Chapter 33 Continued Properties of Light. Law of Reflection Law of Refraction or Snell s Law Chromatic Dispersion Brewsters Angle

Transcription

1 Chapter 33 Continued Properties of Light Law of Reflection Law of Refraction or Snell s Law Chromatic Dispersion Brewsters Angle

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3 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

4 Red Snells Law

5 blue Snells Law

6 Fiber Cable Total Internal Reflection Simulator Halliday, Resnick, Walker: Fundamentals of Physics, 7th Edition - Student Companion Site

7 Why is light totally reflected inside a fiber optics cable? Internal reflection n sin! = n 2 sin! 2 (.509)sin! = (.00)sin 90 =.00! " sin #.509 " deg

8 Total Internal Reflection

9 What causes a Mirage sky eye Index of refraction Hot road causes gradient in the index of refraction that increases as you increase the distance from the road

10 Inverse Mirage Bend

11 Chromatic Dispersion

12 How does a Rainbow get made? Primary rainbow Secondary rainbow Supernumeraries

13 Polarization by Reflection: Brewsters Law

14 Snells Law Example 47. In the figure, a 2.00-m-long vertical pole extends from the bottom of a swimming pool to a point 50.0 cm above the water. What is the length of the shadow of the pole on the level bottom of the pool? Consider a ray that grazes the top of the pole, as shown in the diagram below. Here θ = 35 o, l = 0.50 m, and l 2 =.50 m. The length of the shadow is x + L. x is given by x = l tanθ = (0.50m)tan35 o = 0.35 m. air θ l L is given by water θ 2 L=l 2 tan θ 2 l 2 Use Snells Law to find θ 2 shadow L x

15 Calculation of L According to the law of refraction, n 2 sinθ 2 = n sinθ. We take n = and n 2 =.33 ( & # o ' sin( ' sin 35 2 = sin $ = sin = n! $ 2.33! % " & % # " o L is given by L = l 2 tan! 2 = (.50m) tan o = 0.72m. air water θ l The length of the shadow is L+x. L+x = 0.35m m =.07 m. θ 2 shadow l 2 L x

16 Lecture 4 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

17 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.

18 Law of Reflection! I =! R Mirror B! r! i A

19 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

20 Two light rays and 2 taking different paths between points A and B and reflecting off a vertical mirror B Plane Mirror 2 A Use calculus - method of minimization

21 t = ( h 2 C + y 2 + h (w! y) 2 ) dt dy = 2y h 2 + y +!2(w! y) 2 h (w! y) 2 ) = 0 y h 2 + y = (w! y) 2 h (w! y) 2 ) Write down time as a function of y and set the derivative to 0. sin" I = sin" R! I =! R

22 t = ( v h 2 + y 2 + v 2 h (w! y) 2 ) dt dy = 0 v sin" I = v 2 sin" R n sin! I = n 2 sin! R

23 Mirrors and Lenses Plane Mirrors Where is the image formed

24 Plane mirrors Real side Angle of incidence Normal Virtual side Virtual image Angle of reflection eye i = - p Object distance = - image distance Image size = Object size

25 Problem: Two plane mirrors make an angle of 90 o. How many images are there for an object placed between them? mirror eye 2 object mirror 3

26 Using the Law of Reflection to make a bank shot Assuming no spin Assuming an elastic collision No cushion deformation d d pocket

27 i = - p magnification = 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.

28 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 p + i = f m =!i p

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30 Spherical refracting surfaces Using Snell s Law and assuming small Angles between the rays with the central axis, we get the following formula: n p + n 2 i = n 2! n r

31 Thin Lenses n p + n 2 i = n 2! n r

32 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

33 Thin Lens Equation Lensmaker Equation f = p + i f = (n!)( r! r 2 ) Lateral Magnification for a Lens m =! i p image height m =! object height What is the sign convention?

34 Sign Convention Light Virtual side - V r r 2 Real side - R 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.

35 Rules for drawing rays to locate images from a lens ) 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.

36 Real image ray diagram for a converging lens

37 Object Distance=20 cm and lens focal length is +0 cm Find: Image distance Magnification Height of image Is image erect or inverted Is it real or virtual Draw the 3 rays

38 Object Distance =5 cm and focal length is + 0 cm Find: Image distance Magnification Height of image Is image erect or inverted Is it real or virtual Draw the 3 rays

39 p=object Distance =0 cm

40 Virtual image ray diagram for converging lens

41 Virtual image ray diagram for a diverging lens

42 Example 24(b). Given a lens with a focal length f = 5 cm and object distance p = +0 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays. Virtual side Real side.. F F 2 p = i f! p i = 5! 0 = + 0 i = +0 cm m = y! y = " i p m =! 0 0 =! Image is real, inverted.

43 24(e). Given a lens with the properties (lengths in cm) r = +30, r 2 = -30, p = +0, and n =.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Virtual side Real side.. r 2 p F F 2 r f = & # ( n ' ) $ '! % r r2 " = i f! p m = y! y = " i p f = & $ % (.5 ' ) 30 ' ' 30 #! " = 30 = i 30! 0 =! 5 m =!!5 0 = +.5 f = 30cm i =! 5cm Image is virtual, upright.

44 27. A converging lens with a focal length of +20 cm is located 0 cm to the left of a diverging lens having a focal length of -5 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 Lens f f 2 f f

45 Lens Lens f f 2 f f Ignoring the diverging lens (lens 2), the image formed by the converging lens (lens ) is located at a distance i = f! p = 20cm! 40cm. i = 40cm Since m = -i /p = - 40/40= -, the image is inverted This image now serves as a virtual object for lens 2, with p 2 = - (40 cm - 0 cm) = - 30 cm.

46 Lens Lens f f 2 f f i 2 = f 2! p 2 =!5cm!!30cm i 2 =!30cm. Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i 2 < 0). The magnification is m = (-i /p ) x (-i 2 /p 2 ) = (-40/40)x(30/-30) =+, so the image has the same size orientation as the object.

47 Optical Instruments Magnifying lens Compound microscope Refracting telescope

48 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. Fig (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.

49 Chapter 34 Problem 9 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).

50 (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'?