General Physics 2092
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THE REFRACTION OF LIGHT: LENSES
• Refraction
– As light passes from one medium to another it changes direction at
the interface between the two media.
– This change of direction is known as refraction.
– The relationship between the angle of incidence (θ1) and the angle
of refraction (θ2) . Is given by Snell’s law:
n1 sin θ 1 = n2 sin θ 2
– where n1 and n2 are the indexes of refraction of the media in which
the light is travelling.
n=
Speed of light in a vacuum
Speed of light in the meterial
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THE REFRACTION OF LIGHT: LENSES...
• Total Internal Reflection
– As light passes from a medium of higher index of refraction into a
medium of lower index of refraction an angle of incidence is
reached at which the angle of refraction is 90o.
– This angle is known as the critical angle (θc).
– At all angles greater than the critical angle (θc), the light is totally
reflected back.
sin θ c =
n2
n1
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THE REFRACTION OF LIGHT: LENSES...
• Problem 1
–
–
–
–
Determine the critical angle for light passing from
a) diamond into air and
b) glass into water.
The index of refraction from each substance is as follows: diamond
(2.42), air (1.00), glass (1.50), and water (1.33).
a) Total internal reflection
sin θ c −diamond =
⇒ θ air = 90o
nair
ndiamond
⇒ θ c −diamond = 24o
b) Total internal reflection ⇒ θ water = 90o
sin θ c − glass =
nwater
nglass
⇒ θ c = 63o
Notice that the internal reflection occurs only when light travels from a
medium of higher index of refraction to a medium of lower index of refraction
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THE REFRACTION OF LIGHT: LENSES...
• Dispersion of Light
– The spreading of white light into the full spectrum of visible light
(rainbow colors) is called dispersion.
– The dispersion happens because the index of refraction of a
material depends on the wavelength.
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5
THE REFRACTION OF LIGHT: LENSES...
•
– Rainbows are a spectacular example of dispersion by droplets of
water in the air with the sun at your back.
– Rainbows can be observed when sunlight is bent then reflected off
the back surface of water droplets.
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THE REFRACTION OF LIGHT: LENSES...
• Thin Lenses
– Rays of light parallel to the principal axis of a convex lens
(converging lens) converge at the focal point (F) of the lens after
refraction.
– Rays of light parallel to the principal axis of a concave lens
(diverging lens) diverge after passing through the lens. If the
refracted rays are traced on a straight line back through the lens
they appear to converge at the focal point.
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THE REFRACTION OF LIGHT: LENSES...
• Images Formed by Converging Lens
– The images formed by converging lenses
may be located by using three rays:
– 1) A ray parallel to the axis of the lens
refracts toward the the focal point on the
opposite side of the lens.
– 2) A ray which passes through the focal
point refracts parallel to the axis of the
lens.
– 3) A ray which travels through the center
of the lens continue directly without any
bending.
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THE REFRACTION OF LIGHT: LENSES...
• Images Formed by diverging Lens
– The images formed by diverging lenses
may be located by using three rays:
– 1) A ray parallel to the axis of the lens
refracts away from the axis and appears
to have originated from the focal point
on the left of the lens.
– 2) A ray which points toward the focal
point on the right of the lens refracts
parallel to the axis of the lens (optional).
– 3) A ray which travels through the center
of the lens continue directly without any
bending.
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THE REFRACTION OF LIGHT: LENSES...
• Images Formed by Thin Lenses
– The characteristics of the images formed by thin lenses depend on
the distance from the lens to the object (do) , the focal length (f),
and whether the lens is concave or convex.
1
1
1
=
+
f d o di
• Magnification and Size of Image
– The linear magnification (m) refers to the ratio of the size of the
image (hi) to the size of the object (ho).
– The magnification produced by curved mirror is given by
m=
hi
d
− = − i
ho
do
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THE REFRACTION OF LIGHT: LENSES...
• Sign conventions
– Focal length
• f is positive for converging lens
• f is negative for diverging lens
– Object distance
• do is positive if the object is on the same side where the light is
coming from (real object)
• do is negative if the object is on the opposite side where the light is
coming from (virtual object)***
– Image Distance
• di is positive if the image is on the opposite side where the light is
coming from (real image)
• di is negative if the image is on the same side where the light is
coming from (virtual image)
– magnification
• m is positive if the image upright with respect to the object
• m is negative if the image inverted with respect to the object
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THE REFRACTION OF LIGHT: LENSES...
• Problem 2
– An object 5.0cm high is placed 30cm from a Converging lens of
focal length 20cm.
– a) Draw a ray diagram and locate the position of the image formed.
Draw in the image .
– b) Mathematically determine the image distance from the lens,
magnification, and height of the image.
– c) State the characteristics of the image.
1
1 1
=
+
f d o di
d
m= − i
do
hi = mho
⇒
d i = 60cm
⇒
m = − 2 .0
⇒
hi = − 10.0cm
The image is real, inverted and magnified
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THE REFRACTION OF LIGHT: LENSES...
• Problem 3
– An object 3.0cm high is placed 5.0cm from a Converging lens of
focal length 10cm.
