Chapter 6
(Optics) Images
Reflection
A rough surface causes parallel light rays to be reflected in many different directions.
A smooth surface causes parallel light rays to be reflected in a single direction. This type of
surface looks like a mirror.
Images and Plane Mirrors
An image is a reproduction of an object via light. If the image can form on a surface, it
is a real image and can exist even if no observer is present. If the image requires the
visual system of an observer, it is a virtual image.
Here are some common examples of virtual image.
(a) A ray from a low section of the sky refracts through air that is heated by a road (without
reaching the road). An observer who intercepts the light perceives it to be from a pool of
water on the road. (b) Bending (exaggerated) of a light ray descending across an imaginary
boundary from warm air to warmer air. (c) Shifting of wavefronts and associated bending of
a ray, which occur because the lower ends of wavefronts move faster in warmer air. (d)
Bending of a ray ascending across an imaginary boundary to warm air from warmer air.
Images and Plane Mirrors
As shown in figure (a), a plane (flat) mirror can form a
virtual image of a light source (said to be the object, O) by
redirecting light rays emerging from the source. The image
can be seen where backward extensions of reflected rays
pass through one another. The object’s distance p from the
mirror is related to the (apparent) image distance i from the
mirror by
(a)
Object distance p is a positive quantity. Image distance i for
a virtual image is a negative quantity.
Only rays that are fairly close together can enter the eye after
reflection at a mirror. For the eye position shown in Fig. (b),
only a small portion of the mirror near point a (a portion
smaller than the pupil of the eye) is useful in forming the
image.
(b)
Spherical Mirrors
A spherical mirror is in the shape of a small section of a spherical surface and can be
concave (the radius of curvature r is a positive quantity), convex (r is a negative
quantity), or plane (flat, r is infinite).
We make a concave mirror by curving the mirror’s surface so it is concave (“caved in”
to the object) as in Fig. (b). We can make a convex mirror by curving a plane mirror so
its surface is convex (“flexed out”) as in Fig.(c). Curving the surface in this way (1)
moves the center of curvature C to behind the mirror and (2) increases the field of
view. It also (3) moves the image of the object closer to the mirror and (4) shrinks it.
These iterated characteristics are the exact opposite for concave mirror.
Spherical Mirrors
If parallel rays are sent into a (spherical) concave mirror parallel to the central axis, the
reflected rays pass through a common point (a real focus F ) at a distance f (a positive
quantity) from the mirror (figure a). If they are sent toward a (spherical) convex mirror,
backward extensions of the reflected rays pass through a common point (a virtual focus
F ) at a distance f (a negative quantity) from the mirror (figure b).
For mirrors of both types, the focal length f is related to the radius of curvature r of the
mirror by
where r (and f) is positive for a concave mirror and negative for a convex mirror.
The Mirror equation:
𝟏
𝒔𝒐
+
𝟏
𝒔𝒊
=
𝟐
𝑹
=
𝟏
𝒇
𝑠𝑜 : Object distance from the mirror.
𝑠𝑖 : Image distance from the mirror.
𝑅 : Radius of curvature of the mirror.
𝑓 : Focal length of the mirror =
𝑅
2
Notes:
𝒔𝒐 : is positive when the object is in front of the mirror.
𝒔𝒊 : is positive when the image is real, 1.e., in front of the mirror.
𝒔𝒊 : is negative when the image is virtual, 1.e., behind the mirror.
𝑹 and f: are positive for concave mirror and negative for a convex mirror.
The size of the image :
𝐿𝑖𝑛𝑒𝑎𝑟 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑀 =
=−
𝑆𝑖
𝑆𝑜
=
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑖𝑚𝑎𝑔𝑒
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡
=
𝑖𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑚𝑖𝑟𝑟𝑜𝑟
𝑜𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑚𝑖𝑟𝑟𝑜𝑟
ℎ𝑖
ℎ𝑜
The sign conventions for the given quantities in the mirror equation and magnification
equations are as follows:

f is + if the mirror is a concave mirror

f is - if the mirror is a convex mirror

𝑆𝑖 is + if the image is a real image and located on the object's side of the mirror.

