Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 25

Adding the two Eqs.17.36and17.38gives:A converging lens of focal length 20 cm forms an image of an object of height 30 cm located at a distance 40 cm from the lens.. 2 Assume that the wa

Adding the two Eqs.17.36and17.38gives:

A converging lens of focal length 20 cm forms an image of an object of height

30 cm located at a distance 40 cm from the lens Locate and describe the image.Draw two rays to locate the image

Solution: A converging lens has a positive value for its focal length, i.e.

f = +20 cm To find the image distance when p = 40 cm and f = +20 cm, we

use Eq.17.34as follows:

The image is real and on the back side because i is positive, inverted because M

is negative, and as large as the object, see Fig.17.25

Example 17.9

Repeat Example 17.8 using a diverging lens

Solution: The diverging lens would have f = −20 cm Thus:

592 17 Light Waves and Optics

Consequently, we have: M = −i

p = −(−40/3 cm)

The image is virtual and on the front side because i is negative, upright because

M is positive, and reduced because M is less than unity, see Fig.17.26

Example 17.10

An object is placed 20 cm from a symmetrical lens that has an index of

refrac-tion n = 1.65 The lateral magnification of the object produced by the lens is

M = −1/4 (a) Determine the type of the lens and describe the image (b) What

is the magnitude of the two radii of curvature of the lens?

Solution: (a) Using the lateral-magnification equation, we have:

must be outside the focal point and the image must be inverted and on the back

side of the lens

(b) To find the focal length f of the lens when p = 20 cm and i = +5 cm, we

use Eq.17.34as follows:

From the general lens-makers’ Eq.17.33, f is related to the radii of curvatures R1

and R2 of the two surfaces of the lens and its index of refraction n by the relation:

1

f = (n − 1)

1

R



For a symmetric lens, R1 and R2 have the same magnitude R If R1 is for the

surface where the center of curvature is in the back of the lens, and R2is for thesurface where the center of curvature is in the back of the lens, then using the signconvention of Table17.4, we have R1= +R and R2= −R Thus:

1

f = (n − 1)

1

Two thin coaxial lenses 1 and 2, with focal lengths f1 = +24 cm and f2= +9 cm,

respectively, are separated by a distance L= 10 cm; see part (a) of Fig.17.27 Anobject is placed 6 cm in front of lens 1 Locate and describe the image Draw thenecessary sketches to show how you can reach to the answer

Solution: We first ignore the presence of lens 2 and find the image I1produced

by lens 1 alone, see part (b) Fig.17.27 Equation17.34written for lens 1 leads tothe following steps:

For the second step, we ignore lens 1 and treat the image I1as a virtual object

O1in front of the second lens The distance p1 between the virtual object O1andlens 2 is:

p1= L − i1= 10 cm − (−8 cm) = 18 cm

Equation17.34written for lens 2 leads us to the following:

594 17 Light Waves and Optics

M2= − i

p = −18 cm

18 cm ⇒ M2= −1

The final image is real because i is positive and on the back side of lens 2, inverted because M is negative, and as large as the virtual object I1, see part (c) of Fig.17.27.

The overall magnification of the two lenses is:

M = −4/3 The final image is enlarged because |M| > 1 Notice that when L = 0, we get

M = −i/p as expected from Eq.17.41

17.7 Exercises

Section 17.2 Reflection and Refraction of Light

(1) A beam of light travels in vacuum and has a wavelengthλ = 500 nm The beam

passes through a piece of diamond(n = 2.4) What is the wave’s speed and

wavelength in diamond?

(2) Assume that the wavelength of a yellow beam of light in vacuum isλ = 600 nm,

and that the index of refraction of water is 1.33 (a) What is the speed of this lightwhen it travels in vacuum? (b) What is the speed of this light when it travels

in water? (c) What is the frequency of this light when it travels in vacuum?(d) What is the wavelength of this light when it travels in water? (e) What isthe frequency of this light when it travels in water?

(3) A beam of light having a wavelengthλ = 600 nm is incident perpendicular to a glass plate of thickness d = 2 cm and index of refraction n = 1.5 (a) How long

does it take a point on the beam to pass through the plate? (b) Calculate thenumber of wavelengths in the glass plate

(4) At what angle must a ray of light traveling in air be incident on acetone

(n = 1.38) in order to be refracted at 30◦?

(5) The index of refraction of alcohol is n = 1.4 (a) What is the speed of light in

alcohol? (b) Find the angle of refraction in alcohol assuming light meets theair-alcohol boundary at an angle of incidence of 60◦?

(6) A beam of light in air falls on a liquid surface at an angle of incidence of 55◦.

The liquid has an unknown index of refraction (a) If the beam is deviated by

20◦, what is the value of n? (b) What is the speed of light in this liquid?

