Geometrical Optics

• Reflection • Refraction • Critical angle • Total internal reflection

Wave nature of light Interference Diffraction Polarisation

Lecture 3.3

Geometrical Optics- subset of optics concerning interaction of light with macroscopic material

Geometrical Optics Optics—Branch of Physics, concerning the interaction of light with matter

Dimension larger than a human hair ≈ 50µm

Light can travel through •empty space, • air, •glass, • water, •cornea, •eye lens etc.

Light rays will travel in a straight line if they remain in the same medium

Geometrical Optics ray optics beam of light

Each referred to as a medium

Laws of reflection

3. Incident and reflected rays and normal all lie in the same plane

2. Angles measured with reference to the normal to the surface

Specular reflection

Normal Incident Ray Reflected Ray

θi θr

Metal surface

Smooth surface: reflection at a definite angle --Specular reflection

1. angle of incidence(θi) = angle of reflection(θr)

At the boundary between two media, the light ray can change direction by reflection or refraction

Rough Surface

No unique angle of reflection for all rays Light reflected in all directions

Diffuse reflection

Diffuse reflection

Majority of objects (clothing, plants, people) are visible because they reflect light in a diffuse manner.

Refraction At the surface of a transparent media, glass, water, etc both reflection and refraction occur.

Normal Incident Ray

Reflected Ray

Refracted Ray

Medium 1 (Air)

Medium 2 (glass)

θ1

θ2

θ1

Light ray changes direction going from one medium to another. Which way does it bend and by how much? Is θ2<θ1

or is θ2>θ1

Refraction (deflection from a straight path in passing obliquely from one medium ( such as air) into another (such as glass)

Answer Depends on the speed of light in both media

Refraction

Refraction

Smooth concrete

grass

Analogy: Rolling barrel

speed of light in the material = v

Index of refraction (n) of the medium cnv

=

The amount by which a medium reduces the speed of light is characterised by

Vacuum 1(by definition)

Air 1.0003 Glass 1.52 Water 1.33 Diamond 2.42

Indices of Refraction

Index of refraction

Example Calculate the speed of light in diamond

Speed of light in a vacuum: c = 3x108 ms-1

v =c/n =(3x108 ms-1)/2.42 = 1.24 x108ms-1

Refraction Example

How long does it take light to travel 394cm in glass of refractive index 1.52

8 18 13 10 1.97 10

1.52msv ms

−−×

= = ×

88 1

3.94 2 101.97 10

mt sms

−−= = ×

×

dtv

=

Calculate the speed of light in glass

cvn

=

Normal

θ1

θ2

Incident Ray

Medium 1

Medium 2

θ1

θ2

n2 > n1 n2 < n1

1 1

2 2

Sin vSin v

θθ

=

1 1

2 2

//

Sin c nSin c n

θθ

= 1 2

2 1

Sin nSin n

θθ

=

1 1 2 2n Sin n Sinθ θ=

where v1 and v2 are the speeds of light in media 1 and 2 respectively

Monochromatic light (one colour or frequency)

Refraction

Normal Incident Ray

Law of refraction or Snell’s law (can be derived from Maxwell’s equations)

Normal Incident Ray

Medium 1

Medium 2

θ1

θ2

n2 > n1

n2 < n1

1 2

2 1

Sin nSin n

θθ

=1 1 2 2n Sin n Sinθ θ=

Monochromatic light (one colour or frequency)

n2 > n1

or

Refraction

Incident Ray

Normal incidence θ1 = 0 therefore θ2 = 0. transmitted ray is not deviated independent of the materials on either side of the interface.

Refraction

Incident and refracted rays and the normal are all in the same plane

1 1 2 2n Sin n Sinθ θ=Law of refraction or Snell’s law

Index of refraction changes for different wavelength. This is called dispersion.

Example A laser beam is directed upwards from below the surface of a lake at an angle of 35º to the vertical. Determine the angle at which the light emerges into the air. n1(air) =1.0003 and n2 (water) =1.33

1 1 2 2n Sin n Sinθ θ=0

11.0003 1.33 35Sin Sinθ =

Normal

air

water

θ1 n1

n2

Snell’s law

0

1

10

1

1.33 351.0003

0.76

49.7

SinSin

Sin

θ

θ

θ

=

=

=

35º

If light enters the water at an angle of 49.70, what is its refraction angle in the water?

