be worked out from the geometry shown in Figure 8. The result is that it’s simply proportional the the ratio of focal lengths (with a negative sign), M=− f1 . f2 (7) Note that due to the negative sign, the telescope forms an inverted image. This can be corrected using a device called prism binoculars, shown in Figure 9. These are placed between the objective and the eyepiece and serve to flip the image (making it upright) by way of four internal reflections. Figure 9: Prism binoculars, to make the image formed by a telescope upright. For practical reasons, it’s often more feasible to construct telescopes out of reflecting mirrors rather than lenses. These devices are called reflecting telescopes. Some schemes for constructing reflecting telescopes are shown in Figure 10. In one scheme, (part a in the figure), there is a “cage” inside the telescope which receives the reflected light, typically in a camera. This design is most practical for very large telescopes because the cage blocks incoming light rays. Figure 10: Examples of reflecting telescopes. phys 202 notes, week 10 7 Other designs (parts b and c in the figure) use a secondary mirror to reflect light either along the direction of incoming light (which requires a hole in the primary mirror), or out the side of the telescope (part c). Optical Interference So far, we’ve been working with the ray model of light, which treats light as something traveling along a straight line path. Now we’ll move into talking about some phenomena which rely on the wave nature of light.Recall that light is a form of electromagnetic wave, consisting of #» #» oscillating E and B fields. Interference is the phenomena where two or more waves overlap in space. Related to this is something called the Principle of Superposition: • Constructive interference occurs when waves are in phase. • Destructive interference occurs when waves are 1/2 cycle out of phase. • Double slits and thin films can both form interference patterns. Important equations • Interference When two or more waves overlap, the resulting displacement at any point and any time is found by adding the instantaneous displacements that would be produced at that place and time if the individual waves were each present alone. r2 − r1 = mλ (constructive) r2 − r1 = (m + 1/2) λ (destructive) • Double-slit experiment In other words, you can basically just sum up the individual displacements to get the total result. In the case of light, what we mean by “displacement” is the magnitude of the electric or magnetic field forming the wave. One concept that we need to introduce is that of phase. Phase basically tells whether or not the peaks and troughs making up a wave are lined up or not. If two waves are “in phase”, the displacements are in step with each other. If they are “one cycle out of phase”, the displacements are exactly counter-aligned: the peak of one wave lines up with the trough of the other. Examples are shown in Figure 11. Depending on the phase of two waves, they can interfere either constructively or destructively. Constructive interference (or “reinforcement”) happens when the waves are in phase, resulting in an increase in amplitude. The condition for constructive interference to happen is that the path difference from the two sources must line up with an integer number of wavelengths: r2 − r1 = mλ (m = 0, ± 1, ± 2, ± 3, . . .), Important points d sin θ = mλ (bright) d sin θ = (m + 1/2) λ (dark) ym = Rmλ d (bright) ym = R (m + 1/2) λ d (dark) • Thin film interference 2t = mλ (constructive) 2t = (m + 1/2) λ (destructive) λ = λ0 /nfilm (8) with • The resulting amplitude is the sum of the amplitudes of wave 1 and wave 2; • r1 and r2 are the distances from the wave source to the point of overlap; and • Each of the waves has wavelength λ. Figure 11: Example of in- and out-ofphase waves. phys 202 notes, week 10 8 Figure 12: Wave interference. An example of this is shown in part (b) of Figure 12. The opposite phenomena, destructive interference (or “cancellation”), happens when the waves are exactly a half cycle out of phase. This results in a reduced amplitude. For destructive interference to happen, the path differences align with a half integer multiple of the wavelength, r2 − r1 = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .), (9) with the following being true: • If the two waves have the same amplitude, they completely cancel each other. • r1 and r2 are the same as in Eq. (8). An example of this is shown in part (c) of Figure 12. Double-Slit Interference One of the first experiments showing interference of light waves was the double-slit experiment performed in 1800 by Thomas Young.1 A graphical representation of Young’s experiment is shown in Figure 13. The experiment consists of the following steps: 1. Monochromatic light passes through a single slit S0 , creating a cylindrical wave front on the other side (as a consequence of Huygen’s principle). 