Transcript Heat Flow

Geometrical Optics and Mirrors
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Geometrical Optics
Reflection
Plane Mirrors
Spherical Mirrors
Focal Length
Ray Tracing
Example – real image
Example – inverted image
Sign Rule and Summary
Geometrical Optics
• Formation of Image
– Light reflected from object.
– Some paths enter eye.
• Geometrical Optics
– Reflected light follows straight line paths (Rays)
– Following path of these rays defines image we see
Reflection
• Reflection
– Reflected paths manipulated by mirrors
– Paths focused to form images
• Reflection rules
– Angle of incidence = angle of reflection
– Specular vs diffuse reflection
Plane Mirror
• Scattered light follows path A-B-eye, but appears to
be coming from C-B-eye.
• Object and image distance
• Angle incidence = angle reflection
• Virtual vs. real image
How Tall is the mirror?
• Person want to see entire self in mirror
– 1.5 m to eye level
– 1.6 m to top of head
• Ray coming off shoe, top of head
– Bottom edge of mirror 0.75 m from floor
– Top edge of mirror 1.55 m from floor
– 0.8 m high
Microwave Mirror – Shuttle Tile Fault Detection
• Test panel of 12 tiles on aluminum backplate.
• Interrogate by microwave horns at 45° angle, raster x-y direction.
• Preliminary test – need smaller horns.
www.msi-sensing.com
Spherical and Parabolic Mirrors
• Focusing light rays at a point
– Light originates from infinity, reflects to focal point
– Principal axis, focus, focal length
– Spherical aberration, circular vs. parabolic mirror
Spherical Mirror for small angles
• Examine ray coming from infinity
– Radius from C normal to mirror
– All angles θ equal
– BCF is isosceles triangle (BF = CF)
– For small θ BF=AF and 𝒇 =
𝒓
𝟐
Image formation – Ray diagrams
• Tracing principle rays
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Place arrow principal axis
Tail reflects on itself
Trace principal rays of tip
Everything else falls between
• Principal rays
1.
2.
3.
4.
Comes parallel to axis, reflects
through focal point.
Goes through focal point, emerges
parallel to axis.
Hits mirror at normal incidence,
reflects back on itself.
All converge at I’
Mirror equation
• Two similar triangles
ℎ𝑜
ℎ𝑖
=
𝑑𝑜
𝑑𝑖
ℎ𝑜
ℎ𝑖
=
𝑂𝐹
𝐹𝐴
1
𝑑𝑜
+
1
𝑑𝑖
=
𝑑𝑜 −𝑓
𝑓
• Combining
𝑑𝑜
𝑑𝑖
=
𝑑𝑜 −𝑓
𝑓
=
1
𝑓
𝑚=
ℎ𝑖
ℎ𝑜
=−
𝑑𝑖
𝑑𝑜
Two sets of similar triangles
Example – Concave Mirror
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A 1.5 cm high diamond ring is placed 20 cm from a concave mirror with radius of
curvature 30 cm. Determine
– (a) the position of the image and di = 60 cm
– (b) its size m = -60/20 = -3
– hi = -3 ho = -4.5 cm
– Inverted
– “Real” image
1
𝑑𝑜
1
1
+𝑑 =𝑓
𝑖
𝑑
𝑚 = − 𝑑𝑖
𝑜
Example – Object closer than focal length
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A 1.5 cm high object is place 10 cm from a concave mirror whose radius of
curvature is 30 cm. (a) Draw a principal ray diagram and the position of the image
(b) determine the position of the image and (c) the magnification
– di = -30 cm
– m = --30/20 = +3
– hi = +3 ho = +4.5 cm
– Upright, virtual
1
𝑑𝑜
+
1
𝑑𝑖
=
1
𝑓
𝑚=−
𝑑𝑖
𝑑𝑜
Summary
Convex rear-view mirror
An external rearview car mirror is convex with a radius of curvature 16 m. Determine the
location of the image and its magnification for an object 10 m from the mirror
f = -8 m
di= -4.4m
m = - di/do = -- 4.4/10 = +0.44
1
1 1
+ =
𝑑𝑜 𝑑𝑖 𝑓
𝑚=−
𝑑𝑖
𝑑𝑜