End of ruler

Apparent position of ruler end

air

water

Refraction Real and apparent depth

Ruler partially immersed in water

ruler

2n

1n

d= real depth d’= apparent depth

1 1

2

nd dn

=

2n

1n

d d’

For small angles:

Refraction Real and apparent depth

𝐭𝐭𝐭 (𝜷) =𝑳𝒅

𝐭𝐭𝐭 (𝜶) =𝑳𝒅𝒅

𝐭𝐭𝐭 (𝜶) ≈ 𝒔𝒔𝒔(𝜶)

𝒔𝟏𝒔𝟐

=𝒔𝒔𝒔(𝜷) 𝒔𝒔𝒔(𝜶)

𝒔𝟏𝒔𝟐

=𝒔𝒔𝒔(𝜷) 𝒔𝒔𝒔(𝜶) ≈

𝒕𝒕𝒔 𝜷 𝒕𝒕𝒔 𝜶 =

𝒅𝒅𝒅

Snell’s law:

Geometry:

n1

n2

Setting sun appears flattened (top to bottom) because light from lower part of the sun undergoes greater refraction upon passing through denser air (higher refractive index) in lower part of the Earth’s atmosphere.

Refraction

2

2 1

cSin nSin n

θθ

=

Angle of incident for which refracted ray emerges tangent to the surface is called the critical angle

Refraction Critical Angle

as θ1 is increased θ2 increases

θ1

θ2

n2 < n1

θ2 =900

θc

θc is critical angle

1

2

in this case θ2 = 90o or Sin θ2 =1

2

1c

nSinn

θ =

2

2 1

cc

Sin n SinSin n

θ θθ

= =

1 cθ θ>incident ray undergoes total internal reflection at boundary and cannot pass into the material with the lower refractive index

Refraction Total internal reflection

θ1

θ2

n2 < n1

θ2 =900

θc >θc

θc is critical angle

1

2

Ray undergoes total internal reflection

maximum value of the sine of any angle is one

total internal reflection occurs at interface when n2 <n1

when

Example

1 1 02

1

1.0003sin sin 491.33c

nn

θ − −= = =

Determine the critical angle for water and diamond with respect to air.

1 1 02

1

1.0003sin sin 24.42.42c

nn

θ − −= = =

water

diamond

Refraction

Diamond has large refractive index and consequently small critical angle Light enters from any direction (no TIR on entering) Large number of facets: TIR from facets on back surface: exits from many front facets all of which receive some light at angles <24.40

Diamond Ring

45º

1 1 02

1

1.33sin sin 611.52c

nn

θ − −= = =

Refractive index of glass =1.52 Refractive index of air =1.0003

Total internal reflection at glass air interface if incident angle is >410

Glass prism (right angled isosceles triangle)

What happens the beam if the prism is immersed in water? Refractive index of water =1.33

1 1 02

1

1.0003sin sin 411.52c

nn

θ − −= = =

Example

θc > 45º

Total internal reflection at glass-water interface does not occur

What happens to light ray at the glass-air interface in prism as shown.

Critical angle given by

45º

Refraction

Rainbow formation is due to a combination of refraction and reflection. Incoming white light (broad spectrum of wavelength) is separated into its component colours.

Note: Colour separation due to dispersion (refractive index is different for different wavelengths).

Applications

Fibre optic cables used for telecommunications and for diagnostic tools in medicine

Refraction Total internal reflection

Optical fibre (end on)

Refractive index of core greater than refractive index of clading

Light coupled into core will travel extremely long distances along fibre, undergoing total internal reflection at core-cladding interface and exit only at the other end.

diameter of core 8µm

350

θ4

glass

θ2

θ3

air

air

Light in air is incident on a glass block at an angle of 350 The sides of the glass block are parallel. At what angle does the light emerge into the air from the lower surface of the glass block?

glass block has parallel sides, therefore

θ2 θ3 =

01 2 235n Sin n Sinθ=

01 1 4

04

35

35

n Sin n Sinθ

θ

=

=

Example

Let n1 = refractive index of air & n2 = refractive index of glass

Using Snell’s Law

2 3 1 4n Sin n Sinθ θ=

2 2 2 3n Sin n Sinθ θ=2 3θ θ=

θ1

glass

θ1−θ2

θ2

air

air

Light in air is incident on a glass block at an angle of θ1 and is refracted at an angle θ2.The sides of the block are parallel and a distance T apart. What is the displacement between the entry and exit rays in term of T, θ1 and θ2?

