3/30/15 Chapter # 36 Image Formation • • • • • • • • • • 36.1: Images Formed by Flat Mirrors (6) 36.2: Images Formed by Spherical Mirrors 36.3: Images Formed by Refraction (10) 36.4: Images formed by Thin Lenses (14) 36.5: Lens Aberrations (2) 36.6: The Camera (1) 36.7: The Eye (2) 36.8: The Simple Magnifier (1) 36.9: The Compound Microscope (1) 36.10: The Telescope Forming Images in a plane mirror • The image is upright. Geometrical Construction of an Image • In order to form image in plane mirror me must follow two different rays from the same point (e.g. top). • d i = d0 1 3/30/15 Image Properties • Upright • Reversed right to left • At the same distance behind the mirror as the object is • The same size • Virtual One example more • Virtual Image versus Real Image • Real Image - light rays path through the image. It is always possible to project such image on the screen. • Virtual Image - no light rays path through the image. It is impossible to project such image on the screen. It is exist in our eye only. • Does it mean that virtual image is bad or wrong? • NO! We use virtual image in our mirror every day and a lot of times! 2 3/30/15 Conceptual Question 26-2 • Two flat mirrors meet at right angle. How many images do we have? How Many? Question 26-2 • Questions • Two flat mirrors meet at angle 720. How many images do we have? • Two flat mirrors meet at angle 600. How many images do we have? 3 3/30/15 Mirrors meet at angle 720 • It will be 5 images if mirrors meet at angle 600 Conceptual Question 26-4 • If you watch a clock in a mirror, do the hands rotate clockwise or counterclock wise? Test questions 5, 6, 7 • • • • • • • • • • • • • 5) The image formed in a plane mirror is A) always inverted. B) always upright. C) sometimes inverted. D) sometimes upright. 6) The image formed in a plane mirror is A) always behind the mirror. B) always in front of the mirror. C) sometimes behind the mirror. D) sometimes in front of the mirror. 7) The image formed in a plane mirror is A) at the same distance in front of the mirror as the object is in front of the mirror. B) at a shorter distance in front of the mirror than the distance that the object is in front of the mirror. • C) at the same distance behind the mirror as the object is in front of the mirror. • D) at a shorter distance behind the mirror than the distance the object is in front of the mirror. 4 3/30/15 Test questions 14, 15 • 14) How tall must a plane mirror be in order for you to be able to see your full image in it? • A) the same height as you • B) half of your height • C) 3/4 of your height • D) twice your height • 15) Which one of the following is the correct number for the magnification of a plane mirror? • A) 1.0 • B) 0.5 • C) 1.5 • D) 2.0 Test questions 8, 18, 19 • • • • • • • • • • • • • • • 8) A person approaches a plane mirror at a speed of 6 m/s. How fast does he approach his image? A) 9 m/s B) 10 m/s C) 11 m/s D) 12 m/s 18) A ray of light strikes a plane mirror perpendicular to the mirror. What is the angle of incidence? A) 360° B) 0° C) 180° D) 90° 19) A ray of light strikes a plane mirror perpendicular to the mirror. What is the angle of reflection? A) 0° B) 90° C) 180° D) 360° Spherical Surface • A spherical mirror is a section of a spherical surface. 5 3/30/15 Concave and Convex Spherical Mirrors • Inside reflecting surface - Concave Mirror • Outside reflecting surface - Convex Mirror Question • What type of spherical mirror is this? Center of Curvature and Principle Axis • The center of curvature is the center of the sphere with radius R of which the mirror is a section. • The principle axis is a straight line drawn through the center of curvature and the midpoint of the mirror. 6 3/30/15 Focus • Focus is a point to which a parallel light rays converge (or from which they appear to diverge). f = R/2 Sign convention See also p. 1165-1166 • For concave mirror: f = R/2 > 0 focal point is in the front of the mirror • For convex mirror: f = -R/2 < 0 focal point is behind the mirror Forming Images in spherical mirror • In order to find orientation, size and location of an image in a spherical mirror geometrical optics uses two techniques: • Ray tracing • Mirror equation 7 3/30/15 Ray Tracing • Ray Tracing is a graphical method of finding orientation, size and location of an image. • The are three rays of light with wellknown path: P-ray, F-ray, and C-ray Ray Tracing for Concave and Convex Mirrors • P-ray is drawn parallel to the principal axis and then passes through the focus • F-ray is drawn through the focus and after reflection goes parallel to the principal axis • C-ray is drawn towards the center of curvature and reflected back in the same direction Convex • Image properties: • Virtual • Smaller than object • Upright • Located between the mirror and focal point 8 3/30/15 Ray Tracing and real photo • Question • What type of mirror is this? • Convex mirror always form