Final Exam information - Physicshumanic/p112_lecture17.pdf · Final Exam information • Wednesday, June 6, ... is + if the object is in front of the mirror (real image). d i

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)

Chapter 25

The Reflection of Light: Mirrors

25.5 The Formation of Images by Spherical Mirrors

IMAGING WITH CONCAVE MIRRORS

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.

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:


The principle of reversibility - If the direction of a light ray is reversed, the light retraces its original path.

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).


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).


IMAGING WITH CONVEX MIRRORS

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.

For convex mirrors the image of an object is always virtual, upright, and reduced in size.

25.6 The Mirror Equation and Magnification

length focal=f

distanceobject =od

distance image=id

ionmagnificat=m

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.,


The two right triangles are similar in each case.

fdd io

111=+

o

i

o

i

dd

hh

m −==

Both equations are valid for concave and convex mirrors.

Mirror equation

ho/(-hi) = do/di minus since inverted

ho/(-hi) = (do - f )/f

Magnification equation

Consider the real image produced from a concave mirror.


Summary of Sign Conventions for Spherical Mirrors

mirror. concave afor is +f

mirror.convex afor is −f

mirror. theoffront in isobject theif is +od

mirror. thebehind isobject theif is −od

image). (realmirror theoffront in isobject theif is +id

image). (virtualmirror thebehind isobject theif is −id

m is + for an image upright with respect to the object. m is - for an image inverted with respect to the object.

image

image

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 10.20 cm. Find the location of the image and its size.

Since a concave mirror è f = +R/2 = 10.20/2 = 5.10 cm 1/di = 1/f - 1/do = 1/5.10 - 1/7.10 = 0.055 cm-1 di = 18 cm è real image since positive hi = -(di/do)ho = -(18/7.10)(2.0) = -5.1 cm è magnified and inverted


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.

1cm 037.0cm 661

cm 461111 −−=−

−=−=

ii dfd

cm 27−=id

( ) 41.0cm 66cm 27

=−

−=−=o

i

dd

m

Since a convex mirror, the focal length is negative è f = -46 cm

è image is virtual since negative

è image is upright and reduced

o

Chapter 26

The Refraction of Light: Lenses and

Optical Instruments

26.1 The Index of Refraction

sm1000.3 8×=cLight travels through a vacuum at a speed

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:

vcn ==

material in thelight of Speedin vacuumlight of Speed

26.1 The Index of Refraction

26.2 Snell’s Law and the Refraction of Light

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

2211 sinsin θθ nn =

SNELL’S LAW -- When light strikes an interface between two materials it breaks up into two pieces - one reflected and one refracted (transmitted).


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.


( ) 54.033.146sin00.1sinsin

2

112 ===

nn θ

θ(a)

(b)

332 =θ

( ) 96.000.146sin33.1sinsin

2

112 ===

nn θ

θ

742 =θ


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?


( ) 69.000.131sin33.1sinsin

1

221 ===

nn θ

θ

441 =θ

( ) 313.30.2tan 12 == −θ

First find θ2 from the geometry and then use Snell’s Law to find θ1:


!!"

#$$%

&='

1

2

nnddApparent depth,

observer directly above object

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.

n1 -- medium of object n2 -- medium of observer


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 !!"

#$$%

&='

1

2

nndd

In this case, d’ will be the apparent height of the coin, d is the actual height above the water, n1 = 1.00 for air (object), and n2 = 1.33 for water (the observer), d’ = (50)(1.33/1.00) = 66.5 cm è greater than the actual height


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, θ3 = θ1, but is displaced from it.

1st interface: n1 sin θ1 = n2 sin θ2 2nd interface: n2 sin θ2 = n1 sin θ3 è n1 sin θ1 = n1 sin θ3 è θ3 = θ1

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Final Exam information - Physicshumanic/p112_lecture17.pdf · Final Exam information • Wednesday, June 6, ... is + if the object is in front of the mirror (real image). d i