Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 24

Focal Point of a Spherical Mirror The principal axis or the symmetry axis of a spherical mirror is defined as the axis that passes through its center of curvature C and the center of the

17.3 Total Internal Reflection and Optical Fibers 571

n3 = sin−1 1

1.33 = sin−10.752 = 48.8◦

(b) From the right-angle triangle at the glass-water interface we can find the refracted angleθ3in water to be:

θ3= 90◦− θ c = 41.2◦

Using Snell’s law again at the glass-water interface, we have:

n2sinθ2 = n3sinθ3

n2 = 1.33 × sin 41.2◦

θ2= sin−10.585 = 35.7◦

(c) Since the sides of the glass-walled fish tank are parallel, we can again apply Snell’s law at the air-glass interface to calculateθ1as follows:

n1sinθ1 = n2sinθ2

n1 = 1.5 × sin 35.7◦

θ1= sin−10.875 = 61◦

17.4 Chromatic Dispersion and Prisms

Except in vacuum, the index of refraction depends on the light’s wavelength, i.e its color, seeSect 27.7 Therefore if a beam of light consists of rays of different wave-lengths (as in the case of white light), each ray will refract by a different angle from

a surface This spread of light is called chromatic dispersion, or simply dispersion.

Generally, the index of refraction n decreases with increasing wavelengths This

means that the violet light (with wavelengthλ  425 nm and index n = 1.3435) bends

more than the red light (with wavelengthλ  700 nm and index n = 1.3318) when

passing through the interface between two materials Figure17.9a shows this for a glass block, and Fig.17.9b shows this for a glass prism

The prism of Fig.17.9b is more commonly used to observe color separation of white light because the dispersion at the first surface is enhanced at the second

interface Thus, the violet ray in the white light of Fig.17.9b will emerge from the

right surface with an angle of deviationδVwhich is greater than the angle of deviation

δRof the red ray The differenceδV − δRis known as the angular dispersion, while

δYis the mean deviation of the yellow rays

Air

White light

δ R δ

A

R

V

R

V

the red color (a) Dispersion in a glass block (b) Dispersion in a prism

The general expression of δ for any color turns out to be rather complicated.

However, as the angle of incidence decreases from a large value, the angle of deviation

δ is found to decrease at first and then increase The angle of minimum deviation δm

is found when the ray passes through the prism symmetrically This angle is related

to the angle of the prism A, and its index of refraction n by the relation:

n= sin[(A + δ m)/2]

sin(A/2) −−−−−−−−−−→When A is small n= A + δ m

The most charming example of color dispersion is that of a rainbow To understand the formation of a rainbow we consider a horizontal overhead white sunlight that

is intercepted by spherical raindrops Figure17.10shows refractions and reflection

in two raindrops that explain how light rays from the Sun reach an observer’s eye The first refraction separates the sunlight into its color components Each color is then reflected at the raindrop’s inner surface Finally, a second refraction increases the separation between colors, and these color rays finally make it to the observer’s eye Using Snell’s law and geometry, we find that the maximum deviation angles of red and violet are about 42 and 40◦, respectively The rainbow that you can see is a personal one because different observers receive light from different raindrops

17.4 Chromatic Dispersion and Prisms 573

White light

White light

Raindrop

To

observer

Part of a rainbow

40 °

42 °

Violet li ght

Violet li

ght Red li ght

Red li

ght

used to explain the rainbow’s formation for the case of the red and violet colors only

Example 17.4

A monochromatic light ray is incident from air (with index n1= 1) onto an equi-lateral glass prism (with index n2= 1.5) and is refracted parallel to one of its faces

(i.e we have a symmetric ray), see Fig.17.11 (a) What is the angle of incidence

θ1at the first face? (b) What is the subsequent angle of incidence at the second face? (c) Is the light ray totally reflected at the second face? If not, find the angle

of minimum deviation of the light ray Then check that Eq.17.12holds

Fig 17.11

Air

60 °

60 °

1

q

2

q

m d

1

Solution: (a) The path of a symmetric light ray going through the prism (of apex

angle 60◦) and back out again into the air is shown.

Using elementary geometry, this figure shows that the angle of refractionθ2

can be found as follows:

θ2+ 60◦= 90◦

Therefore, using Snell’s law:

n1sinθ1 = n2sinθ2

n2sinθ2 n1



= sin−1



1.5 × sin 30◦ 1



= sin−1(0.75) = 48.59◦

(b) Again, by simple geometry the horizontal light ray inside the prism must

be incident on the second face with an angleθ

1= θ2= 30◦. (c) We know that if the incident angle is greater than the critical angle, then total internal reflection must occur Therefore, we first calculate the critical angle

as follows:

θc= sin−1n1

n2

= sin−1 1

1.5

= sin−10.666

= 41.8◦

Since θ

1< θc, then the light ray refracts at the second face, and total internal

reflection will not occur.

