Chapter 26 The Refraction of Light: Lenses and Optical Instruments 1
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Chapter 26
The Refraction of Light: Lenses and
Optical Instruments
1
26.1 The Index of Refraction
sm1000.3 8×=cLight travels through a vacuum at a speed
Light travels through materials at a speed less than its speed in a vacuum.
DEFINITION OF THE INDEX OF REFRACTION n (or refractive index)
The index of refraction of a material is the ratio of the speed of light in a vacuum to the speed of light in the material:
vcn ==
material in thelight of Speedin vacuumlight of Speed
2
“Refraction” refers both to (1) the crossing of light from one medium to another medium with different refractive index, and (2) the change of direction that occurs on crossing the boundary
26.2 Snell’s Law and the Refraction of Light
SNELL’S LAW OF REFRACTION When light travels from a material with one index of refraction (n1) to a material with a different index of refraction (n2) , the angle of incidence (θ 1) is related to the angle of refraction (θ 2) by
2211 sinsin θθ nn =
3
Note light rays are reversible, so the assignment of 1,2 to incident and refracted rays , respectively, is arbitrary.
http://www.youtube.com/watch?v=stdi6XJX6gU
The angles are always measured from the normal to the interface
4
θ 1 (deg) θ 2 (deg) sinθ 2 sinθ 1
0 0 0.00 0.0015 10 0.17 0.2630 22 0.37 0.5045 30 0.50 0.7160 40 0.64 0.8775 45 0.71 0.97
y = 1.37x
0.00
0.20
0.40
0.60
0.80
1.00
0.00 0.20 0.40 0.60 0.80 1.00
sinθ1 vs. sinθ2
θ 1 (deg) θ 2 (deg) sinθ 2 sinθ 1
0 0 0.00 0.0015 12 0.21 0.2630 23 0.39 0.5045 33 0.54 0.7160 42 0.67 0.8775 45 0.71 0.97
y = 1.32x
0.00
0.20
0.40
0.60
0.80
1.00
0.00 0.20 0.40 0.60 0.80 1.00
sinθ1 vs. sinθ2
In-Class experimental verification of Snel’s Law
21
212211 sinsinsinsin θθθθ
nnnn =→=
Measuring θ2 for θ1 =0,15,30,45,60, and 75°, and plotting sinθ1 vs. sinθ2, we expect a straight line graph with slope of n2/n1 = n(water)/n(air) = 1.33
7:30 am Class
8:35 am Class
5
normal
normal
Refracted wave-fronts
Incident wave-fronts
Incident ray
Refracted ray
boundary
boundary
θ1
θ2
A close-up of one frame from ripple tank refraction video
Refraction is a wave phenomenon (one that we didn’t study in Ch. 16 and 17)
Medium 1: (lower right)
Speed v1 Frequency f Wavelength λ1 Incident angle θ1
Medium 2: (upper left)
Speed v2 Frequency f Wavelength λ2 Incident angle θ2
Note: the frequency is preserved in refraction
6
θ1
θ1
θ2
θ2 b
λ1
∆1
∆2
λ2
normal
normal
boundary
boundary Incident wave-fronts
Refracted wave-fronts b
11sin:1 λθ =∆
b: distance along the boundary between two successive wave-fronts (in both media)
The two triangles share a common hypotenuse b
2
1
2
1
sinsin
λλ
θθ
=→
fvvf 1111 =→= λλ
b2
2sin:2 λθ =∆
fv22 =λAnd also
2
1
2
1
sinsin
vv
=θθ
For light: 2211 / , / ncvncv ==
1
2
2
1
2
1
//
sinsin
nn
ncnc
==θθ
2211 sinsin θθ nn =
General Form of Snel’s Law
And zooming in:
Derivation of Snel’s Law
7
cm 2.51 =λMeasured:
cm 7.32 =λMeasured:
2
1
1
2
1
2
//
vv
vcvc
nn
==
405.1cm 7.3cm .25
2
1
1
2 ===λλ
nn
2
1
2
1
λλ
λλ
==ff
°= 421θMeasured:
12
12 sinsin θθ
nn
=
476.042sin405.11
=°=
Prediction:
°== − 4.28 476.0sin 12θ
°±= 1282θMeasured:
Video/Graphical Experiment
26.2 Snell’s Law and the Refraction of Light
APPARENT DEPTH Example : Finding a Sunken Chest
The searchlight on a yacht is being used to illuminate a sunken chest. At what angle of incidence should the light be aimed?
