Download F1 The ray approximation in optics assumes that light travels from

F1
The ray approximation in optics assumes that light travels from one point to another along a narrow path called a
ray that may be represented by a directed line (i.e. a line with an arrow on it). In a uniform medium
(where the refractive index is the same everywhere) the rays are straight lines, though they may be kinked at
mirror surfaces or upon passing through the interface between two different media and they may bend
continuously in non-uniform media where the refractive index changes smoothly with position.
The ray approximation gives an adequate description of light transmission provided diffraction effects can be
ignored. This happens when the transverse dimensions of the optical systems or obstacles through or around
which the light is passing are large compared with the wavelength of light.
FLAP P6.2
Rays and geometrical optics
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S570 V1.1
F2
A ray from the object to the eye is
constructed by drawing the incident and
reflected parts of the ray in the same plane
as the normal at the point of incidence,
and by making the angles of incidence
and reflection equal (see Figure 3).
From the geometry of the (similar)
triangles shown in Figure 24, we can see
that the virtual image will be 201cm
behind the mirror and located along the
extended normal from the object to the
mirror.
P
normal
P'
O
θi
θi θR
A
θi θR
mirror
mirror
Figure 34A ray of light reflected
from a plane mirror.
N
A
I'
Figure 244See Answer T2.
FLAP P6.2
Rays and geometrical optics
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S570 V1.1
F3
(a) We use Snell’s law and assume the refractive index of air is 1.00:
sin θ1 1. 00
=
sin θ 2 1.50
therefore
sin θ1 = 0.667sin θ 2
The maximum value of sin1θ2 is 1.00 and at this limiting value
θ1c = arcsin (0.667) = 41.8°
This is the critical angle. When θ 1c > 41.8° there can be no refracted ray and total internal reflection occurs.
(b) Light rays are confined in optical fibres by total internal reflection. These devices are used extensively in
modern optical communication systems. See also Subsections 4.3 and 4.4.
FLAP P6.2
Rays and geometrical optics
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S570 V1.1
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route
through the module and to proceed directly to Ready to study? in Subsection 1.3.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the
Closing items.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it
here.
FLAP P6.2
Rays and geometrical optics
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S570 V1.1
R1
If sin1θ = 0.45, then θ = arcsin1(0.45). To three significant figures this gives: θ = 26.7°.
Consult trigonometry in the Glossary for further information.
FLAP P6.2
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R2
The relation between wavelength λ, speed c and frequency f of a wave is c = f0λ.
The values for blue and red light are as follows:
blue light
λ = 4501nm, therefore f = 6.67 × 10141Hz
red light
λ = 6501nm, therefore f = 4.62 × 10141Hz
Consult the relevant terms in the Glossary for further information.
FLAP P6.2
Rays and geometrical optics
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S570 V1.1
R3
The rise and fall of a wave creates a vibration or oscillation at any point through which the wave passes.
The term phase is generally used to describe the stage that an oscillator has reached in its repetitive oscillatory
cycle, and is often expressed as a fraction of 2π since 2π radians represents one complete cycle. Two waves are
said to be in phase at a certain point if the oscillations they would individually cause at that point are exactly in
step, rising and falling together with equal phases at each moment. Under appropriate conditions, two waves
arriving simultaneously at a given point may cause a joint disturbance at that point which is equal to the sum of
the disturbances they would cause individually; under these circumstances the waves are said to interfere when
they meet.
Consult the relevant terms in the Glossary for further information if required, but few other ‘wave’ concepts are
needed in this module.
FLAP P6.2
Rays and geometrical optics
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S570 V1.1
T1
See Figure 23. At the first reflection, point A, the reflected angle θR1
is equal to the incident angle θi1. The mirrors are at right angles so
that at the second reflection, point B, θ i 2 = 90° − θ R 1 and
θR2 = θi2 = 90° − θR1. The angle of the ray relative to the horizontal
after the second reflection is given by:
α = 90° − θR2 = 90° − (90° − θR1) = θR1 = θi1.
The ray is finally reflected back parallel but the direction of travel is
reversed. This is the basic principle of operation for a bicycle reflector
and also for the reflectors placed on the Moon by the Apollo 11
astronauts (used to reflect laser beams sent from Earth and thereby to
measure the distance from the Earth to the Moon1—1the lunar ranging
experiment).
FLAP P6.2
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A
θi1
θR1
θi2 θ
R2
B
Figure 234See Answer T1.
α
mirror
T2
See Figure 24. Consider two rays from O1—1one along the normal to the
mirror and the other at any angle of incidence θ1i. The ray ON is reflected
back towards O so the
virtual image must lie somewhere along the normal. For the ray reflected at
the arbitrary point A, the angle of incidence equals the angle of reflection:
θ0i = θR. Hence the angle NI′A = θR. But the angle NOA = θ0i = θR so that the
triangles ONA and I′NA are identical. This means that side I′N is equal to
side ON for any value of θi. All rays reflected in the mirror from O appear to
diverge from I′ with I′N = ON, so the image appears to be at the same
distance from the mirror as the object.
O
θi
θi θR
mirror
N
A
I'
Figure 244See Answer T2.
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T3
The ray diagram for refraction through a
µ1
µ2
µ1
parallel-sided glass block is shown in Figure
25.
At A (the point where the light enters the
θ1B
B
block):
θ2B
µ1 sin θ 2 A
=
A
θ2A
µ 2 sin θ1A
θ1A
At B (the point where the light leaves the
block):
µ 2 sin θ1B
=
µ1 sin θ 2B
Multiplying these two equations together, and Figure 254See Answer T3.
noting that θ2A = θ2B gives us
sin θ 2 A sin θ1B
sin θ1B
1=
×
=
sin θ1A sin θ 2B sin θ1A
i.e. sin1θ1B = sin1θ1A and θ 1B = θ1A; the ray emerging from the block is parallel to the ray entering the block.
FLAP P6.2
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S570 V1.1
µ
T4
We use the data given in Figure 8 for the variation of refractive index of
a form of glass called ‘fused quartz’.
Snell’s law gives us
µ glass
sin θ air
=
sin θ glass
µ air
400
At λ = 4501nm, µglass = 1.465,
sin θ glass =
sin 40°
= 0. 4388
1. 465
1.46
1.45
and we put µair = 1.00. At λ = 6501nm, µglass = 1.46,
sin 40°
hence sin θ glass =
= 0. 4403 and θ glass = 26.12°
1. 46
hence
1.47
and
θ glass = 26. 02°
The difference in angles is 26.12° − 26.02° = 0.10°
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500
600
λ/nm
Figure 84The variation of the refractive
index with wavelength for ‘fused quartz’
glass. (The wavelength is measured in a
vacuum).
T5
Using Snell’s law for the glass–air interface:
µair1 sin1θair = µglass1sin1θglass
We can set µair = 1.00 and then find:
µglass = sin1(35°)/sin1(22°) = 0.574/0.375 = 1.53.
The refractive index of a material is defined as the ratio of the speed of light in a vacuum to the speed of light in
the material, i.e. µglass = c/vglass, so we have:
vglass = c/µglass = (3.00 × 108 /1.53)1m1s–1 = 1.96 × 108 1m1s−1
FLAP P6.2
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T6
The real and apparent depth due to refraction are shown in Figure 26. If we
take one ray normal to the interface along the line ON and another ray at
angle θ1 along OA. Snell’s law gives us:
sin θ1 µ 2
=
sin θ 2
µ1
But when θ1 and θ 2 are small:
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µ1
N
sin1θ1 ≈ tan1θ1 and sin1θ2 ≈ tan1θ2
tan θ1 µ 2
so that
=
tan θ 2
µ1
NA I ′N µ 2
therefore
=
NA ON µ1
ON µ 2
thus
=
as required in Equation 3
I ′N µ1
real depth
ON µ 2
=
=
apparent depth I ′N µ1
θ1
A
θ1
I′
θ2
θ2
O
(Eqn 3)
S570 V1.1
Figure 264See Answer T6.
µ2
From Equation 3
real depth
ON µ 2
=
=
apparent depth I ′N µ1
(Eqn 3)
we have:
µ
real depth
1.33
= water =
µ air
apparent depth
1. 00
i.e. real depth = (1.33 × 1.0)1m = 1.331m
FLAP P6.2
Rays and geometrical optics
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THE OPEN UNIVERSITY
S570 V1.1
T7
The critical angle is given by Equation 5:
µ 
θ c = arcsin  2 
 µ1 
(Eqn 5)
µ

