Final Exam information

Transcription

1 Final Exam information Wednesday, June 6, 2012, 9:30 am - 11:18 am Location: in recitation room Comprehensive (covers all course material) 35 multiple-choice questions --> 175 points Closed book and notes Make up your own equation sheet (same rules as midterm)

2 Chapter 25 The Reflection of Light: Mirrors

3 25.5 The Formation of Images by Spherical Mirrors IMAGING WITH CONCAVE MIRRORS To find the image of an object placed in front of a concave mirror, there are several types of rays which are particularly useful è ray tracing: This ray is initially parallel to the principal axis and passes through the focal point. This ray initially passes through the focal point, then emerges parallel to the principal axis. This ray travels along a line that passes through the center and so reflects back on itself.

4 25.5 The Formation of Images by Spherical Mirrors If the object is placed between F and C, the image is real, inverted and magnified. If the object is placed at a distance greater than C from the mirror, the image is real, inverted and reduced in size. A real image is one where light is actually passing through the image (it can be projected onto a screen). The principle of reversibility - If the direction of a light ray is reversed, the light retraces its original path.

5 25.5 The Formation of Images by Spherical Mirrors When an object is placed between the focal point F and a concave mirror, The image is virtual, upright, and magnified (as in the case of images from flat mirrors, a virtual image is one from which light appears to be emanating but through which light does not pass, e.g. it cannot be projected onto a screen).

6 25.5 The Formation of Images by Spherical Mirrors IMAGING WITH CONVEX MIRRORS For convex mirrors the image of an object is always virtual, upright, and reduced in size. Ray 1 is initially parallel to the principal axis and appears to originate from the focal point. Ray 2 heads towards the focal point, emerging parallel to the principal axis. Ray 3 travels toward the center of curvature and reflects back on itself.

7 25.6 The Mirror Equation and Magnification So far we have discussed concave and convex mirrors qualitatively and graphically. We now want to derive two simple equations which provide quantitative relationships among the quantities we have defined to describe mirrors, i.e., f focal length d o d i object distance image distance m magnification

8 25.6 The Mirror Equation and Magnification Consider the real image produced from a concave mirror. The two right triangles are similar in each case. h o /(-h i ) d o /d i minus since inverted m h h i o d d i o Magnification equation h o /(-h i ) (d o - f )/f 1 d o + 1 d i 1 f Mirror equation Both equations are valid for concave and convex mirrors.

9 25.6 The Mirror Equation and Magnification Summary of Sign Conventions for Spherical Mirrors f f is + for a is for a concave mirror. convex mirror. d o d o is + if the object is in front of the mirror. is if the object is behind the mirror. d i is + if the image object is in front of the mirror (real image). d i is if the image object is behind the mirror (virtualimage). m is + for an image upright with respect to the object. m is - for an image inverted with respect to the object.

10 Example. A Real Image formed by a Concave Mirror. A 2.0 cm high object is placed 7.10 cm from a concave mirror whose radius of curvature is cm. Find the location of the image and its size. Since a concave mirror è f +R/ / cm 1/d i 1/f - 1/d o 1/5.10-1/ cm -1 d i 18 cm è real image since positive h i -(d i /d o )h o -(18/7.10)(2.0) -5.1 cm è magnified and inverted

11 25.6 The Mirror Equation and Magnification Example. A Virtual Image Formed by a Convex Mirror A convex mirror is used to reflect light from an object placed 66 cm in front of the mirror. The focal length of the mirror is 46 cm in back of the mirror. Find the location of the image and the magnification. Since a convex mirror, the focal length is negative è f -46 cm d f i do i 46 cm 66 cm cm d i 27 cm è image is virtual since negative m d d i o ( 27 cm) 66 cm 0.41 è image is upright and reduced

12 Chapter 26 The Refraction of Light: Lenses and Optical Instruments

13 26.1 The Index of Refraction Light travels through a vacuum at a speed c m s Light travels through materials at a speed less than its speed in a vacuum. DEFINITION OF THE INDEX OF REFRACTION The index of refraction of a material is the ratio of the speed of light in a vacuum to the speed of light in the material: Speed of light in vacuum n Speed of light in the material c v

14 26.1 The Index of Refraction

15 26.2 Snell s Law and the Refraction of Light SNELL S LAW -- When light strikes an interface between two materials it breaks up into two pieces - one reflected and one refracted (transmitted). SNELL S LAW OF REFRACTION When light travels from a material with one index of refraction to a material with a different index of refraction, the angle of incidence is related to the angle of refraction by n 1 sinθ1 n2 sin θ 2

16 26.2 Snell s Law and the Refraction of Light Example 1 Determining the Angle of Refraction A light ray strikes an air/water surface at an angle of 46 degrees with respect to the normal. Find the angle of refraction when the direction of the ray is (a) from air to water and (b) from water to air.

17 26.2 Snell s Law and the Refraction of Light (a) sinθ n sinθ1 n ( 1.00) sin θ 33 2 (b) sinθ n sinθ1 n ( 1.33) sin θ 2 74

18 26.2 Snell s Law and the Refraction of Light APPARENT DEPTH Example 2 Finding a Sunken Chest The searchlight on a yacht is being used to illuminate a sunken chest. At what angle of incidence should the light be aimed?

19 26.2 Snell s Law and the Refraction of Light First find θ 2 from the geometry and then use Snell s Law to find θ 1 : ( ) 31 θ tan 1 2 sinθ n sinθ n ( 1.33) sin θ 1 44

20 26.2 Snell s Law and the Refraction of Light Because light from the chest is refracted away from the normal when the light enters the air, the apparent depth of the image is less than the actual depth. Simpler case -- look directly above the object. Apparent depth, observer directly above object & n d' d $ % n n 1 -- medium of object n 2 -- medium of observer 2 1 #! "

21 26.2 Snell s Law and the Refraction of Light Example. On the Inside Looking Out A swimmer is under water and looking up at the surface. Someone holds a coin in the air, directly above the swimmer s eyes at a distance of 50 cm above the water. Find the apparent height of the coin as seen by the swimmer (assume n 1.33 for water). Use the equation d' & d $ % n n 2 1 #! " In this case, d will be the apparent height of the coin, d is the actual height above the water, n for air (object), and n for water (the observer), d (50)(1.33/1.00) 66.5 cm è greater than the actual height

22 26.2 Snell s Law and the Refraction of Light THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL When a ray of light passes through a pane of glass that has parallel surfaces and is surrounded by air, the emergent ray is parallel to the incident ray, θ 3 θ 1, but is displaced from it. 1 st interface: n 1 sin θ 1 n 2 sin θ 2 2 nd interface: n 2 sin θ 2 n 1 sin θ 3 è n 1 sin θ 1 n 1 sin θ 3 è θ 3 θ 1