refraction oflight

8/11/2019 Refraction ofLight

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Higher Physics Unit 3

3.2 Refraction of Light


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Refraction

What is Refraction?

Refraction is when the speed and direction of light changes, aslight moves from one material to another.


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a

m

a = angle of incidence in air (larger)m = angle of incidence in material

When light passes from air into a material the ratio is constant.m

a

sin

sin


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The absolute refractive index, n, of a medium is given by:

where 1 is the angle in a vacuum (air is used as an approximation) and

2 is the angle in the medium.

Material Refractive Index n

glass 1.5perspex 1.47

water 1.33

diamond 2.4

2

1

sin

sinn =

m

a

sin

sinn =

*** NOT ON DATA SHEET ***


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Example

A ray of light shines into a block of perspex. Calculate angle x

nsinsin

m

a =

1.47

xsin

20sin=

1.47

20sinxsin =

= 13.5x

200

x


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P&N Tutorial Booklets

Q3.22 3.26


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Outcome 3

Refractive Index of a Perspex Block

normala

pB

A

C

Place the block on white paper and tracearound its outline.

Draw in the normal at the midpoint B.

With incident angle a = 100, measure

the angle p, the refracted angle inthe perspex.

Repeat for other values of incident angle.

Use an appropriate format to determine the refractive index of the perspex block.


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Refractive Index

The refractive index can also be found using:

m

a

v

vn =

m

a

n =

speed of light in air (3x108 ms-1)

speed of light in material (ms-1)

wavelength of light in air (m, nm)

wavelength of light in material (m, nm)


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Example 1

Show that the refractive index of glass 1.50.

-18a ms103v =

-18m ms102v =

?n =

m

a

v

vn =

8

8

102

103

=

1.5n =


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Example 2

The refractive index of water is 1.33.Calculate the speed of light in the water.

-18a ms103v =

?vm =

1.33n =

m

a

vvn=

m

8

v

1031.33

=

1.33103v

8

m

=

18m ms102.26v

=


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Frequency

Light of wavelength 600nm in air is shone through glass ofrefractive index 1.5.

Calculate: a) speed of light in the glassb) wavelength of light in the glass

c) frequency of light in the air

d) frequency of light in the glass

2 x 108 ms-1

400nm

5 x 1014 Hz

5 x 1014 Hz


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Conclusion

The frequency of a wave is determined by its source and does notchange in different media.

fair = fmaterial

Velocity and wavelength change.FREQUENCY DOES NOT CHANGE

Summary

When light is refracted:


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Snells Law

Refractive index depends on the frequency of the incident light.

Refraction occurs because a wave travels at different speeds indifferent media. As the light changes speed, it changes direction.

The refractive index is equal to the ratio of the speeds, giving:

but as frequency is constant this cancels to:

2

1

2

1

2

1

f

f

v

v

sin

sinn ===

2

1

2

1

2

1

v

v

sin

sinn ===


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Q3.27 3.30


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Critical Angle

When a ray of light is shone into a semi-circular glass block, mostof the light is refracted however this is a partial reflection.

normal

small

incidentangle partial reflection

refracted light


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If the angle within the medium m is increased, a point is reached wherethe angle in air a is 90.

The angle in the medium that causes this is the CRITICAL ANGLE.

normal

largeincidentangle

partial reflection

refracted light


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On increasing the angle within the medium m beyond the criticalangle, all the light is reflected.

This is known as TOTAL INTERNAL REFLECTION.

normal

largerincidentangle

reflected light


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Refraction or Total Internal Reflection

Comparing the size of the incident angle (i) with the critical angle (c)determines whether refraction or total internal reflection occurs.

ci

reflectionpartialandrefraction

reflectioninternaltotal


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normalap

B

C

Critical Angle of a Perspex Block

Make measurements of various incident angles p and thecorresponding refracted angle a to determine the criticalangle c for the perspex block.


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Critical Angle Formula

When the angle in the mediumis equal to the critical angle,the angle in air is 900

So applying Snells Law:

normal

c

900

m

a

sin

sinn =

csin

90sinn =

csin

1n =

But because sin 90 = 1

or

n

1sin c =


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Example

The refractive index of glass is 1.5. Calculate the critical angle.

n

1sin c =

1.51sin c =

.....666.0sin c =

0c 8.41 =

( )0.666.....sin 1c

=


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Example 2

A swimming pool is illuminated by a lamp built into the bottom of thepool.

Three rays of light from the same point in the lamp are incident on thewater-air boundary with angles of incidence of 30, 40 and 50 asshown.

The refractive index of the water in the pool is 1.33.


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(a) Draw a diagram to show clearly what happens to each ray at theboundary.

Indicate on your diagram the sizes of appropriate angles.

All necessary calculations must be shown.

Critical Angle

n

1sin c =

1.33

1sin c =

( )0.752sin 1c

=

= 48.8c

ci reflectioninternaltotal


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42 59


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(b) An observer stands at the side of the pool and looks into thewater.

Explain, with the aid of a diagram, why the image of the lampappears to be at a shallower depth than the bottom of the pool.

To an observer the light appears to originate from where thebroken rays cross.

This produces a virtual image of the lamp at the apparentlyshallower depth.


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Refraction & Total Internal Reflection in Nature


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Q3.31 3.36

refraction oflight

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