2. The wave front travels to two separate slits S1 and S2 which are at different vertical distances from S0 . This creates two separate cylindrical wave fronts on the other side. 3. These two wavefronts travel to a screen where they are observed. Often referred to as “Young’s DoubleSlit Experiment”. 1 phys 202 notes, week 10 9 Figure 13: Thomas Young’s double-slit experiment. The result, observed on the screen in the last step, is a series of bright and dark bands. The bright bands occur when the light from S1 and S2 traveled some integer multiple of the wavelength, satisfying Eq. (8) and giving constructive interference. Likewise, the dark bands occur when the distances satisfy Eq. (9) (half integer multiple of λ) and give destructive interference. The path lengths in the bottom of Figure 13 can be analyzed to figure out the angles for which constructive and destructive interference occur. From part (c) in the figure, the difference in path lengths is given by r2 − r1 = d sin θ, (10) where d is the distance from the slits to the screen. Combining with Eq. (8), we can figure out the angles for which constructive interference occurs. These are given by d sin θ = mλ (m = 0, ± 1, ± 2, ± 3, . . .). (11) Similarly, we can work out the location of destructive interference from Eqns. (10) and (9): d sin θ = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .). (12) Figure 14: Fringes deom the double slit experiment. phys 202 notes, week 10 The bands that form on the screen are often called fringes. A picture of what these look like is shown in Figure 14. Sometimes it’s useful to have an equation for the vertical positions of these fringes, ym . This can be worked out from Eq. (11) and the geometry of the problem, mλ (m = 0, ± 1, ± 2, ± 3, . . .), (13) d where R is the distance from the slits to the screen and d is the distance between the slits. Note that Eq. (13) gives the position of the bright fringes. Similarly, the position of the dark fringes is given by ym = R ym = R (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .), d (14) Thin Film Interference Another way to create interference patterns is to pass light through a thin film of refracting material, as shown in the top of Figure 15. When this happens, part of the light is reflected directly from the top of the film to the observer’s eye, while the rest is refracted; reflected at the bottom of the film; and then refracted to the observer. This process creates a path difference resulting in interference. Since the path length is fixed, we get either constructive or destructive interference depending on the wavelength of the light. The formation of constructive or destructive interference is described by the following equations: Constructive 2t = mλ (m = 0, ± 1, ± 2, ± 3, . . .) (15) Destructive 2t = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .), (16) where t is the thickness of the film. Note that the wavelength λ in Eqns. (15) and (16) is the wavelength in the film medium. This is related to the incoming wavelength λ0 by λ = λ0 /nfilm . (17) This is why rainbow-like patterns form on a thin film, such as the oil slick shown in Figure 15(b). In this case, the incoming light is made up of many different colors (each a different wavelength). Some of the colors have the right wavelength to give constructive interference; others give destructive interference. Those colors interfering constructively are the ones that we see as the bright rainbow spots. In the example above, we discussed thin film interference of light of all colors. Figure 16 shows another situation where thin film interference can occur on monochromatic (single color/wavelength) light. It Figure 15: Thin film interference. 10 phys 202 notes, week 10 11 involved stacking two refracting materials close together with an angled gap. The thickness of the gap at any point is labeled t. The series of events leading to interference patterns is: 1. Some of the incoming light is reflected back to the observer by the top surface. 2. The rest travels through thickness t before reflecting off the bottom surface. 3. In the second reflection, the light undergoes a half-cycle phase shift. This happens whenever light reflects off a surface with nb > na . 4. The reflected light travels back to the observer, with path length difference 2t and phase shift of 1/2. 5. Interference patterns occur, but the patterns are the opposite of what we would expect if there wasn’t a phase change. In other words, Eqns. (15) and (16) are interchanged in terms of whether they refer to constructive or destructive interference. Note that in general, whether Eqns. (15) and (16) apply to constructive or destructive interference depends on whether or not the first and second reflections induce a phase shift. To summarize, • Neither reflection induces a phase shift – Eq. (15) constructive – Eq. (16) destructive • Only one reflection induces a phase shift – Eq. (16) constructive – Eq. (15) destructive • Both reflections induce a phase shift – Eq. (15) constructive – Eq. (16) destructive. Figure 16: Thin film reflection of monochromatic light due to a gap between refracting materials.