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Lecture 14 Images Chapter 34 •Geometrical Optics •Fermats Principle -Law of reflection -Law of Refraction •Plane Mirrors and Spherical Mirrors •Spherical refracting Surfaces •Thin Lenses •Optical Instruments -Magnifying Glass, Microscope, Refracting telescope • Polling Questions Left over from Lecture 13 and Chapter 33 1) Show electric and magnetic fields are out of phase in demo. 2) Reflection and Refraction 1) Red or green laser through smoky block 2) Show maximum bending angle 3) Total internal reflection 1) Show it with block 1) Dispersion through prism 2) Brewsters Law 1) Show using polaroid paper Reflection and Refraction of Light Dispersion: Different wavelengths have different velocities and therefore different indices of refraction. This leads to different refractive angles for different wavelengths. Thus the light is dispersed. The frequency does not change when n changes. v f changes when medium changes f does not change when medium changes Snells Law n1 sin 1 n2 sin 2 Red 1.00sin 1 1.509 1.00sin 90 sin 2 1.509 2 41.505deg sin 2 n1=1.000 n2=1.509 θ1 θ2 v1 sin 1 v2 sin 2 Snell’s Law at Work Fiber Cable Why is light totally reflected inside a fiber optics cable? Internal reflection n1 sin 1 n2 sin 2 (1.509)sin 1 (1.00)sin 90 1.00 1 sin 1 1 1.509 41.505 deg Equilateral prism dispersing sunlight late afternoon. Sin θrefr= 0.866nλ n increases with frequency Blue light bends more than red 9 How much light is polarized when reflected from a surface? Polarization by Reflection: Brewsters Law Fresnel Equations The parallel and perpendicular components of the electric vector of the reflected and transmitted light wave is plotted below when light strikes a piece of glass. Now the intensity of polarized light is proportional to the square of the amplitude of the oscillations of the electric field. So, we can express the intensity of the incoming light as I0 = (constant) E 2. E What causes a Mirage eye sky 1.09 1.09 1.08 Index of refraction 1.08 1.07 1.07 1.06 Hot road causes gradient in the index of refraction that increases as you increase the distance from the road Inverse Mirage Bend Geometrical Optics:Study of reflection and refraction of light from surfaces The ray approximation states that light travels in straight lines until it is reflected or refracted and then travels in straight lines again. The wavelength of light must be small compared to the size of the objects or else diffractive effects occur. Law of Reflection I R Mirror B r i A 1 Fermat’s Principle Using Fermat’s Principle you can prove the Reflection law. It states that the path taken by light when traveling from one point to another is the path that takes the shortest time compared to nearby paths. JAVA APPLET Show Fermat’s principle simulator http://www.phys.hawaii.edu/~teb/java/ntnujava/index.html Two light rays 1 and 2 taking different paths between points A and B and reflecting off a vertical mirror B Plane Mirror 2 A 1 Use calculus - method of minimization t C1 ( h12 y 2 h22 (w y)2 ) dt dy 2y h y 2 1 2 y h12 y 2 Write down time as a function of y and set the derivative to 0. sin I 2(w y) h (w y) ) 2 2 I R 2 0 (w y) h22 (w y)2 ) sin R t( 1 1 h12 y 2 h22 (w y)2 ) v1 v2 dt 0 dy 1 sin I v1 1 sin R v2 n1 sin I n 2 sin R Mirrors and Lenses Plane Mirrors Where is the image formed Plane mirrors Angle of Real side incidence Normal Angle of reflection Virtual side Virtual image i=-p eye Object distance = - image distance Image size = Object size Problem: Two plane mirrors make an angle of 90o. How many images are there for an object placed between them? mirror eye object 2 mirror 1 3 Using the Law of Reflection to make a bank shot Assuming no spin Assuming an elastic collision No cushion deformation d d pocket i=-p magnification = 1 What happens if we bend the mirror? Concave mirror. Image gets magnified. Field of view is diminished Convex mirror. Image is reduced. Field of view increased. Rules for drawing images for mirrors • Initial parallel ray reflects through focal point. •Ray that passes in initially through focal point reflects parallel from mirror •Ray reflects from C the radius of curvature