why can the image of the building be seen inside the plane mirror ? p.43
TRANSCRIPT
Why can the image ofthe building be seeninside the plane mirror ?
P.43
Light ray from the objectenter the eye.(Directly or indirectly)
Object canbe seen
ObjectP.43
Luminousobjects
Emit theirown light
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Non-luminousobjects
Can be seen when lightis reflected fromluminous objects
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Light travels in straight line
Light rayLine along which lightenergy is transferred
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Lightbeam
Cone of light
Pencil of light
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Incidentray
Reflectedray
Refractedray
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Incidentray
Reflectedray
Refractedray
Transparentobject
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Incidentray
Reflectedray
Refractedray
Opaqueobject
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All the lightis absorbed
BlackReflection
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All the lightis reflected
WhiteAbsorption
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Ray box
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Experiment :Laws of reflection
0o
Normal
Angle ofreflection
Angle ofincidence
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What is the relation between i & r ?
i r
0o
15o
30o
45o
60o
75o
i r
0o 0o
15o 15o
30o 30o
45o 45o
60o 60o
75o 75o
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Plane mirror
Incidentray
Reflectedray
Normal
i r
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Laws ofreflection
Incident ray, reflected ray& normal lie on thesame plane
i = r
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30o
What is r ?
60o 60o
30o
P.43 Q1(a)
55o
What is r ?
35o
35o
55o
P.43 Q1(b)
45o
45o
45o
45o
45o
45o
P.44 Q2
60o
50o40o
40o50o
70o
20o
P.44 Q2
20o
40o
P.44 Q2
30o65o
25o 65o
85o
5o
5o
P.44 Q2(b)
45o
45o45o
45o
45o
45o 45o
P.44 Q2(c)
70o
10o10o
P.44 Q5
20o
60o
Regular reflection
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Rough surface
Light rays are scattered
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Diffuse reflection
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Lightrays
Copy the shapeof the object Image
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Real image
Image is formedon the screen
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Virtualimage
Image is formedbehind the screen
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Realimage
Converge at theposition of the image
Light rays passthrough the real image
Real image can beformed on the screen
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Virtualimage
Diverge from theposition of the image
Light rays do not passthrough the real image
Virtual image cannotbe formed on the screen
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P.45 Q1
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u v
Experiment :Locating an imageby ray tracing
What is the relationship between u & v ?P.45
Two trianglesare congruent
u v
Same areas &same bases
Sameheights
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Image formed by a plane mirror
Object distance u = image distance v
Same size
Upright
Virtual
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P.46 Q1
B
D
C
A
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1.4m
0.1m
1m
0.05 + 0.7 mP.46
0.7m
0.7m
0.05m0.05m
Two pairs ofCongruenttriangles
The minimum sizeof the mirror is half of the heightof the lady
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1m1m
1.6m0.8m
Two trianglesare similar
The minimum length of the mirroris half the height of the girl
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P.46 Q1(a)
2 m/s 2 m/s
P.46 Q1(b)
4 m/s2 m/s
P.47 Q2(a)
2.3 m
(b) We can see the chart behind the boy.
Left hand Right hand
Lateral inversion
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Left & righthand sidesare interchanged
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What will be seen in the rear mirror?
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90o
3 images
I1 & I2 arelaterally inverted
I12 or I21
are not
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M2
M1
I2I12
I1 O
I21
I12 & I21
overlap
I2 is the imageof O on M2
I1 is the imageof O on M1
I12 is the imageof I1 on M2
I21 is the imageof I2 on M1
I12 is formed by 2 reflection of lighttherefore no lateral inversion
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I212
M2
M1
60oI121 & I212
overlap5 imagesare formed
I12 & I21 are notlaterally inverted
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Infinite numberof images
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M2 M1
5cm2cm
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Dressing
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Decoration
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Space seems to be larger
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The image of the bearcan be seen indirectly
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Bus driver cansee passengerson the top deck
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I2
I1
O
M2
M110cm
20cm
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