1.33 
= 62.5°
θ c = arcsin  water  = arcsin 

1.50 
 µ glass 
At incident angles greater than 62.5° the ray is totally internally reflected. Replacing the water by air gives
θc = arcsin1(1.00/1.50) = 41.8°.
FLAP P6.2
Rays and geometrical optics
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S570 V1.1
T8
Since it is continually entering layers of lower and lower refractive index it must be refracted smoothly and
progressively further and further from the normal to the Earth’s surface. The ray will follow a smooth curve and
eventually will be bent back towards the surface, following a symmetric path and arriving at the surface some
distance from its departure point. The atmosphere here acts like a mirror.
FLAP P6.2
Rays and geometrical optics
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THE OPEN UNIVERSITY
S570 V1.1
T9
Since the light ray is travelling along the direction of the changing refractive index it is always travelling along
the normal, with an angle of incidence equal to 90°, and so is not refracted at all1—1although its speed through
the medium will change with the refractive index.
FLAP P6.2
Rays and geometrical optics
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THE OPEN UNIVERSITY
S570 V1.1
T10
See Figure 20. If the angles of incidence and reflection are to be equal
then, with NA = x0:
x d−x
=
a
b
i.e. xb = ad − ax
Gathering terms involving x on the left-hand side,
so
P′
P
θi θR
a
b
x
N
d
A
N′
mirror
Figure 204A conceivable route for light
rays passing from P to P′ via a reflection
at A; but is it the route the rays will
actually follow?
x(a + b) = ad
ad
x=
a+b
Putting in the values for a, b and d we find
10 cm × 20 cm
x=
= 11.8 cm
(10 + 7) cm
A careful drawing, using this value of x, gives a total true path length of 26.31mm. Nearby paths, with
NA = 9.81cm or 13.81cm, are about 1.331mm longer, confirming that the true path is a minimum path.
FLAP P6.2
Rays and geometrical optics
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1