8
26.2 Snell’s Law and the Refraction of Light
APPARENT DEPTH
Example : Finding a Sunken Chest The searchlight on a yacht is being used to illuminate a sunken chest. At what angle of incidence should the light be aimed?
9
31)606.0(tanm 3.3m 0.2tan 11
2 ==
= −−θ
Desired angle of refraction:
( ) 685.000.1
31sin33.1sinsin1
221 ===
nn θθ
Solve for desired angle of incidence:
441 =θ
26.2 Snell’s Law and the Refraction of Light
=′
1
2
nndd
Apparent depth, observer directly above object
10
d’
d θ1
θ1
θ2
θ2
b Zooming in
'
tan , tan 21 db
db
== θθ2
1
tantan'
θθ
=→dd
2211 sinsin θθ nn =
Note in this case, water is medium 1, and the air is medium 2 1n
2n
2
1
1
2
sinsin
θθ
=→nn
Overhead observer small and 21 θθ
θθθθ ≈≈→ tansin small
2
1
1
2
2
1 and 'θθ
θθ
≈≈nn
dd
1
2 'nn
dd
≈→
For objects in water seen from air above:
ddd43
33.100.1
=≈′
26.3 Total Internal Reflection
(a) When light passes from a medium of larger refractive index into one of smaller refractive index (n1 > n2), the refracted ray bends away from the normal: . θ2 > θ1
Critical angle: ,90sinsin 21 °= nn cθ
11
Total Internal Reflection θ 1: incident angle θ 2: refracted angle
(b) As one increases θ1, eventually θ2 reaches its maximum: 90°. The incident angle that gives sinθ 2 is known as the critical angle : θ1=θC (for θ2=90°)
(c) for θ1> θC, the incident ray undergoes total internal reflection. There is no refracted ray at all.
→> 21 nn1
2sinnn
c =θ
Application of Total Internal Reflection: Optical fibers are coated with a cladding of higher refractive index than the fiber itself. The internal reflections keeps the light in the fiber even when the fiber is bent. Makes the end of fibers seem luminous
For glass to air
°== − 425.10.1sin 1
Cθ
1n
2n
26.4 Polarization and the Reflection and Refraction of Light
1
21tannn
B−=θ
Brewster’s law
12
When the incident angle is equal to
(b) The reflected ray is completely polarized in the horizontal plane.
(a) The reflected ray is perpendicular to the refracted ray
For air to glass: °== − 560.15.1tan 1
Bθ
If the incident ray is linearly polarized in the incident plane 100% transmission
Dispersion The index of refraction of visible light changes slightly with wavelength. In glass n is an decreasing function of λ (and an increasing function of f )
1n
2n
26.5 The Dispersion of Light: Prisms and Rainbows
13
The reflection is NOT total internal reflection
In seeing a rainbow, we are looking at sunlight that has gone through a refraction-reflection-refraction interaction in the water droplets suspended in air
Common Misconception
14
30.0 cm
30.0 cm
50.0 cm
x
θ1
θ2
n1=1.50
n2=1.00
A
B
C
Fermat’s Principle A ray originates from point A (0,0) inside a piece of glass, and crosses the glass-air boundary at point B (x, 30.0 cm), and is detected by a sensor at point C (50.0 cm, 60.0 cm).
Glass has refractive index n1=1.50
Find x, where the ray crosses the boundary An interesting solution: Fermat’s Principle: The ray from A to C takes the path of least time
2222 )30()50(BC ,)30(AB +−=+= xx
cm/ns 0.3000.1
0.30 ,cm/ns 0.2050.1
0.30 cm/ns, 30.02
21
1 =======ncv
ncvc
30)30()50(
20)30(BCAB
2222
21
+−+
+=+=
xxvv
t
Minimum at x=17.0 cm
493.0)30()17(
17sin 221 =
+=θ
740.0)30()33(
33sin 222 =
+=θ
!!! 500.1493.0740.0
sinsin
2
1
1
2
nn
===θθ
26.6 Lenses
Refractive Lenses are typically formed by grinding and polishing the surface of glass or special plastics to spherical and/or planar shapes. They come in two general types: With a converging lens,
paraxial rays that are parallel to the principal axis converge to the focal point.
