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Geometric Optics April 6, 2014 Chapter 32 1 Law of Refraction ! The law of refraction can be expressed as n1 sinθ1 = n2 sinθ 2 ! where n1 and θ1 are the index of refraction and angle from the normal in the first medium and n2 and θ2 are the index of refraction and angle from the normal in the second medium ! The law of refraction is also called Snell’s Law ! We will always assume the index of refraction of air is one ! The situation of light incident on various media from air is common so we will write the formulas for refraction of light incident on a surface with index of refraction nmedium as sinθ air nmedium = sinθ medium April 6, 2014 Chapter 32 2 Total Internal Reflection ! Consider the situation where light travels across the boundary between two media where the first medium has an index of refraction greater than the second medium, n1 > n2 ! In this case the light is bent away from the normal ! As one increases the angle of incidence from medium 1, θ1, one can see that the angle of the transmitted light, θ2, can approach 90° ! When θ2 reaches 90°, total internal reflection takes place instead of refraction ! The critical angle θc for total internal reflection is n2 sinθ1 sinθ c n2 = = ⇒ sinθ c = n1 sinθ 2 sin90° n1 April 6, 2014 Chapter 32 3 Color Different colors are associated with light of different wavelengths. The longest wavelengths are perceived as red light and the shortest as violet light. The table below is a brief summary of the visible spectrum of light. Dispersion The slight variation of index of refraction with wavelength is known as dispersion. Shown is the dispersion curves of two common glasses. Notice that n is larger when the wavelength is shorter, thus violet light refracts more than red light. Consequences of Dispersion Red light passing through a prism White light passing through a prism Each wavelength has a different index of refraction and so refracts at a different angle Prism Index of refraction for blue light is larger than for red light, so blue light bends more than red light Rainbows Red light from high in the sky reaches your eye; violet light from lower in the sky does the same Water droplets Sometimes there can be a double rainbow, when there are two internal reflections, but the order of colors will be reversed. Sprinkler rainbow conspiracy April 6, 2014 Chapter 32 9 Lenses ! When light is refracted crossing a curved boundary between two different media, the light rays follow the law of refraction at each point on the boundary ! The angle at which the light rays cross the boundary is different along the boundary, so the refracted angle is different at different points along the boundary ! A curved boundary between two optically transparent media is called a lens ! Light rays that are initially parallel before they strike the boundary are refracted in different directions depending on the part of the lens they strike ! Depending of the shape of the lens, the light rays can be focused or caused to diverge April 6, 2014 Chapter 33 11 Lens Maker’s Formula ! If the front surface of the lens is part of a sphere with radius R1 and the back surface is part of a sphere with radius R2, then we can calculate the focal length f of the lens using the lens-makers formula (Derivation 33.1 in book) ⎛ 1 1⎞ 1 = (n −1) ⎜ − ⎟ f ⎝ R1 R2 ⎠ ! The curvatures R1 and R2 have different signs depending on whether their center point is on the same side as object (negative) or on the opposite side (positive) • For a convex lens, R1 is positive and R2 is negative • For a concave lens, R1 is negative and R2 is positive ! If we have a lens with the same radii on the front and back of the lens so that |R1| = |R2|= R, we get 1 2(n −1) = f R April 6, 2014 Chapter 33 12 Lens Maker’s Formula April 6, 2014 Chapter 33 13 Lens Focal Lengths ! Unlike mirrors, lenses have a focal length on both sides ! Light can pass through lenses while mirrors reflect the light allowing no light on the opposite side ! The focal length of a convex (converging) lens is defined to be positive ! The focal length of a concave (diverging) lens is defined to be negative April 6, 2014 Chapter 33 14 Convex Lenses ! A convex (converging) lens is shaped such that parallel rays will be focused by refraction at the focal distance f from the center of the lens ! In the drawing on the right, a light ray is incident on a convex glass lens ! At the surface of the lens, the light ray is refracted toward the normal ! When the ray leaves the lens, it