𝑆𝑖 is - if the image is a virtual image and located behind the mirror.

ℎ𝑖 is + if the image is an upright image (and therefore, also virtual)

ℎ𝑖 is - if the image an inverted image (and therefore, also real)
Example 1: An object is located 30.0 cm from a converging mirror with a focal length of 5.0 cm.
a) Where will the image be formed?
Known information: 𝒇 = 𝟓. 𝟎 𝒄𝒎, 𝑺𝒐 = 𝟑𝟎. 𝟎 𝒄𝒎, 𝒉𝒐 = 𝟒. 𝟎 𝒄𝒎
1
𝑠𝑜
+
1
𝑠𝑖
=
1
⇒
𝑓
1
𝑠𝑖
1
1
𝑓
𝑠𝑜
= −
=
1
5.0
−
1
30
⇒
1
𝑠𝑖
= 0.16667 ⇒ 𝑠𝑖 = 6.0 𝑐𝑚
b) What is the magnification?
𝑆𝑖
𝑆𝑜
𝑀= −
= −
6.0
30.0
= −0.2 The image is smaller and inverted - negative sign.
c) If the object is 4.0 cm tall, how tall is the image ?
ℎ𝑖
ℎ𝑜
𝑆
𝑆
6.0
= − 𝑆 𝑖 ⇒ ℎ𝑖 = − 𝑆 𝑖 . ℎ𝑜 = − 30.0 . 4.0 ∴ ℎ𝑖 = −0.8 𝑐𝑚
𝑜
𝑜
The image is inverted because 𝐡𝐢 is negative .
Example 2: Determine the image distance and image height for a 5.00-cm tall object
placed 45.0 cm from a concave mirror having a focal length of 15.0 cm.
Answer: Si = 22.5 cm and hi = -2.5 cm
Use 1/f = 1/So + 1/Si where f = 15 cm and So = 45 cm
Then use hi / ho = - Si / So where ho = 5 cm, do = 45 cm, and Si = 22.5 cm
Example 3: A diverging mirror with a focal length of (-5.0) cm produces an image of an
object that is 15.0 cm from the mirror.
a) What is the distance of the image?
1
1
𝑜
𝑖
1
1
1
1
1
1
.𝑠 + 𝑠 = 𝑓 ⇒ 𝑠 = 𝑓 − 𝑠 = −5.0 − 15.0 ⇒ 𝑠𝑖 = −3.8 𝑐𝑚
𝑜
𝑖
The negative sign means that the image is 3.8 cm behind the mirror.
b) What is the magnification?
𝑀= −
𝑆𝑖
𝑆𝑜
=−
−3.8
15.0
= 0.25
The image is smaller (M < 1) and erect (M is positive).
c) If the object is 20.0 cm high, how high is the image?
ℎ𝑖
ℎ𝑜
=−
𝑆𝑖
𝑆𝑜
⇒ ℎ𝑖 = −
𝑆𝑖
.
𝑆𝑜
ℎ𝑜 = −
−3.8
15.0
. 20.0 ∴ ℎ𝑖 = 5.1 𝑐𝑚
The image should be positive because diverging mirrors always have an upright image.
Example 4: Determine the image distance and image height for a 5.00-cm tall object placed 30.0 cm from a concave mirror
having a focal length of 15.0 cm.
Answer: Si = 30.0 cm and hi = -5.0 cm
Use 1 / f = 1 / So + 1 / Si where f =15 cm and So = 45 cm
Then use hi / ho = - Si / So where ho = 5 cm, So = 45 cm, and Si = 30.0 cm
Example 5: Determine the image distance and image height for a 5.00-cm tall object placed 20.0 cm from a concave mirror
having a focal length of 15.0 cm.
Answer: Si = 60.0 cm and hi = -15.0 cm
Use 1 / f = 1 / So + 1/Si where f =15 cm and So=45 cm
Then use hi / ho = -Si / So where ho = 5 cm, So = 45 cm, and Si = 60.0 cm
Example 6:Determine the image distance and image height for a 5.00-cm tall object placed 10.0 cm from a concave mirror
having a focal length of 15.0 cm.
Answer: Si = -30.0 cm and hi = +15.0 cm
Use 1 / f = 1 / So + 1 / Si where f = 15 cm and So = 10.0 cm