Using the geometry shown in the figure, we can find for this special case that the angle of minimum deviation is given by the following relation:

δm = 2(θ1− θ2)

= 2(48.59◦− 30◦)

= 37.18◦ Substituting A= 60◦andδm = 37.18◦in Eq.17.12gives:

n2= sin[(A + δ m)/2]

sin(A/2)

= sin[(60◦+ 37.18◦)/2]

sin(60◦/2)

= 0.75

0.5 = 1.5

17.4 Chromatic Dispersion and Prisms 575

This value of n2 obtained from Eq.17.12satisfies the given value of index of refraction of the prism

17.5 Formation of Images by Reflection

Mirrors gather and redirect light rays to form images of objects by reflection

To explain this, we will use the ray approximation model in terms of geometric optics, in which light travels in straight lines

17.5.1 Plane Mirrors

A plane mirror is a plane surface that can reflect a beam of light in one direction

instead of either scattering it in many directions or absorbing it

Figure17.12a shows how a plane mirror can form an image of a point object O located at a distance p from the mirror In this figure, we consider two diverging rays leaving O and strike the mirror and then are reflected to the eye of an observer The

rays appear to diverge from point I behind the mirror Thus, point I is the image of

point O The geometry of the figure indicates that the image I is opposite to object

O and is located at a distance as far behind the mirror as the object is in front of the

mirror

(b)

Mirror

p

I

q

q

Mirror

I i

h ′

h

q q

q q

O

p

(a) Front side Back side Front side Back side

mirror (a) An image formed for a point object (b) An image formed by an extended object, where the

object is an upright arrow of height h

Figure17.12b shows how a plane mirror can form an image of an extended object

O The object in this figure is an upright arrow of height h placed at a distance p from

the mirror The full image can be inferred by locating the images of selected points

on the object One of the two rays at the tip of the arrow follows a horizontal path

to the mirror and reflects back on itself The second ray follows an oblique path and reflects according to the laws of reflection, as shown in the figure Using geometry

we find that the image I is upright, opposite to the object, and located behind the

mirror at a distance equal to the object’s distance in front of the mirror In addition, the height of the object and its image are equal Also, the geometry of Fig.17.12b

indicates that h/h = i/p.

The image I in both parts of Fig.17.12is called a virtual image because no light

rays pass through it In addition, the value of i is considered to be negative since the image is behind the mirror and the value of his considered to be positive since the

image is upright.

We define the lateral magnification M of a horizontal overhead image as follows:

M= Image height Object height =h

We can use the relation h/h = i/p and the sign convention to write the lateral

mag-nification M as follows:

M= h

h = −i

For plane mirrors, M = 1, since his positive and equal to h, or i is negative and has

a magnitude equal to p The image formed by a plane mirror is upright but reversed.

The reversal of right and left is the reason why the word AMBULANCE is printed

as “ ” across the front of ambulance vehicles People driving in front

of such an ambulance can see the word “AMBULANCE” immediately evident when looking in their rear-view mirrors and make way

17.5.2 Spherical Mirrors

A spherical mirror is simply a mirror in the shape of a small section of the surface of

a sphere that has a center C and radius R When light is reflected from the concave

17.5 Formation of Images by Reflection 577

surface of the mirror, the mirror is called a concave mirror However, when light is reflected from the convex surface of the mirror, the mirror is called a convex mirror.

Focal Point of a Spherical Mirror

The principal axis (or the symmetry axis) of a spherical mirror is defined as the

axis that passes through its center of curvature C and the center of the mirror c, see

Fig.17.13 We consider the reflection of light coming from an infinitely far object

O located on the principal axis of a concave or convex spherical mirror Because of

the great distance between the object and the mirror, the light rays reach the mirror

parallel to its principal axis.

c

C F

R

f

Concave mirror

c

Front side Back side Front side Back side

Principal axis Principal

axis

Convex mirror

mirror (b) The same rays will diverge from a convex mirror and appear to come from a virtual focal point

When parallel rays reach the surface of the concave mirror of Fig.17.13a, they

will reflect and pass through a common point F If we place a card at F, a point image

would appear at F Therefore, this point is called the real focal point However, in

the case of the convex mirror of Fig.17.13b, the parallel rays reflect from the mirror

and appear to diverge from a common point F behind the mirror If we could place

a card at F, no image would appear on the card Therefore, this point is called the

virtual focal point The distance f from the center of the mirror to the focal point

(real or virtual) is called the focal length of the mirror.