is refracted away from the normal ! Let us now study the case of several horizontal light rays incident on a convex lens April 6, 2014 Chapter 33 15 Thin Lenses ! These rays will be focused to a point a distance f from the center of the lens on the opposite side from the incident rays ! The rays enter the lens, get refracted at the surface, traverse the lens in a straight line, and get refracted when they exit the lens April 6, 2014 Chapter 33 16 Thin Lenses ! Let’s represent the thin lens approximation by drawing a black dotted line at the center of the lens ! Instead of following the detailed trajectory of the light rays inside the lens, we draw the incident rays to the centerline and then on to the focal point ! This real-life lens is a thick lens and there is displacement between the refraction at the entrance and exit surfaces ! We will treat all lenses as thin lenses and treat the lens as a line at which refraction takes place April 6, 2014 Chapter 33 17 Images with Convex Lenses ! Convex lenses can be used to form images ! We will show the geometric construction of the formation of an image using a convex lens with focal length f ! We place an object standing on the optical axis ! This object has a height ho and is located a distance do from the center of the lens such that do > f April 6, 2014 Chapter 33 18 Images with Convex Lenses ! We start with a ray along the optical axis of the lens that passes straight through the lens that defines the bottom of the image ! A second ray is then drawn from the top of the object parallel to the optical axis, which is focused through the focal point on the other side of the lens April 6, 2014 Chapter 33 19 Images with Convex Lenses ! A third ray is drawn through the center of the lens that is not refracted in the thin lens approximation ! A fourth ray is drawn from the top of the object through the focal point on the same side of the lens that is then directed parallel to the optical axis April 6, 2014 Chapter 33 20 Images with Convex Lenses ! The three rays starting at the top of the object intersect, locating the top of the image ! Any two of these three rays from the top of the object can be used to locate the top of the image ! In this case, with do > f, a real, inverted, and enlarged image with height hi < 0 is formed at a distance di > 0 from the center of the lens April 6, 2014 Chapter 33 21 Images with Convex Lenses ! Now let us consider the image formed by an object with height ho placed a distance do from the center of the lens such that do < f ! The first ray again is drawn from the bottom of the object along the optical axis ! The second ray is drawn from the top of the object parallel to the optical axis and is focused through the focal point on the opposite side ! A third ray is drawn through the center of the lens April 6, 2014 Chapter 33 23 Images with Convex Lenses ! A fourth line is drawn such that it originates from the focal point on the same side of the lens and is then focused parallel to the optical axis ! A virtual image is formed on the same side of the lens as the object by extrapolating the three rays from the top of the object back until they intersect ! Red-and-black dashed lines represent the extrapolated rays ! In this case, with do < f, a virtual, upright, and enlarged image is formed with hi > 0 and di < 0 April 6, 2014 Chapter 33 24 Images formed with Concave Lenses ! We show the formation of an image using a concave lens ! We place an object standing on the optical axis represented by the green arrow ! This object has a height ho and is located a distance do from the center of the lens such that do > f ! We again start with a ray along the optical axis of the lens that passes straight through the lens that defines the bottom of the image ! A second ray is then drawn from the top of the object parallel to the optical axis April 6, 2014 Chapter 33 25 Images formed with Concave Lenses ! This ray is refracted such that the extrapolation of the diverging ray passes through the focal point on the other side of the lens ! A third ray is drawn through the center of the lens that is not refracted in the thin lens approximation ! This ray is extrapolated back along its original path ! The image is formed at the intersection of these two rays ! The image is virtual, upright, and reduced April 6, 2014 Chapter 33 26 The Lens Equation ! The images formed by lenses are described by the lens equation 1 1 1 + = do di f ! This equation is the same relationship between focal length, image distance, and