Example

d

1 2sin( ) dl

θ θ− =

l 1 2sin( )d l θ θ= −T

2cosTl

θ=

1 22

sin( )cos

Td θ θθ

= −1 2 1 2

2

(sin cos cos sin )cos

Td θ θ θ θθ

= −

1 1 2(sin cos tan )d T θ θ θ= −

12 1

2

nSin Sinn

θ θ=

2cos Tl

θ =

1 012

20

2

sin ( 35 )

22

n Sinn

θ

θ

−=

=0 0 05 (sin 35 cos35 tan 22 )

1.2d cmd cm

= −=

θ1

T = 5 cm, n =1.52, θ1 = 350

A HeNe laser has a wavelength of 633 nm in air (assume n=1) and 474 nm in the aqueous humor inside an eyeball. Calculate the index of refraction of the aqueous humor and the speed and frequency of the light in the substance.

0

0

633633 1.34474

nnmnnm

λλλ

=

= = =

8 18 13 10 2.25 10

1.34c x msv x msn

−−= = =

8 114

9

2.25 10 4.75 10474 10

v x msf x Hzx mλ

−

−= = =

8 114

0 90

3.00 10 4.75 10633 10

c x msf x Hzx mλ

−

−= = =

Refractive index

Speed in aqueous humor

Frequency of the light inaqueous humor

Frequency of the light in air

Example

0fcnv f

λλ

= =

Infrared Ultra violet Wavelength

Visible spectrum

v f λ=Electromagnetic wave

Transverse wave V: velocity f: frequency L: wavelength

Light: Electromagnetic wave

Electromagnetic wave

Electromagnetic Waves

Geometrical Optics Light represented by rays Traveling in straight lines

Not strictly correct

Diffraction Light waves deviate from straight path and “spread out” as they pass by obstacle or through an opening.

Width ≤ λ

Width > λ

Light has a wave nature

All waves subject to diffraction e.g. light, sound, water etc.

Waves

Electromagnetic Waves Wave nature of light

First proof---Thomas Young 1801

Superimposed 2 light beams and saw constructive and destructive interference

Beams obtained by passing sunlight through two closely spaced narrow slits

Interference pattern (bright & dark regions

Slit widths ≈ λ r1

r2

r2 = r1+ nλ constructive interference (bright) r2 = r1+ (n+½)λ destructive interference (dark)

where n is an integer

laser

Destructive interference

Electromagnetic Waves Diffraction and Resolution

Microscopes, telescopes, cameras, eyes Circular apertures (diameter d)

For circular aperture Minimum angle of resolution min

1.22d

λθ =

s1

s2 s2

s1

θ θ

Rayleigh criterion

Electromagnetic Waves

For circular aperture Minimum angle of resolution min

1.22d

λθ =

Rayleigh criterion

Two point like sources viewed through A circular aperture of size d

FM (88-104MHz)

AM (525-1610KHz)

Radio Reception in mountainous area Diffraction

Electromagnetic Waves

X-ray diffraction

λ = 3 m

λ = 200 m

Longer wavelength waves diffracted around and between mountains -better reception

X-rays λ 0.1nm

Atomic spacing in crystalline solids

X-ray diffraction used to investigate internal structure of important biological molecules - example, proteins and DNA

Unpolarised light viewed along direction of propagation

Polarised Light

polarised light viewed along direction of propagation

Light source

Light beam Polariser

Light waves vertically polarised

Schematic representation

Polaroid filter Unpolarised

light Polarised light

Polarisation – orientation of transverse wave

Polarised Light Schematic representation

Vertically polarised light wave

Unpolarised Incident beam

Horizontal polariser

Vertical polariser

Unpolarised light

Polarised light

Polarising filter

Light can become polarised by •Reflection •refraction •scattering

Unpolarised incident light

Polarised reflected light

Polarised incident light

Polarised reflected light

Polarised incident light No reflected

light

Polarised Light

?

?

Polarised Light Applications 3D movies

2 slightly different images projected on screen

2 cameras, a short distance apart, photograph original scene

Each image linearly polarised in mutually perpendicular direction

3D glasses have perpendicular polarisation axis

Each eye sees a different image associated with different viewing angle from each camera

Brain perceives the compound image as having depth or three dimensions.

Polarisation of light : application

Demineralised enamel is polarisation sensitive

shading may be seen, indicating the early stages of caries at the tooth’s surface

Application to dentistry Early detection of caries

Demineralised enamel viewed directly with unpolarised light No information

Polarised light incident on the dental tissue

Visual, mechanical probing, x rays???