Why can the image of the building be seen inside the plane mirror ?

Why can the image ofthe building be seeninside the plane mirror ?

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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

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55o

What is r ?

35o

35o

55o

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45o

45o

45o

45o

45o

45o

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60o

50o40o

40o50o

70o

20o

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20o

40o

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30o65o

25o 65o

85o

5o

5o

P.44 Q2(b)

45o

45o45o

45o

45o

45o 45o

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70o

10o10o

P.44 Q5

20o

60o

Smooth surface

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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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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

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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

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4 m/s2 m/s

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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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