object distance that we had for mirrors ! To treat all possible cases for lenses we must define conventions for distances and heights ! The focal length of a converging lens is positive ! The focal length of a diverging lens is negative ! The object distance and object height for a single lens are positive ! If the image is on the opposite side of the lens then the image distance is is positive and the image is real ! If the image is on the same side of the lens as the object, the image distance is negative and the image is virtual April 6, 2014 Chapter 33 27 The Lens Equation ! The magnification formula is the same for lenses as mirrors hi di m= =− ho do ! If the image is upright, then the image height and magnification are positive ! If the image is inverted, the the image height is negative and the magnification is negative ! For converging lenses where do > f, we always get a real, inverted image on the opposite side ! For converging lenses where do = f, we get an image at infinity ! For converging lenses where do < f, we always get a virtual, upright, and enlarged image on the same side April 6, 2014 Chapter 33 28 The Lens Equation ! The following table summarizes the results for all values of do for a converging lens ! For a diverging lens, the image is always virtual, upright, and reduced April 6, 2014 Chapter 33 29 Mirrors and Lenses The Lens Equation ! Instrument makers often describe lenses in terms of the power of the lens rather than the focal length of the lens ! The power of a lens is given in diopters 1 D= f expressed in meters f 1 1 1 + = do di f ! Eyeglasses are usually categorized in diopters ! Typical reading glasses have D = +1.50 diopters ! The focal length of these lenses is 1 1 f= = = 0.667 m −1 D 1.50 m ! More later in this chapter about vision corrections April 6, 2014 Chapter 33 31 The Human Eye ! The extremes over which distinct vision is possible are called the far point and near point ! The far point of a normal eye is infinity ! The near point of a normal eye depends on the ability of the eye to focus, which changes with age ! Several common vision defects result from incorrect focal distances ! In the case of myopia (nearsightedness), the image is produced in front of the retina ! In the case of hyperopia (farsightedness), the image would be produced behind the retina if it could, but the retina absorbs the light before an image can be formed April 6, 2014 Chapter 33 32 The Human Eye Normal-sighted Nearsighted Farsighted ! Myopia can be corrected using diverging lenses • Negative focal length, negative power ! Hyperopia can be corrected using converging lenses • Positive focal length, positive power • Reading glasses April 6, 2014 Chapter 33 33 Corrective Lenses for Myopia PROBLEM ! What is the power of the corrective lens for a myopic (nearsighted) person whose uncorrected far point is 15 cm? SOLUTION ! The corrective lens must form a virtual, upright image 15 cm in front of the lens of an object located at infinity ! The object distance is ∞ and the image distance is –15 cm 1 1 1 + = = −6.7 diopters ∞ −0.15 m f ! The required lens is a diverging lens with a power of –6.7 diopter (a focal length of -0.15 m) April 6, 2014 Chapter 33 34 Corrective Lenses for Hyperopia PROBLEM ! What is the power of the corrective lens for a hyperopic (far-sighted) person whose uncorrected near point is 75 cm? SOLUTION ! The corrective lens must form a virtual, upright image at the uncorrected near point Virtual image at uncorrected near point Close object ! The object distance is 25 cm, the image distance is –75 cm 1 1 1 + = = +2.7 diopters 0.25 m −0.75 m f ! The required lens is a converging lens with a power of +2.7 diopter (a focal length of +0.37 m) April 6, 2014 Chapter 33 35 The Microscope ! The simplest microscope is a system of two lenses ! The first lens is a converging lens of short focal length, fo, called the objective lens ! The second lens is another converging lens of greater focal length, fe, called the eyepiece ! The object to be magnified is placed just outside the focal length of the objective lens so that do,1 ≈ fo ! This arrangement allows the objective lens to form a real, inverted, and enlarged image of the object ! This image then becomes the object for the eyepiece lens ! This intermediate image is placed just inside the focal length fe of the eyepiece lens, so that the eyepiece lens produces a virtual, upright, and enlarged image of the intermediate image and do,2 ≈ fe April 6, 2014 Chapter 33 36 April 6, 2014 Chapter 33 37