– a) Draw a ray diagram and locate the position of the image formed.
Draw in the image .
– b) Mathematically determine the image distance from the lens,
magnification, and height of the image.
– c) State the characteristics of the image.
1
1 1
=
+
f d o di
d
m= − i
do
hi = mho
⇒
d i = − 10cm
⇒
m = + 2.0
⇒
hi = + 6.0cm
The image is virtual, upright and magnified
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THE REFRACTION OF LIGHT: LENSES...
• Problem 4
– An object 3.0cm high is placed 30cm from a diverging lens of
focal length 20cm.
– a) Draw a ray diagram and locate the position of the image formed.
Draw in the image .
– b) Mathematically determine the image distance from the lens,
magnification, and height of the image.
– c) State the characteristics of the image.
1
1 1
=
+
f d o di
d
m= − i
do
hi = mho
⇒
d i = − 12cm
⇒
m = + 0.40
⇒
hi = + 1.2cm
The image is virtual, uprightand diminished
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THE REFRACTION OF LIGHT: LENSES...
• The Human Eye
– Light passes through a transparent outer membrane called the
cornea and the lens.
– The image is formed in the back of the eye on the retina.
– For distant objects, the ciliary muscles relax, the lens becomes
thinner and the focal length greater and the image focuses on the
retina.
– For nearby objects, the muscles contract causing the curvature of
lens to increase therefore decreasing the focal length and the image
focuses on the retina.
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THE REFRACTION OF LIGHT: LENSES...
• Common Defects of the Eye
– Nearsightedness (myopia)
– In nearsightedness a person can see nearby objects clearly but
objects at a distance are blurred (long eye eyeball).
– A diverging lens is used to converge the rays so that the image
comes to focus at the retina.
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THE REFRACTION OF LIGHT: LENSES...
• Common Deffects of the Eye
– Farsightedness (hyeropia)
– In farsightedness a person can see distant objects clearly but
nearby objects are blurred (short eyeball).
– A converging lens is used to converge the rays so that the image
comes to focus at the retina.
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THE REFRACTION OF LIGHT: LENSES...
• Problem 5
– A student requires reading glasses of +2.5diopters to read the print
on this page when the page is held 25cm from her eyes. Determine
the minimum distance that she would have to hold the newspaper
in order to read the print without her eyeglasses. Assume that the
distance from the lens to eye is negligible.
P=
1
f
⇒
f = + 0.40m = 40cm
With her glass on, the print is held 25 cm from her eyes and the
image is formed at the near point
1
1 1
=
+
f d o di
⇒
d i = − 67cm
Without her glasses, the student would have to hold the newspaper at least
67 cm from her eyes in order to read the print.
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THE REFRACTION OF LIGHT: LENSES...
• The Magnifying Glass
– A simple magnifying glass is a
converging lens which is used to
produce enlarged image of an
object on the retina of the eye.
– A virtual enlarged image is
produced when the object is placed
within the focal point.
– The angular magnification is given
by
M =
θ′
θ
≈
N
do
– where N is the distance from the
eye to the near point.
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THE REFRACTION OF LIGHT: LENSES...
• The Astronomical Telescope.
– A telescope consists of a converging lens with a long focal length
(fo) called the objective and converging with a short focal length
(fe) as the eyepiece.
– As a result the first image is located near the focal point of the
objective and therefore the second image is magnified, inverted
and virtual.
M =
θ′
θ
≈
fo
fe
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THE REFRACTION OF LIGHT: LENSES...
• Problem 6
– Two converging lenses are 20 cm apart. The focal length of the
lenses are 5.0 cm and 10 cm respectively. An object 3.0 cm height
is placed 15 cm in front of the 5.0 cm lens.
– a) Draw an accurate ray diagram and locate the final image formed
by the combination.
– b) Mathematically determine the position and size of the final
image. 1
1
1
=
+
⇒ d i1 = 7.5cm
The first image is
f1
d o1 d i1
d
h
m1 = − i1 = i1
d o1 ho1
⇒
hi1 = − 1.5cm
real, inverted and
diminished
The image produced by the first lens becomes the object for the second lens.
d o 2 = 20 − 7.5 = 12.5cm
m2 = −
di2
do2
1
1
1
=
+
⇒ d i 2 = 50cm
f2 do2 di2
h
= i2
⇒ hi = + 6.0cm
ho 2
The final image is
real, upright and
magnified
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THE REFRACTION OF LIGHT: LENSES...
• Problem 7
– During a laboratory experiment, a student position two lenses to
form a compound microscope. The objective and eyepiece have
focal length of 0.40 cm and 3.0 cm respective. The lenses 4.9 cm
apart and an object is place 0.50 cm in front of the objective lens.
a) Draw a ray diagram and locate the position of the final image.
– b) Mathematically determine the position and of the final image.
1
1
1
=
+
f o1 d o1 d i1
⇒
d i1 = 2.0cm
The image produced by the objective lens becomes the object for the second lens.
d i1 = 4.9 − 2.0 = 2.9cm
1
1
1
=
+
f e do2 di2
⇒
d i 2 = − 87cm
The final image is virtual and formed 87 cm in front of the eyepiece
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