image which is smaller than object. So, this is concave mirror. Concave: take a look at the FIG 9 3/30/15 Mirror Equation • Mirror equation is an exact mathematical expression between the object distance and the image distance for a given mirror. h0 h =− i d0 di h0 d0 = −hi di € m= hi d =− i h0 d0 € € h0 d0 − R = −hi R− di Mirror Equation d0 d 0 − R = di R− di € R 1 1 1− − 1 d0 R d0 = = di R− di 1− di R € 1 1 1 1 − = − di R R d 0 ⇒ 1 1 2 1 +€ = = di d0 R f € 1 1 − R d0 1= 1 1 − di R € Concave Mirror: final table 1 1 1 + = di d 0 f € ⇒ di = d0 f d f , h i = − h 0 i = − h0 d0 − f d0 d0 − f d0 di |hi| m 2f < d0<∞ f < di < 2f |hi| < h0 m<1 real, inverted, reduced f < d0 < 2f 2 f < di |hi| > h0 m>1 real, inverted, enlarge d0 < f di < 0 |hi| > h0 m>1 virtual, upright, enlarge di = d 0 f d0 − f ⇒ xi = x0 f , h i = − h0 x 0 −1 d0 − f ⇒ hi −1 = h0 x 0 −1 € 10 3/30/15 Convex Mirror: final table 1 1 1 + =− di d 0 f € ⇒ di = − d0 f d f , h i = − h 0 i = h0 d0 + f d0 d0 + f d0 di hi m ∞ > d0 >0 |di| < f hi < h 0 m<1 virtual, upright, reduced What type of mirror is this? • Can we use spherical mirror equation for plane mirror? f →∞ ⇒ 1 1 + = 0 ⇒ di = − d0 di d 0 • € 11 3/30/15 Problems 21, 22, 23, 24 • An object with a height of 42 cm is placed 2.0 m in front of a concave mirror with a focal length of 0.50 m. (a) Determine the approximate location and size of the image using a ray diagram. (b) Is the image upright or inverted? • Find the location and magnification of the image produced by the mirror in Problem 21 using the mirror and magnification equations. • An object with a height of 42 cm is placed 2.0 m in front of a convex mirror with a focal length of -0.50 m. (a) Determine the approximate location and size of the image using a ray diagram. (b) Is the image upright or inverted? • Find the location and magnification of the image produced by the mirror in Problem 23 using the mirror and magnification equations. Problems 26, 30, 35, 39 • A concave mirror produces a virtual image that is three times as tall as the object. (a) If the object is 22 cm in front of the mirror, what is the image distance? (b) What is the focal length of this mirror? • An object is placed between C and F in front of a concave mirror with a focal length of 10 cm. What is the position of its image in relation to the mirror and what are its characteristics? • An object is placed 70 cm in front of a concave mirror with a focal length of 35 cm. What is the magnification? Test Problem 88 • An object is placed in front of a convex mirror at a distance larger than twice the focal length of the mirror. The image will appear • A) in front of the mirror. • B) inverted and reduced. • C) upright and reduced. • D) upright and enlarged. 12 3/30/15 Test problems 26-43, 44 • John's face is 20 cm in front of a concave shaving mirror. If he observes his image to be twice as big and erect, what is the focal length of the mirror? • John's face is 20 cm in front of a concave shaving mirror of focal length 30 cm. How large an image does he observe? Nice picture. Isn’t it? • Conceptual question 26-4 • After passing through the prism, is the ray (a) deflecting upward, (b) still horizontal, (c) deflected downward ? 13 3/30/15 Lenses: devices used to focus light and form image • Converging lenses take parallel rays of light and bring them together at a focus. • Diverging lenses cause parallel rays of light to spread out as if diverging from a point source. Convex (converging) Lens • We can treat convex lens as two prisms placed back-to-back. After refraction a parallel rays of light are brought together at a focal point. Convex Lens • 14 3/30/15 Concave (diverging) Lens • We can treat concave lens as two prisms placed point-to-point. When diverging rays from a lens are extended back, they appear to originate at a focal point. Concave Lens • Images formed by Lenses • In order to find orientation, size and location of an image formed by lens geometrical optics uses two techniques: • Ray tracing • Thin lens equation 15 3/30/15 Ray tracing • Ray Tracing is a graphical method of finding orientation, size and location of an image. • The are three rays of light with well-known path: P-ray, F-ray, and M-ray Ray Tracing for Diverging and Converging Lenses • P-ray is drawn parallel to the principal axis. After refraction it passes through the focus. • F-ray is drawn through focus and then it passes parallel to the principal axis • M-ray goes through the midpoint of the lens and then continues in its original direction. Examples of ray tracing • 16 3/30/15 Thin lens equation • For convex lenses • For concave lenses 1 1 1 + = di d 0 f f di = d 0 d0 − f hi d f m= =− i =− h0 d0 d0 − f € 1 1 1 + =− di d 0 f f di = − d 0 d0 + f hi d f m= =− i = h0 d0 d0 + f • f>0 for convex lenses; f<0 for concave lenses • m>0 for upright image; m<0 for inverted image € • d>0 when image is on opposite side; d<0 otherwise Converging