For concave and convex mirrors, the following relation relates the focal length f

to the radius of curvature R:

f =R

17.5.2.1 Concave Mirrors

Sharp and Blurred Images

Rays that diverge from any point on an object and make small angles with the principal

axis (called paraxial rays) will reflect from the spherical concave mirror and intersect

at one image point See Fig.17.14a for a point object on the principal axis On the other hand, rays that diverge from the same point and make large angles with the

principal axis will reflect and intersect at different image points, see Fig.17.14b This

condition is called spherical aberration.

c

Small angles incidence

I O

c

O

I2 I1

Large angles incidence

reflect from the spherical concave mirror and meet at the same point image I (b) When rays diverge from O at large angles with the principal axis, they reflect from the spherical concave mirror and meet at different points I1, I2,

The Mirror Equation

The relationship between an object’s distance p, its image distance i, and the focal length f of a concave mirror can be found when light rays make small angles with

the principal axis (paraxial rays) Figure17.15a shows two rays (leaving an object O

of height h) reflected to form an image I of height h The first ray strikes the mirror

at its center c and is reflected The second ray passes through the focal point F and

reflects parallel to the principal axis

From the purple triangles of Fig.17.15a, we see that:

tanθ = h

p = h

h = i

From the yellow triangles of Fig.17.15b, we see that:

p − f =

h

17.5 Formation of Images by Reflection 579

c F

f

O

I

h′

i p

h′

(b)

c F

f

h

O

I

h′

i

p

h′

q q

a

a

(a)

h

tip of an arrow (b) Demonstration of the geometry produced by only the second ray

By comparing Eqs.17.16and17.17, we find that:

i

Dividing both sides of this equation by pif, we get:

1

p +1

i = 1

Equation17.19 is known as the mirror equation for spherical mirrors, and this

expression holds when we interchange p and i, i.e when we can replace the object

O by the image I and vice versa For a given value of f, we notice the following for concave mirrors:

• When p > f , the image distance i is positive A positive value of i means that the

image is real and inverted See Fig.17.16a,b for images smaller or larger than the

object

• When p < f , the mirror equation is satisfied by a negative value of the image

dis-tance i The negative image disdis-tance means that the image is virtual When we

extend two rays from the object we find that the virtual image is upright and enlarged , see Fig.17.16d

If we use this sign convention in the lateral magnification Eq.17.13, then we can also

write M as follows:

M= h

h = −i

We get an upright image for positive values of M and an inverted image for negative

values of M as shown in Fig.17.16

F

O

(b) (a)

Front

c F

O I

Front

c F

O

(c)

Front

C

c

F

O

(d)

Front

image

I

(i = +∞)

Real image

Real image

I

and C (c) The object at F (d) The object inside the focal point F and its virtual upright image I

17.5.2.2 Convex Mirrors

Convex mirrors like those shown in Fig.17.17are called diverging mirrors The

images formed by these types of mirrors are virtual because the reflected rays appear

to originate from an image behind the mirror Furthermore, the images are always

upright and smaller than the object Because of this feature, these types of mirrors

are often used in stores to prevent shoplifting

C

F

R

Virtual focal point

Principal

axis

O

Virtual image

I

Convex mirror

f

c

than the object

17.5 Formation of Images by Reflection 581

We can use Eqs.17.19and17.20for either concave or convex spherical mirrors

if we stick to the sign conventions presented in Table17.2 This table gives the sign

conventions for the quantities f, i, h, and M.

Quantity Symbol Positive values when Negative values when Focal length f The mirror is concave The mirror is convex

Image location i The image is in front of The image is in behind the

mirror (real image) mirror (virtual image)

Image height h The Image is upright The Image is inverted Magnification M The Image is upright The Image is inverted

aThe object location p and its height h are both positive

Example 17.5

A concave mirror has a focal length of 10 cm Locate and describe the image

formed by an object having distances: (a) p = 25 cm, (b) p = 15 cm, (c) p = 10 cm, and (d) p= 5 cm

Solution: Concave mirrors have a positive focal length, i.e f = +10 cm (a) To

find the image distance, we use Eq.17.19as follows:

1

p +1

i = 1

25 cm+1

25 cm 1

i =25 cm− 10 cm

The positive sign of i indicates that the image is real and located on the front

side of the mirror The magnification of the image can be determined using

Eq.17.20as:

M= −i

p = −50/3 cm

3

The negative sign of M indicates that the image is inverted In addition, the image

is reduced (66 7% of the size of the object) because the absolute value of M is

less than unity, see Fig.17.16a

(b) When p= 15 cm, the mirror and magnification equations give:

1

15 cm+1