15
With a diverging lens, paraxial rays that are parallel to the principal axis appear to originate from the (virtual) focal point.
−
26.6 Lenses
16
For lenses, the object and the viewer are on opposite sides (for mirrors they were on the same side) :
For ray tracing we commonly use three rays.
[1] from tip of object, parallel to the principle axis on the object sides, through the lens and then through the real focal point on the viewer side
[2] from tip of object, through the real focal point on the object side, then through the lens and emerge parallel to the principle axis on the viewer side
[3] from tip of object straight through the middle of the lens
4
Converging lenses are analogous to concave mirrors.
26.7 The Formation of Images by Lenses Image Formation by a Converging Lens
17
Case 1: (Example: ordinary camera taking a picture of a distant object)
Object is placed beyond 2F (object side): 2F for a converging lens is analogous to C Shrunken, inverted REAL image located between F and 2F on observer side.
fd 20 > fdf i 2<<
Real images are always inverted and are always on the viewer side of the lens
fdd io
111=+
o
i
o
i
dd
hhm −==and
The same equations that applies to mirrors also apply to lenses
fdo 2>fdo 2
11<→
oi dfd111
−=ffdi 2
111−>→
fdi 211
>→ fdi 2 <→
0>od 01>→
od oi dfd111
−=fdi
11<→ fdi >→ fdf i 2 <<→
26.7 The Formation of Images by Lenses
18
Case 2: (Example: projection of a downward nearby object onto a upward REAL image on screen)
fdf 20 << fdi 2>
enlarged, inverted REAL image beyond 2F on observer side. Object placed between F (object side) and 2F (analogous to C for mirrors)
fdo 2<fdo 2
11>→
oi dfd111
−=ffdi 2
111−<→
fdi 211
<→ fdi 2 >→
enlarged, upright VIRTUAL image on object side
Object placed inside F (object side)
fdo <<0fdo
11>→ 0111
<−=→oi dfd
0 <→ id
fd <0
0<id Case 3: (Example: magnifying glass)
1>−
=−=oo
i
dff
ddm
26.8 The Thin-Lens Equation and the Magnification Equation
Example: The Real Image Formed by a Camera Lens A 1.70-m tall person is standing 2.50 m in front of a camera. The
camera uses a converging lens whose focal length is 0.0500 m.
(a) Find the image distance and determine whether the image is real or virtual.
(b) Find the magnification and height of the image on the film.
1m 6.19m 50.2
1m 0500.0
1111 −=−=−=oi dfd
(a)
m 0510.0=id real image
(b) 0204.0m 50.2m 0510.0
−=−=−=o
i
ddm
( )( ) m 0347.0m 70.10204.0 −=−== oi mhh 19
fdd io
111=+
o
i
o
i
dd
hhm −==and
26.6 Lenses
20
Diverging Lenses are analogous to convex mirrors
Ray-tracing for diverging lenses
[1] from tip of object, parallel to the principle axis on the object sides, through the lens and bending outward as if it came from the virtual focal point on the object side
[2] from tip of object, toward the real focal point on the viewer side, then through the lens and emerge parallel to the principle axis on the viewer side
[3] from tip of object straight through the middle of the lens
4
26.7 The Formation of Images by Lenses
Image Formation by a Diverging Lens
21
00 >d
0<id
0<f
Virtual images are always upright and are always on the object side of the lens
A diverging lens always forms a shrunken, upright, VIRTUAL, image.
A diverging lens has a VIRTUAL focus on the object side of the lens
0 ,0 <> fd 011111<−−=−=→
ooi dfdfd0 <→ id
oo
i
dff
ddm
−=−=
odff−−
−=
odff+
= 1 <→ m
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