Then use hi / ho = -Si / So where ho = 5 cm, So = 45 cm, and Si = -30.0 cm
Refraction of Light
Index of Refraction:
𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒗𝒂𝒄𝒖𝒎𝒆
𝒄
𝒏 = 𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 = 𝒗
Snell’s Law :
The critical angle is given by
𝑛𝑖 sin 𝜃𝑖 = 𝑛𝑡 sin 𝜃𝑡
𝒏
𝜽𝒄 = 𝐬𝐢𝐧−𝟏 ( 𝒏𝟐 )
𝒊
𝒊𝒇 𝒏𝒊 < 𝒏𝒕 𝒄𝒂𝒍𝒍𝒆𝒅 𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏
𝒊𝒇 𝒏𝒊 > 𝒏𝒕 𝒄𝒂𝒍𝒍𝒆𝒅 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏
Example 1: Sunlight strikes the surface of a lake. A diver sees the Sun at an angle of 42.0°
with respect to the vertical. What angle do the Sun’s rays in air make with the vertical?
Solution:
𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2 ⇒ 1.00 sin 𝜃1 = 1.333 sin42°
∴ sin 𝜃1 = 0.8920 ∴ 𝜃1 = sin−1 0.8920 = 63.1°
Example 2: Light travels through a vacuum at a speed of 2.998 × 108
following media:
a.
Water (n = 1.333).
b.
Crown glass (n = 1.52).
c.
Cubic zirconia (n = 2.16).
d.
Diamond (n = 2.419).
𝑚
𝑠
. Determine the speed of light in the
Answer:
a.
2.249 × 108
b.
1.97 × 108
c.
1.39 × 10
d.
1.239 × 108
𝑚
𝑠
m
s
m
8
s
m
s
Example 3: The speed of light in a certain glass is 1.91 × 108
glass?
Solution:
𝑐
𝑣
𝑛= =
𝑚
𝑠
𝑚
8
1.91×10
𝑠
2.998×108
= 1.57
𝑚
𝑠
. What is the refractive index of the
Example 4: What is the critical angle for light passing from glass (n=1.54) to water
(n=1.33)?
Solution:
𝑛𝑖 sin 𝜃𝑖 = 𝑛𝑡 sin 𝜃𝑡 ⇒ 𝑛𝑖 sin 𝜃𝑐 = 𝑛𝑡 sin90°
sin 𝜃𝑐 =
𝑛𝑡
𝑛𝑖
=
1.33
1.54
= 0.864 ∴ 𝜃𝑐 = 59.7°
Example 5: Determine the critical angle of the following materials when surrounded by
air:
°
1. Teflon (n = 1.38) 46.4
°
2. Pyrex glass (n = 1.47) 42.9
°
3. Polycarbonate glass (n = 1.59) 39.0
°
4. Sapphire gemstone (n = 1.77) 34.4
5. Diamond (n = 2.42) 25.4°
Summary
• Thin Lens:
Real and Virtual Images
• If the image can form on a surface, it is
a real image and can exist even if no
observer is present. If the image
requires the visual system of an
observer, it is a virtual image.
Image Formation
• Spherical mirrors, spherical refracting
surfaces, and thin lenses can form
images of a source of light—the object
— by redirecting rays emerging from
the source.
• Spherical Mirror:
Eq. 34-3 & 4
• Spherical Refracting Surface:
Eq. 34-8
Eq. 34-9 & 10
Optical Instruments
• Three optical instruments that extend
human vision are:
1. The simple magnifying lens, which
produces an angular magnification mθ
given by
Eq. 34-12
2. The compound microscope, which
produces an overall magnification M
given by
Eq. 34-14
3. The refracting telescope, which
produces an angular magnification
mu given by
Eq. 34-15