Lens: final table 1 1 1 + = di d 0 f € ⇒ di = d0 f d f , h i = − h 0 i = − h0 d0 − f d0 d0 − f d0 di |hi| m 2f < d0<∞ f < di < 2f |hi| < h0 |m| < 1 real, inverted, reduced f < d0 < 2f 2 f < di |hi| > h0 |m| > 1 real, inverted, enlarged d0 < f di < 0 hi > h 0 m>1 virtual, upright, enlarged Diverging Lens: final table 1 1 1 + =− di d 0 f € ⇒ di = − d0 f d f , h i = − h 0 i = h0 d0 + f d0 d0 + f d0 di hi m 0 < d0<∞ 0 < di < f hi < h 0 m<1 virtual, upright, reduced 17 3/30/15 Test problems 64, 70 • The index of refraction of a certain medium is 1.5. What is the speed of light in that medium? • If a lens is submerged in water, what happens to its focal length? (a) decreases, (b) stays the same, (c) becomes negative, (d) increases • 75) An object is placed 10 cm from a convex lens of focal length 20 cm. What is the magnification? • 76) An object is placed 60 cm from a convex lens of focal length 10 cm. What is the magnification? • 78) An object is placed 16 cm from a concave lens of focal length 20 cm. What is the magnification? Test problem • Which of the following statements are true? • A) For total internal reflection to take place, light must go from a medium of lower index of refraction to a medium of higher index of refraction. • B) A convex lens always produces a virtual image. • C) A concave lens always produces a virtual image. • D) The index of refraction for a given material is directly proportional to the frequency of light. See also Example 34.7 page 1173 The Camera • The main parts are a lens equipped with shutter, light-tight enclosure, and light-sensitive film (CCD array) that records an image. 18 3/30/15 Projector - “camera in reverse” • A carbon-arc lamp shines through the film (or Liquid Crystal Display), and the projection lens forms a real, enlarged, inverted image of the film. Problem • An ordinary 35 mm color slide has a picture area of 24x36 mm. What focal length would a projection lens need in order to project a 1.2 m x 1.8 m image of this picture on a screen 5.0 m from the lens? The Human Eye • Air 1.00 • Cornea 1.38 • Aqueous humor 1.33 • Lens 1.40 • Vitreous humor 1.34 • Retina consists of rods and cones 19 3/30/15 Human Eye picture • The amount of light is controlled by iris. In bright light it can be 1mm. At night 7mm. Retinal rods and cones • Sensitivity of rods and cones • 20 3/30/15 Image in the eye • Image is real, reduced, inverted. • Our eye can not focus an image if the object is closer then near point distance, N. For young people with normal eye N is 25 cm. Accommodation in the human eye • Ciliary's muscles change the shape of the lens in order to see object closer. This process is named accommodation. Nearsightedness • Refractive power is the ability of the lens to refract light. • Refractive power: • 1 refractive power = f −1 [m ] = [diopter] € 21 3/30/15 Farsightedness • For diverging lens R.P. < 0 • For converging lens R.P. > 0 The Maximum Angular size of the image • If object is closer to the eye the image at the retina becomes greater. • For (a) h θ≈ • For (b) Maximum possible θ≈ h0 N 0 d0 h θ$ ≈ 0 d0$ € € The Magnifying Glass • We put object at focus. So, angular size is θ " ≈ h0 f • Angular magnification: M≈ € θ # h0 f N = = θ h0 N f € 22 3/30/15 Greater Magnification • M≈ θ# N N = = θ p d0 € Problem 27-4 • What is the angular magnification when image produced (a) at infinity, (b) at the near point? N = 26 cm, f = 4.3 cm. • At infinity: M= N 26cm = =6 f 4.3cm • At near point: € M= N 26cm = =7 d0 3.7cm 1 1 1 + = di d 0 f 1 1 1 = − d0 f (−di ) € € Test 27 - 19 • The power of a lens is 4.0 diopters. What is the focal length of this lens? • What is the power of a lens that has a focal length of -40 cm? • The focal length of a magnifying glass is 15 cm. If the near point of a person is 25 cm, what is the magnification of this glass for this person when his eyes are focused at infinity (at near point)? 23 3/30/15 The compound microscope • Compound Microscope Magnification • Objective is a lens with a short focal distance. Image: real, inverted, enlarged. • Eyepiece is a magnifier which that views the image of the objective. mobj = − M =− di d ≈− i d0 f obj M eye = − N f eye di N f obj f eye € Test Problems 50, 68 • The focal lengths of the objective and the eyepiece in a microscope are 0.29 cm and 2.50 cm, respectively. An object is placed at 0.30 cm from the objective lens and the image of this object is viewed with the eyepiece adjusted for minimum eyestrain. What is the final magnification of the microscope? • A) -190 • B) -300 • C) -310 • D) -470 • The distance between the object and the eyepiece of a compound microscope is 18.0 cm. The focal length of its objective lens is 0.80 cm and the eyepiece has a focal length of 2.3 cm. What is the total magnification of the microscope? • A) -120 • B) -200 • C) -360 • D) -480 24 3/30/15 Why in this way? • 25
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