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Chapter 26
The Refraction of Light: Lenses and Optical Instruments
Chapter 26 The Refraction of Light: Lenses and Optical Instruments26.1 The Index of Refraction
26.2 Snellβs law and the Refraction of Light
26.3 Total Internal reflection
26.6 Lenses
26.7 The formation of Images by lenses
26.8 The Thin-Lens equation and the Magnification equation
26.9 Lenses in combination
26.10 The Human eye
26.14 Lens aberrations
26.1 The Index of Refraction
β’ Light travels through a vacuum at a speed β’ Light travels through materials at a speed less than
its speed in a vacuum.β’ The change in speed as a ray of light goes from one
material to another causes the ray to deviate from its incident direction
β’ This is called refraction
β’ 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:
π=πππππππ hπππ π‘ πππ£πππ’π’π
πππππππ hπππ π‘ ππ hπ‘ ππππ‘πππππ=ππ£
26.2 Snellβs Law and the Refraction of Light
β’ When light strikes the interface, part of the light is reflected with the angle of incidence equalling the angle of reflection
β’ When a ray enters the second material and changes direction (is refracted) it behaves in one of the followinga) When light travels from a less
dense medium to a more dense one, the refracted light ray is bent toward the normal
b) When light travels from a more dense medium to a less dense one, the refracted light ray is bent away the normal
β’ The refraction that occurs at the interface between two materials obeys the Snellβs law of refraction
β’ The law states that1) the refracted ray, the incident ray, and the
normal to the interface all lie in the same plane
2) The angle of refraction is related to the angle of incidence according to
π1 sinπ1=π2 sin π2
Example 1 Determining the Angle of Refraction
A light ray strikes an air/water surface at an angle of 46 degrees with respect to the normal. Find the angle of refraction when the direction of the ray is (a)from air to water and (b)from water to air.
Solution
π1 sinπ1=π2 sin π2
β΄sin π2=π1sin π1
π2
sin π2=(1.00 ) (sin 46 )
1.33ΒΏ0.54
(a)
β΄π2=33 Β°
APPARENT DEPTH
β’ When an object lies under water, it appears to be closer to the surface than it actually is apparent depth
69.0
00.1
31sin33.1sinsin
1
221
n
n
441
313.30.2tan 12
Example 2 Finding a Sunken Chest
The searchlight on a yacht is being used to illuminate a sunkenchest. At what angle of incidence should the light be aimed?
Apparent depth, observer directly above object
πβ²=π (π2π1 )
Conceptual Example 4 On the Inside Looking Out
A swimmer is under water and looking up at the surface. Someoneholds a coin in the air, directly above the swimmerβs eyes. To theswimmer, the coin appears to be at a certain height above the water. Is the apparent height of the coin greater, less than, or the same as its actual height?
THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL (STUDY)
THE DERIVATION OF SNELLβS LAW
26.3 Total Internal Reflection
β’ If the incident ray is at the critical angle , the angle of refraction is 90Β°
β’ The critical is obtained from Snellβs lawβ’ When the incidence angle exceeds the critical
angle, all the incidence light ray is reflected back into the material from which it came
β’ This phenomenon is known as total internal reflection
sin ππ=π2sin 90 Β°
π1=π2π1
hπ€ ππππ1>π2
Example 5 Total Internal ReflectionA beam of light is propagating through diamond and strikes the diamond-air interface at an angle of incidence of 28 degrees. (a)Will part of the beam enter the air or will
there be total internal reflection? (b)Repeat part (a) assuming that the diamond is
surrounded by water.
4.2442.2
00.1sinsin 1
1
21
n
nc(a)
(b) 3.3342.2
33.1sinsin 1
1
21
n
nc
Conceptual Example 6 The Sparkle of a DiamondThe diamond is famous for its sparkle because the light coming fromit glitters as the diamond is moved about. Why does a diamond exhibit such brilliance? Why does it lose much of its brilliance when placed under water?
Fiber optics
β’ Application of total internal reflection occurs is fiber optics
β’ Consists of cylindrical inner core that carries the light and the outer concentric shell, the cladding
β’ Little light is lost as a result of absorption by the core
β’ Light can travel many kilometres before its intensity diminishes appreciably
Endoscopy
β’ Optical fiber cables are used in medicine in endoscopy
β’ Used to peer inside the body
β’ Doctor is using bronchoscope to examine the lungs of a patient
Colonscope is used to examine the interior of the colon for diagnosing colon cancer in its early stage
β’ Optical fibres have made arthroscopic surgery possible
β’ Insert the instrument and the cable into a joint, with only a tiny incision and minimal damage to surrounding tissues
26.4 Polarization and the Reflection and Refraction of Light
Brewsterβs law
β’ When light is incident on a non-metallic surface at the Brewster angle, the reflected light is completely polarized parallel to the surface
β’ The reflected and refracted rays are perpendicular to each other
tanππ΅=π2π1
26.5 The Dispersion of Light: Prisms and Rainbows (Study)
The net effect of a prism is to change the direction of a light ray.Light rays corresponding to different colors bend by different amounts.
Conceptual Example 7 The Refraction of Light Depends on Two Refractive Indices
It is possible for a prism to bend light upward,downward, or not at all. How can the situationsdepicted in the figure arise?
26.6 Lensesβ’ Lenses refract light in such a way that an image of
the light source is formed.β’ With a converging (convex) lens, paraxial rays that
are parallel to the principal axis converge to the focal point.
β’ With a diverging (concave) lens, paraxial rays that are parallel to the principal axis appear to originate from the focal point.
26.7 The Formation of Images by Lenses
RAY DIAGRAMS for convex (converging) lens
β’ When the object is placed a distance further than twice the focal length from the lens, the image is real, inverted and smaller than the object.
