Chapter 26
The Refraction of Light:
Lenses and Optical
Instruments
Chapter 26 The Refraction of Light: Lenses and
Optical Instruments
26.1 The Index of Refraction
26.2 Snell’s 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
• Light travels through a vacuum at a speed 𝑐 = 3 × 108 𝑚/𝑠
• Light travels through materials at a speed less than its speed in a
vacuum.
• The change in speed as a ray of light goes from one material to
another causes the ray to deviate from its incident direction
• This is called refraction
• The index of refraction 𝑛 of a material is the ratio of the speed
of light in a vacuum to the speed of light in the material:
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚
𝑐
𝑛=
=
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑣
26.2 Snell’s Law and the Refraction of Light
• When light strikes the interface,
part of the light is reflected with the
angle of incidence equalling the
angle of reflection
• When a ray enters the second
material and changes direction (is
refracted) it behaves in one of the
following
a)
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
• The refraction that occurs at the interface between two
materials obeys the Snell’s law of refraction
• The law states that
1) the refracted ray, the incident ray, and the normal to
the interface all lie in the same plane
2) The angle of refraction 𝜃2 is related to the angle of
incidence 𝜃1 according to
𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2
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
(a)
𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2
𝑛1 sin 𝜃1
∴ sin 𝜃2 =
𝑛2
1.00 sin 46
sin 𝜃2 =
1.33
= 0.54
∴ 𝜃2 = 33°
APPARENT DEPTH
• When an object lies under water, it appears to be closer to the
surface than it actually is apparent depth
Example 2 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?
 2  tan 1 2.0 3.3  31
n2 sin  2 1.33sin 31
sin 1 

 0.69
n1
1.00
1  44
Apparent depth, observer
directly above object
𝑛2
𝑑 =𝑑
𝑛1
′
Conceptual Example 4 On the Inside Looking Out
A swimmer is under water and looking up at the surface. Someone
holds a coin in the air, directly above the swimmer’s eyes. To the
swimmer, 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 SNELL’S LAW
26.3 Total Internal Reflection
• If the incident ray is at the critical angle 𝜃𝑐 , the angle of
refraction is 90°
• The critical is obtained from Snell’s law
• When the incidence angle exceeds the critical angle, all the
incidence light ray is reflected back into the material from which
it came
• This phenomenon is known as total internal reflection
𝑛2 sin 90° 𝑛2
sin 𝜃𝑐 =
=
𝑛1
𝑛1
𝑤ℎ𝑒𝑟𝑒 𝑛1 > 𝑛2
Example 5 Total Internal Reflection
A 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.
 n2 
 1.00 