of mirror reflects along itself. • Ray that reflects from mirror at little point c is reflected symmetrically 1 1 1 r f 2 p i f i Lateral magnification = ratio of image height/object height m p One simple way to measure focal length of a concave mirror? Spherical refracting surfaces Using Snell’s Law and assuming small Angles between the rays with the central axis, we get the following formula: n1 n2 n2 n1 p i r Chapter 34 Problem 32 In Figure 34-35, A beam of parallel light rays from a laser is incident on a solid transparent sphere of index of refraction n. n1 n2 n2 n1 p i r Fig. 34-35 (a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere? (b) What index of refraction, if any, will produce a point image at the center of the sphere? Enter 'none' if necessary. n1 n2 n2 n1 p i r Apply this equation to Thin Lenses where the thickness is small compared to object distance, image distance, and radius of curvature. Neglect thickness. Converging lens Diverging lens Simple Lens Model 1 1 1 f p i Thin Lens Equation Lensmaker Equation 1 1 1 (n 1)( ) f r1 r2 Lateral Magnification for a Lens What is the sign convention? i m p image height m object height Sign Convention Light Virtual side - V Real side - R r1 r2 p i Real object - distance p is pos on V side (Incident rays are diverging) Radius of curvature is pos on R side. Real image - distance is pos on R side. Virtual object - distance is neg on R side. Incident rays are converging) Radius of curvature is neg on the V side. Virtual image- distance is neg on the V side. Rules for drawing rays to locate images from a lens 1) A ray initially parallel to the central axis will pass through the focal point. 2) A ray that initially passes through the focal point will emerge from the lens parallel to the central axis. 3) A ray that is directed towards the center of the lens will go straight through the lens undeflected. Real image ray diagram for a converging lens Virtual image ray diagram for converging lens Virtual image ray diagram for a diverging lens Example Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays. Virtual side . F1 p Real side . F2 Example Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays. Virtual side Real side . . F1 F2 p 1 1 1 i f p m 1 1 1 1 i 5 10 10 m i 10 cm y i y p 10 1 10 Image is real, inverted. Example Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Real side Virtual side . F1 r1 r2 p . F2 Example Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Real side Virtual side . F1 r1 r2 p . F2 y i y p 1 1 1 n 1 f r1 r2 1 1 1 i f p 1 1 1 1 1.5 1 f 30 30 30 1 1 1 1 i 30 10 15 m f 30cm i 15cm Image is virtual, upright. m 15 1.5 10 Example A converging lens with a focal length of +20 cm is located 10 cm to the left of a diverging lens having a focal length of -15 cm. If an object is located 40 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. Treat each lens Separately. Lens 1 Lens 2 +20 -15 f1 f2 f1 f2 40 10 Lens 1 Lens 2 +20 -15 f1 f2 f 1 f2 40 40 10 30 Ignoring the diverging lens (lens 2), the image formed by the converging lens (lens 1) is located at a distance 1 1 1 1 1 . i1 40cm i1 f1 p1 20cm 40cm Since m = -i1/p1= - 40/40= - 1 , the image is inverted This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm. Lens 1 Lens 2 +20 -15 f1 f2 f 1 f2 40 40 10 1 1 1 1 1 i2 f 2 p2 15cm 30cm 30 i2 30cm. Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i2 < 0). The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image has the same size orientation as the object. Optical Instruments Magnifying lens Compound microscope Refracting telescope Galileo - converging + diverging lens Keplerian - converging + converging lens Reflecting Telescope Chapter 34 Problem 91 Figure 34-43a shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Figure 34-43b). A "normal" eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Figure 34-43c). (a) Suppose that for the parallel rays of Figure 34-43a and Figure 34-43b, the focal length f of the effective thin lens of the eye is 2.52 cm. For an object at distance p = 48.0 cm, what focal length f' of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f'?

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