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Why can the image ofthe building be seeninside the plane mirror ?
P.43
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Light ray from the objectenter the eye.(Directly or indirectly)
Object canbe seen
ObjectP.43
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Luminousobjects
Emit theirown light
P.43
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P.43
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Non-luminousobjects
Can be seen when lightis reflected fromluminous objects
P.43
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Light travels in straight line
Light rayLine along which lightenergy is transferred
P.43
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Lightbeam
Cone of light
Pencil of light
P.43
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Incidentray
Reflectedray
Refractedray
P.43
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P.43
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Incidentray
Reflectedray
Refractedray
Transparentobject
P.43
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Incidentray
Reflectedray
Refractedray
Opaqueobject
P.43
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All the lightis absorbed
BlackReflection
P.43
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All the lightis reflected
WhiteAbsorption
P.43
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P.43
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P.43
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P.43
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P.43
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P.43
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Ray box
P.43
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P.43
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P.43
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P.43
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P.43
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P.43
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P.43
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Experiment :Laws of reflection
0o
Normal
Angle ofreflection
Angle ofincidence
P.43
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P.43
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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
P.43
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P.43
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Plane mirror
Incidentray
Reflectedray
Normal
i r
P.43
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Laws ofreflection
Incident ray, reflected ray& normal lie on thesame plane
i = r
P.43
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30o
What is r ?
60o 60o
30o
P.43 Q1(a)
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55o
What is r ?
35o
35o
55o
P.43 Q1(b)
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45o
45o
45o
45o
45o
45o
P.44 Q2
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60o
50o40o
40o50o
70o
20o
P.44 Q2
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20o
40o
P.44 Q2
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30o65o
25o 65o
85o
5o
5o
P.44 Q2(b)
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45o
45o45o
45o
45o
45o 45o
P.44 Q2(c)
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70o
10o10o
P.44 Q5
20o
60o
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Regular reflection
P.45
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Rough surface
Light rays are scattered
P.45
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Diffuse reflection
P.45
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Lightrays
Copy the shapeof the object Image
P.45
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Real image
Image is formedon the screen
P.45
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P.45
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Virtualimage
Image is formedbehind the screen
P.45
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Realimage
Converge at theposition of the image
Light rays passthrough the real image
Real image can beformed on the screen
P.45
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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
P.45
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P.45 Q1
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P.45
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u v
Experiment :Locating an imageby ray tracing
What is the relationship between u & v ?P.45
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Two trianglesare congruent
u v
Same areas &same bases
Sameheights
P.45
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P.45
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Image formed by a plane mirror
Object distance u = image distance v
Same size
Upright
Virtual
P.45
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P.46 Q1
B
D
C
A
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P.46 Q2
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P.46 Q2
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1.4m
0.1m
1m
0.05 + 0.7 mP.46
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0.7m
0.7m
0.05m0.05m
Two pairs ofCongruenttriangles
The minimum sizeof the mirror is half of the heightof the lady
P.46
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1m1m
1.6m0.8m
Two trianglesare similar
The minimum length of the mirroris half the height of the girl
P.46
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P.46 Q1(a)
2 m/s 2 m/s
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P.46 Q1(b)
4 m/s2 m/s
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P.47 Q2(a)
2.3 m
(b) We can see the chart behind the boy.
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Left hand Right hand
Lateral inversion
P.49
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Left & righthand sidesare interchanged
P.49
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P.49
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What will be seen in the rear mirror?
P.49
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P.49
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90o
3 images
I1 & I2 arelaterally inverted
I12 or I21
are not
P.49
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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
P.49
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I212
M2
M1
60oI121 & I212
overlap5 imagesare formed
I12 & I21 are notlaterally inverted
P.49
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P.49
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P.49
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Infinite numberof images
P.49
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M2 M1
5cm2cm
P.49
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Dressing
P.49
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Decoration
P.49
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Space seems to be larger
P.49
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The image of the bearcan be seen indirectly
P.50
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P.50
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Bus driver cansee passengerson the top deck
P.50
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P.50
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P.50
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P.50
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P.50
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I2
I1
O
M2
M110cm
20cm
P.50