  sin 1 
  24.4
 2.42 
 n1 
(a)
 c  sin 1 
(b)
 n2 
1  1.33 



 c  sin    sin 
  33.3
 2.42 
 n1 
1
Conceptual Example 6 The Sparkle of a Diamond
The diamond is famous for its sparkle because the light coming from
it 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
• Application of total internal reflection occurs is fiber optics
• Consists of cylindrical inner core that carries the light and the
outer concentric shell, the cladding
• Little light is lost as a result of absorption by the core
• Light can travel many kilometres before its intensity
diminishes appreciably
Endoscopy
• Optical fiber cables are
used in medicine in
endoscopy
• Used to peer inside the
body
• 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
• Optical fibres have
made arthroscopic
surgery possible
• 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
• When light is incident on a non-metallic surface at the Brewster
angle, the reflected light is completely polarized parallel to the
surface
• The reflected and refracted rays are perpendicular to each other
Brewster’s law
𝑛2
tan 𝜃𝐵 =
𝑛1
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 situations
depicted in the figure arise?
26.6 Lenses
• Lenses refract light in such a way that an image of the light
source is formed.
• With a converging (convex) lens, paraxial rays that are parallel
to the principal axis converge to the focal point.
• 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
• 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.
• When the object is placed between F and 2F, the real image is
inverted and larger than the object.
• 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
• 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
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
ℎ𝑖
𝑑𝑖
𝑚=
=−
ℎ𝑜
𝑑𝑜
Summary of Sign Conventions for Lenses
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
ℎ𝑖
𝑑𝑖
𝑚=
=−
ℎ𝑜
𝑑𝑜
NOTE THAT:
• 𝑓 is positive for a converging (convex) lens and negative for a
diverging (concave) lens
• 𝑑𝑜 is positive if the object is to the left of the lens and negative
if it is to the right of the lens
• 𝑑𝑖 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)
• 𝑚 is positive for an upright image and negative for an inverted
image
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.
1
1
1
=
+
𝑓 𝑑𝑜 𝑑 𝑖
∴ 𝑑𝑖 = 0.0510𝑚
ℎ𝑖
𝑑𝑖
𝑚=
=−
ℎ𝑜
𝑑𝑜
1 1 1
= −
𝑑𝑖 𝑓 𝑑0
1
1
1
=
−
𝑑𝑖 0.05𝑚 2.5𝑚
The image is real since 𝑑𝑖 is +ve
−𝑑𝑖 ℎ𝑜
ℎ𝑖 =
𝑑𝑜
∴ ℎ𝑖 = −0.0347 m
−0.0510𝑚 1.70𝑚
ℎ𝑖 =
2.50𝑚
26.9 Lenses in Combination
• 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 distance
1
1
1
= −
𝑑𝑖2 𝑓𝑒 𝑑𝑜2
1
1
1
= −
𝑑𝑖1 𝑓𝑜 𝑑𝑜1
1
1
1
=
−
𝑑𝑖1 15.0𝑚𝑚 24.1𝑚𝑚
1
= 0.0252𝑚𝑚−1
𝑑𝑖1
∴ 𝑑𝑖1 = 39.7 𝑚𝑚
𝑑𝑜2 = 61.0𝑚𝑚 − 𝑑𝑖1 = 61.0𝑚𝑚 − 39.7𝑚𝑚 = 21.3 𝑚𝑚
1
1
1
= −
𝑑𝑖2 𝑓𝑒 𝑑𝑜2
1
= −0.0077 𝑚𝑚−1
𝑑𝑖2
𝑑𝑖2 = −130𝑚𝑚
1
1
1
=
−
𝑑𝑖2 25.5 𝑚𝑚 21.3 𝑚𝑚
26.10 The Human Eye
• Far point:- is the location of the farthest object on which a fully
relaxed eye can focus ( located nearly at infinity)
• 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)
• Accommodation:- the process in which the lens changes its focal
length to focus on objects at different distances
NEARSIGHTEDNESS (Myopia)
• The person can focus on near objects but not on distant object
• The far point is no longer at infinity but some distance closer to
the eye
• The focal length of the eye is shorter than it should be and the
distant object forms an image in front of the retina
This defect is corrected by the use of a diverging (concave) lens
• The lens creates an image of the distance object at the far point
of the nearsighted eye.
Example 12 Eyeglasses for the Nearsighted Person
A nearsighted person has a far point located only 521 cm from the
eye. 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 glasses
so the person can see distant objects.
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
𝑑𝑜 = ∞
𝑑𝑖 = 521𝑐𝑚
1 1
1
= +
𝑓 ∞ −519𝑐𝑚
𝑓 = −519𝑐𝑚
FARSIGHTEDNESS
THE REFRACTIVE POWER OF A LENS – THE DIOPTER
• The extent to which rays of light are refracted depends on
its focal length
• Optometrists who prescribe correctional lenses and the
opticians who make the lenses do not specify the focal
length.
• Instead they use the concept of refractive power.
1
𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑑𝑖𝑜𝑝𝑡𝑒𝑟𝑠 =
𝑓 (𝑖𝑛 𝑚𝑒𝑡𝑒𝑟𝑠)
• A converging lens has a positive refractive power, and a
diverging lens has a negative refractive power
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?
Using the lens equation
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
• Can read the magazine when the image formed by
the lens is at the near point of the eye
1
1 1
∴
= −
𝑑𝑜 𝑓 𝑑𝑖
𝑑𝑖 = −142 𝑐𝑚
1
1
1
=
−
𝑑𝑜 35.1 𝑐𝑚 −142 𝑐𝑚
∴ 𝑑𝑜 = 28.1 𝑐𝑚
26.14 Lens Aberrations
• In a converging lens, spherical aberration prevents light rays
parallel to the principal axis from converging at a single point.
• Lens aberrations limit the formation of perfectly focused or
sharp images by the optical lens
• Spherical aberration can be reduced by using a variable-aperture
diaphragm.
Chromatic aberrations
• The index of refraction of material from which the material is
made varies with lens
• Different colours are focused at different points along the
principal axis
• Chromatic aberration is greatly reduced by the use of compound
lens
• The lens combination is known as an achromatic lens
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
∴ 𝑑𝑖 = −40.6𝑐𝑚
At age 40
1
1
1
=
+
𝑓 𝑑 𝑜 𝑑𝑖
1 1 1
= −
𝑑𝑖 𝑓 𝑑𝑜
1
1
1
=
−
𝑑𝑖 65𝑐𝑚 25𝑐𝑚
At age 45
1 1 1
= −
𝑑𝑖 𝑓 𝑑𝑜
1
1
1
=
−
𝑑𝑖 65𝑐𝑚 29𝑐𝑚
∴ 𝑑𝑖 = −52.4𝑐𝑚
∴ 𝑑𝑖 = −40.6𝑐𝑚
∴ 𝑑𝑖 = −52.4𝑐𝑚
The man’s near point has shifted by:
52.4𝑐𝑚 − 40.6𝑐𝑚 = 11.8𝑐𝑚
b)
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
1
1
1
=
+
𝑓 25 −52.4𝑐𝑚
𝑑𝑜 = 25𝑐𝑚
𝑓 = 47.8𝑐𝑚
THE END