180 likes | 200 Views
Physics 1161: Lecture 17. Reflection and Refraction of Light. Today’s Lecture will cover textbook sections 26-3 – 26-5, 26-8. q i. q r. Last Time. Today. q 1. Next time. n 1. n 2. q 2. Overview. Reflection:. q i = q r. Flat Mirror: image equidistant behind. Spherical Mirrors:
E N D
Physics 1161: Lecture 17 Reflection and Refraction of Light • Today’s Lecture will cover textbook sections 26-3 – 26-5, 26-8
qi qr Last Time Today q1 Next time n1 n2 q2 Overview Reflection: qi = qr Flat Mirror: image equidistant behind Spherical Mirrors: Concave or Convex Refraction: n1 sin(q1)= n2 sin(q2) Flat Lens: Window Spherical Lenses: Concave or Convex Absorption
#1 #2 #3 Concave Mirror Principal Rays 1) Parallel to principal axis reflects through f. 2) Through f, reflects parallel to principal axis. 3) Through center. O f Image is (in this case): Real or Imaginary Inverted or Upright Reduced or Enlarged c **Every other ray from object tip which hits mirror will reflect through image tip
Mirror Equation do I di O f c • do = distance object is from mirror: • Positive: object in front of mirror • Negative: object behind mirror • di = distance image is from mirror: • Positive: inverted image (in front of mirror) • Negative: upright image (behind mirror) • f = focal length mirror: • Positive: concave mirror • Negative: convex mirror (coming soon)
Where in front of a concave mirror should you place an object so that the image is virtual? Mirror Equation: • Close to mirror • Far from mirror • Either close or far • Not Possible • Concave mirror: f > 0 • Object in front of mirror: do > 0 • Virtual image means behind mirror: di < 0 • When do < f then di <0 : virtual image.
Magnification Equation q do do q Angle of incidence I di ho q di hi q Angle of reflection O • ho = height of object: • Positive: always • hi = height of image: • Positive: image is upright • Negative: image is inverted • m = magnification: • Positive / Negative: same as for hi • < 1: image is reduced • > 1: image is enlarged
Example Solving Equations A candle is placed 6 cm in front of a concave mirror with focal length f=2 cm. Determine the image location. di = + 3 cm (in front of mirror) Real Image!
Magnification A 4 inch arrow pointing down is placed in front of a mirror that creates an image with a magnification of –2. • What is the size of the image? • 2 inches • 4 inches • 8 inches 4 inches Magnitude gives us size. • What direction will the image arrow point? • Up 2) Down (-) sign tells us it’s inverted from object
C C C F F F Object Image Image Object Object Image 3 Cases for Concave Mirrors Upright Enlarged Virtual Inside F Inverted Enlarged Real Between C&F Inverted Reduced Real Past C
#1 I #2 #3 Convex Mirror Rays 1) Parallel to principal axis reflects through f. 2) Through f, reflects parallel to principal axis. 3) Through center. O f c Image is: Virtual (light rays don’t really cross) Upright (same direction as object) Reduced (smaller than object) (always true for convex mirrors!):
Virtual Image! Image is Upright! Example Solving Equations A candle is placed 6 cm in front of a convex mirror with focal length f=-3 cm. Determine the image location. Determine the magnification of the candle. If the candle is 9 cm tall, how tall does the image candle appear to be? di = - 2 cm (behind mirror) m = + 1/3 hi = + 3 cm Physics 1161: Lecture 17, Slide 11
Mirror Summary • Angle of incidence = Angle of Reflection • Principal Rays • Parallel to P.A.: Reflects through focus • Through focus: Reflects parallel to P.A. • Through center: Reflects back on self • |f| = R/2
qi qr q1 n1 n2 q2 Light Doesn’t Just Bounce It Also Refracts! Reflected: Bounces (Mirrors!) qi = qr Refracted: Bends (Lenses!) n1 sin(q1)= n2 sin(q2) Physics 1161: Lecture 17, Slide 13
Speed of light in vacuum Speed of light in medium Index of refraction Index of Refraction 186,000 miles/second: it’s not just a good idea, it’s the law! so always!
Indices of Refraction Physics 1161: Lecture 17, Slide 15
Snell’s Law When light travels from one medium to another the speed changes v=c/n, but the frequency is constant. So the light bends: n1 sin(q1)= n2 sin(q2)
Example Snell’s Law Practice 1 r Usually, there is both reflection and refraction! A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle qr = 60. The other part of the beam is refracted. What is q2? q1 =qr =60 sin(60) = 1.33 sin(q2) n1 q2 = 40.6 degrees n2 normal
Apparent depth: d apparent fish d actual fish Apparent Depth n2 n1 50