Textbook sections 26-3 – 26-5, 26-8 Physics 1161: Lecture 17 Reflection & Refraction

R f 1) 2) 3) p.a. Preflight 17.1 Which ray is NOT correct? 36% 45% 18%

R f 1) 2) 3) p.a. Preflight 17.1 Ray through center should reflect back on self. Which ray is NOT correct?

Preflight 17.3 The image produced by a concave mirror of a real object is: 1)Always Real 2)Always Virtual 3)Sometimes Real, Sometimes Virtual 39% 30%

Where in front of a concave mirror should you place an object so that the image is virtual? 1.Close to mirror 2.Far from mirror 3.Either close or far 4.Not Possible

Where in front of a concave mirror should you place an object so that the image is virtual? 1.Close to mirror 2.Far from mirror 3.Either close or far 4.Not Possible

Solving Equations A candle is placed 6 cm in front of a concave mirror with focal length f=2 cm. Determine the image location. R f p.a. Preflight 17.2 Compared to the candle, the image will be: Larger Smaller Same Size 29% 12% 59%

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! R f p.a. Preflight 17.2 Compared to the candle, the image will be: Larger Smaller Same Size

Physics 1161: Lecture 17, Slide 9 3 Cases for Concave Mirrors Inside F C F ObjectImage C F Object Image C F Object Image Between C&F Past C Inverted Enlarged Real Upright Enlarged Virtual Inverted Reduced Real

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?

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!

Preflight 17.4 The image produced by a convex mirror of a real object is ? 1)Always real 2)Always virtual 3)Sometimes real and sometimes virtual 24% 61% 15%

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 14 Indices of Refraction

Snell’s Law n1n1 n2n2 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 ) 11 22 1) n 1 > n 2 2) n 1 = n 2 3) n 1 < n 2 Preflight 17.6 Compare n 1 to n % 58 %

Snell’s Law n1n1 n2n2 When light travels from one medium the speed changes. If the angle of incidence is greater than 0, the light bends. During this process, the frequency remains constant. n 1 sin(  1 )= n 2 sin(  2 ) 11 22 1) n 1 > n 2 2) n 1 = n 2 3) n 1 < n 2  1 <  2 sin  1 < sin  2 n 1 > n 2 Preflight 17.6 Compare n 1 to n 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 ? 11 rr Usually, there is both reflection and refraction!

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!

Which way should you move object so image is real and diminished? 1.Closer to the lens 2.Farther from the lens 3.A converging lens can’t create a real, diminished image. F F Object P.A.

Which way should you move object so image is real and diminished? 1.Closer to the lens 2.Farther from the lens 3.A converging lens can’t create a real, diminished image. F F Object P.A.

ImageObject Image Object Image 3 Cases for Converging Lenses This could be used as a projector. Small slide on big screen This is a magnifying glass This could be used in a camera. Big object on small film Upright Enlarged Virtual Inverted Enlarged Real Inverted Reduced Real Inside F Past 2F Between F & 2F

1) Rays parallel to principal axis pass through focal point. 2) Rays through center of lens are not refracted. 3) Rays toward F emerge parallel to principal axis. Diverging Lens Principal Rays F F Object P.A. Image is (always true): Real or Imaginary Upright or Inverted Reduced or Enlarged

1) Rays parallel to principal axis pass through focal point. 2) Rays through center of lens are not refracted. 3) Rays toward F emerge parallel to principal axis. Diverging Lens Principal Rays F F Object P.A. Image is virtual, upright and reduced. Image

Which way should you move the object to cause the image to be real? 1.Closer to the lens 2.Farther from the lens 3.Diverging lenses can’t form real images F F Object P.A.

Which way should you move the object to cause the image to be real? 1.Closer to the lens 2.Farther from the lens 3.Diverging lenses can’t form real images F F Object P.A.