Today’s Lecture will cover textbook sections 26-3 – 26-5, 26-8 Physics 1161: Lecture 17 Reflection and Refraction of Light

Overview Reflection: Refraction: Absorption Flat Mirror: image equidistant behind Spherical Mirrors: Concave or Convex Flat Lens: Window Spherical Lenses: Concave or Convex Today Last Time Next time ii rr 11 22 n2n2 n1n1  i =  r n 1 sin(  1 )= n 2 sin(  2 )

O Concave Mirror Principal Rays f c 1) Parallel to principal axis reflects through f. #1 3) Through center. #3 Image is (in this case): Real or Imaginary Inverted or Upright Reduced or Enlarged **Every other ray from object tip which hits mirror will reflect through image tip 2) Through f, reflects parallel to principal axis. #2

O I Mirror Equation c dodo didi d o = distance object is from mirror: Positive: object in front of mirror Negative: object behind mirror d i = 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) f

Where in front of a concave mirror should you place an object so that the image is virtual? Mirror Equation: 1)Close to mirror 2)Far from mirror 3)Either close or far 4)Not Possible When d o < f then d i <0 : virtual image. Virtual image means behind mirror: d i < 0 Concave mirror: f > 0 Object in front of mirror: d o > 0

O I Magnification Equation dodo dodo hoho Angle of incidence didi hihi Angle of reflection didi h o = height of object: Positive: always h i = height of image: Positive: image is upright Negative: image is inverted m = magnification: Positive / Negative: same as for h i < 1: image is reduced > 1: image is enlarged    

Solving Equations A candle is placed 6 cm in front of a concave mirror with focal length f=2 cm. Determine the image location. d i = + 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? 1)2 inches 2)4 inches 3)8 inches What direction will the image arrow point? 1)Up2) Down (-) sign tells us it’s inverted from object Magnitude gives us size. 4 inches

3 Cases for Concave Mirrors Inside F C F ObjectImage C F Obje ct Image C F Obje ct Image Between C&F Past C Inverted Enlarged Real Upright Enlarged Virtual Inverted Reduced Real

O Convex Mirror Rays c 1) Parallel to principal axis reflects through f. 2) Through f, reflects parallel to principal axis. #2 I 3) Through center. #3 Image is: Virtual (light rays don’t really cross) Upright (same direction as object) Reduced (smaller than object) (always true for convex mirrors!): f #1

Physics 1161: Lecture 17, Slide 11 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? m = + 1/3 h i = + 3 cm Image is Upright! d i = - 2 cm (behind mirror) Virtual Image!

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

Physics 1161: Lecture 17, Slide 13 Light Doesn’t Just Bounce It Also Refracts! Reflected: Bounces (Mirrors!) Refracted: Bends (Lenses!) ii rr 11 22 n2n2 n1n1  i =  r n 1 sin(  1 )= n 2 sin(  2 )

Speed of light in medium Index of refraction Speed of light in vacuum so Index of Refraction 186,000 miles/second: it’s not just a good idea, it’s the law! always!

Physics 1161: Lecture 17, Slide 15 Indices of Refraction

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: n 1 sin(  1 )= n 2 sin(  2 )

n1n1 n2n2 Snell’s Law Practice normal 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  r = 60. The other part of the beam is refracted. What is  2 ? sin(60) = 1.33 sin(  2 )  2 = 40.6 degrees  1 =  r =  11 rr Usually, there is both reflection and refraction!

n2n2 n1n1 d d Apparent depth: Apparent Depth 50 actual fish apparent fish