 # Physics 123 23. Light: Geometric Optics 23.1 The Ray Model of Light 23.2 Reflection - Plane Mirror 23.3 Spherical Mirrors 23.5 Refraction - Snell’s law.

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Physics 123

23. Light: Geometric Optics 23.1 The Ray Model of Light 23.2 Reflection - Plane Mirror 23.3 Spherical Mirrors 23.5 Refraction - Snell’s law 23.7 Converging and Diverging Lenses

Reflection ii rr  i =  r Angle of incidence equals the angle of reflection

Image in a Plane Mirror

d o = d i

Spherical Mirrors Concave mirror Convex mirror

Concave Mirror radius = r focus = f r f f = r / 2

Concave Mirror A parallel ray reflects through the focus

Concave Mirror Solar Cooker!

Concave Mirror A ray through the focus reflects parallel

Concave Mirror Image is inverted, real, reduced

Concave Mirror Equation 1 / d o + 1 / d i = 1 / f

Concave Mirror Problem A 3 cm high candle is located 5 cm from a concave mirror whose radius of curvature is 20 cm. What are the characteristics of the image?

Concave Mirror Problem f = 10 cm 1/d o + 1/ d i = 1/f 1/5 + 1/ d i = 1/10 1/ d i = - 1/10 d i = - 10 cm The image is 10 cm behind the mirror, virtual, upright, and magnified 2X m = - d i / d o

Concave Mirror Problem Makeup mirror! If the object is within the focus of the concave mirror the image is enlarged (magnified), upright but virtual … it’s all in your head!!!

Convex Mirror

Image will ALWAYS be reduced, virtual, upright

Jurassic Park! Rear View Mirror Objects in the mirror are closer than they appear!

Convex Mirror Equation 1 / d o + 1 / d i = 1 / f Note: f is negative d i is negative

Convex Mirror Problem A 3 cm high candle is located 5 cm from a convex mirror whose radius of curvature is 20 cm. What are the characteristics of the image?

Convex Mirror Problem f = - 10 cm 1/d o + 1/ d i = 1/f 1/5 + 1/ d i = - 1/10 1/ d i = - 3/10 d i = - 3.3 cm The image is 3.3 cm behind the mirror, virtual, upright, and reduced in size m = - d i / d o m = 1 / 3

Refraction ii rr

Index of Refraction n = sin  i / sin  r

Refraction Problem The index of refraction of glass is 1.5. A ray of light is incident on a glass pane at an angle of 60 0. Calculate the angle of refraction.

Refraction Problem n = sin  i / sin  r 1.5 = sin 60 0 / sin  r 1.5 = 0.866 / sin  r sin  r = 0.866 / 1.5 sin  r = 0.58  r = 35 0

Converging (convex) Lens

Converging Lens When the object is outside the focus, the image is real and inverted

Converging Lens When the object is inside the focus, the image is virtual, upright, and enlarged.

Diverging (concave) Lens

The image is ALWAYS reduced, upright, virtual

Lens Problem A 35 mm slide projector uses a converging lens with f = 20 cm. The screen is 3 m away. How far is the slide from the lens?

Lens Problem 1/d o + 1/ d i = 1/f 1/d o + 1/ 300 = 1/20 1/d o = 1/20 - 1/ 300 d o = 21. 4 cm m = - d i / d o m = - 300 / 21.4 m = -14 Image is real, inverted, magnified

That’s all folks!

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