β’ When the object is placed between F and 2F, the real image is inverted and larger than the object.
β’ When the object is placed between F and the lens, the virtual image is upright and larger than the object.
RAY DIAGRAMS for diverging (concave) lens
β’ Regardless of the position of a real object, a diverging lens always forms an image that is upright, virtual and smaller than the object.
26.8 The Thin-Lens Equation and the Magnification Equation
π=hπhπ
=βππ
ππ
1π=1ππ
+1ππ
Summary of Sign Conventions for Lenses
NOTE THAT:β’ is positive for a converging (convex) lens and
negative for a diverging (concave) lensβ’ is positive if the object is to the left of the lens and
negative if it is to the right of the lens β’ is positive for the image formed to the right of the
lens (real image) and negative if the image is formed to the left of the lens (virtual image)
β’ is positive for an upright image and negative for an inverted image
1π=1ππ
+1ππ
π=hπhπ
=βππ
ππ
Example 9 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.1π=1ππ
+1ππ
1ππ
=1πβ1π0
1ππ
=1
0.05πβ
12.5π
β΄ππ=0.0510π
π=hπhπ
=βππ
ππh π=
(βππ ) (hπ )ππ
h π=(β0.0510π ) (1.70π )
2.50π
The image is real since is +ve
m
26.9 Lenses in Combination
β’ The image produced by one lens serves as the object for the next lens.
Example 11.The objective and eyepiece lenses (of a compound microscope) have focal lengths of 15.0mm and 25.5 mm respectively. A distance of 61.0 mm separates the lenses. The microscope is used to examine objects placed 24.1 mm from the objective lens. Determine the final image distance1ππ2
=1π πβ1ππ2
1ππ1
=1π π
β1ππ1
1ππ1
=1
15.0ππβ
124.1ππ
1ππ1
=0.0252ππβ 1
β΄ππ 1=39.7ππ
ππ2=61.0ππβππ 1=61.0ππβ39.7ππ=21.3ππ
1ππ2
=1π πβ1ππ2
1ππ2
=1
25.5ππβ
121.3ππ
1ππ2
=β0.0077ππβ 1
ππ 2=β130ππ
26.10 The Human Eye
β’ Far point:- is the location of the farthest object on which a fully relaxed eye can focus ( located nearly at infinity)
β’ Near point:- the point nearest the eye at which an object can be placed and still produce a sharp image at the retina (located 25 cm from the eye)
β’ Accommodation:- the process in which the lens changes its focal length to focus on objects at different distances
NEARSIGHTEDNESS (Myopia)
β’ The person can focus on near objects but not on distant object
β’ The far point is no longer at infinity but some distance closer to the eye
β’ The focal length of the eye is shorter than it should be and the distant object forms an image in front of the retina
β’ The lens creates an image of the distance object at the far point of the nearsighted eye.
This defect is corrected by the use of a diverging (concave) lens
Example 12 Eyeglasses for the Nearsighted PersonA nearsighted person has a far point located only 521 cm from theeye. Assuming that eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lens of the glassesso the person can see distant objects.
1π=1ππ
+1ππ
1π=1β
+1
β519ππ
π =β519ππ
ππ=β ππ=521ππ
FARSIGHTEDNESS
THE REFRACTIVE POWER OF A LENS β THE DIOPTER
β’ The extent to which rays of light are refracted depends on its focal length
β’ Optometrists who prescribe correctional lenses and the opticians who make the lenses do not specify the focal length.
β’ Instead they use the concept of refractive power.
π ππππππ‘ππ£ππππ€ππ (πππππππ‘πππ )= 1π (πππππ‘πππ )
β’ A converging lens has a positive refractive power, and a diverging lens has a negative refractive power
Using the lens equation
1π=1ππ
+1ππ
β’ Can read the magazine when the image formed by the lens is at the near point of the eye
β΄ 1ππ
=1πβ1ππ
1ππ
=1
35.1ππβ
1β142ππ
β΄ππ=28.1ππ
ππ=β142ππ
Exercise 26-77 (Ex 26-76 in the 8th Ed)Your friend has a near point of 142 cm, and she wears contact lenses that have a focal length of 35.1 cm.How close can she hold a magazine and still read it clearly?
26.14 Lens Aberrations
β’ In a converging lens, spherical aberration prevents light rays parallel to the principal axis from converging at a single point.
β’ Lens aberrations limit the formation of perfectly focused or sharp images by the optical lens
β’ Spherical aberration can be reduced by using a variable-aperture diaphragm.
β’ Chromatic aberration is greatly reduced by the use of compound lens
β’ The lens combination is known as an achromatic lens
Chromatic aberrations
β’ The index of refraction of material from which the material is made varies with lens
β’ Different colours are focused at different points along the principal axis
1π=1ππ
+1ππ
1ππ
=1πβ1ππ
1ππ
=1
65ππβ
125 ππ
β΄ππ=β40.6ππAt age 40
At age 45
1ππ
=1πβ1ππ
1ππ
=1
65ππβ
129ππ
β΄ππ=β52.4ππ
Exercise 26-121 (Ex 26-119 in the 8th Ed)At age 40, a certain man requires contact lenses, with f= 65cm, to read a book held 25cm from his eyes. At age 45, while wearing these contacts he must now hold a book 29cm from his eyes(a)By what distance has his near point changed?(b)What focal-length lenses does he require at
age 45 to read a book at 25cm
β΄ππ=β40.6ππ β΄ππ=β52.4ππ
The manβs near point has shifted by:52.4ππβ40.6ππ=11.8ππ
b)
1π=1ππ
+1ππ
1π=125
+1
β52.4ππ
ππ=25πππ =47.8ππ
THE END