596 17 Light Waves and Optics(7) A beam of light in air strikes a glass plate at an angle of incidence of 53◦ If

the thickness of the glass plate is 2 cm and its index of refraction is 1.6, whatwill be the lateral displacement of the beam after it emerges from the glass?(8) A beam of light in air falls on water at an angle of incidence of 45◦and then

passes through a glass block before it emerges out to air again The surfaces

of water and glass are parallel and their indexes of refraction are 1.33 and1.63, respectively (a) What is the angle of refraction in water? (b) What is theangle of refraction in glass? (c) Show that the incoming and outgoing beamsare parallel (d) At what distance does the beam shift from the original if thethickness of water and glass are both equal to 1 cm?

Section 17.3 Total Internal Reflection and Optical Fibers

(9) Diamond has a high index of refraction n = 2.42 To some extent, Diamond’s

“brilliance” is attributed to its total internal reflection Find the critical anglefor the diamond-air surface

(10) A beam of light passes from glass to water The index of refraction of glassand water are 1.52 and 1.333, respectively (a) What is the critical angle of

incidence in glass? (b) If the angle of incidence in glass is 45◦, what is the

angle of refraction in water?

(11) As it travels through ice, light has a speed of 2.307 × 108m/s (a) What is theindex of refraction of ice? (b) What is the critical angle of incidence for lightgoing from ice to air? (c) If the angle of incidence in ice is 45◦, what is the

angle of refraction in air?

(12) As the sun sets, its rays are nearly tangent to the surface of water, see Fig.17.28.The index of refraction of water is 1.33 (a) At which angle from the normalwould the fish in the figure see the sun? (b) Refraction at the water-air boundarychanges the apparent position of the Sun What is the apparent direction of theSun with respect to the fish (measured above the horizontal)?

(13) Figure17.29 shows a sketch of an Optical fiber cable that has a length

L = 1.51 m, diameter of D = 251 µm, and index of refraction n = 1.3 A ray

of light is incident on the left end of the cable at an angle of incidenceθ1= 45◦.

(a) What is the critical angle of incidence for light going from inside the cable

to air? (b) Find the angle of refractionθ2and the length Does the angle θ

fulfill the condition of total internal reflection? (c) How many reflections doesthe light ray make before emerging from the other end?

Fig 17.28 See Exercise (12) Direction of the sun

as seen by the fish

L

Air

1θ

2θ

Fig 17.29 See Exercise (13)

(14) Using the figure of Exercise 13, show that the largest angle of incidence θ1

for which total internal reflection occuring at the top surface is given by therelation sinθ1=(n2/n1)2− 1 Now find the value of this angle using the data

of Exercise 13

Section 17.4 Chromatic Dispersion and Prisms

(15) Find the difference in time needed for two short pulses of light to travel 12 kmthrough a fiber optics cable, assuming that the cable’s index of refraction for apulse of wavelength 700 nm is 1.5 and 1.53 for a pulse of wavelength 400 nm

(16) A monochromatic light ray is incident from air (n1 = 1) onto one of the faces

of an equilateral prism that has an index of refraction n2 = 1.5, see Fig.17.30

If the angle of incidenceθ1is 40◦, then at what angle from the normal would

this ray leave the prism?

598 17 Light Waves and Optics

Fig 17.30 See Exercise (16)

Air n

Glass n

60 60

°

(17) A narrow beam of white light is incident from air onto a plate of fused quartz

at an angle of incidenceθ1= 60◦; see Fig.17.31 The index of refraction of

quartz for violet and red light is nV = 1.470 and nR= 1.458, respectively Find

the angular widthδV −δRbetween the violet and red light rays inside the quartz

Fig 17.31 See Exercise (17)

(18) A prism has an index of refraction n = 1.5 and an apex angle A = 30◦ The

prism is set for minimum deviation by allowing a ray of monochromatic light

to pass through it symmetrically, as shown in Fig.17.32 (a) Find the angle ofminimum deviationδm (b) Find the value of the angle of incidence θ1.

Fig 17.32 See Exercise (18)

Air

Glass

1θ

m

δ

30°

Section 17.5 Formation of Images by Reflection

(19) The height h of a man is 200 cm The top of his hat t, his eyes e, and his feet

f are marked by dots on Fig.17.33 In order for the man to be able to see his

entire length in a vertical plane mirror, he needs a mirror of height H, as shown The figure also shows two paths, one for the light ray leaving his hat t and entering his eyes e, and another for the light ray leaving his feet f and again entering his eyes e (a) Find the height H of the mirror (b) Use two rays to

make a geometric sketch for the location and the height of the man’s image

Fig 17.33 See Exercise (19)

e

f

H

a b

(21) A concave mirror has a focal length f = +0.2 m An object of height 3 cm is

placed 0.1 m along its principal axis Locate and describe the image formed bythe mirror

(22) Repeat Exercise 21 using a convex mirror

(23) Assume a spherical concave mirror has a positive focal length | f | Use the

mirror equation 1/p + 1/i = 1/| f | to determine where an object must be placed

if the image created has the same size as the object, i.e when M = | − i/p| = 1.

(24) Assume a spherical convex mirror has a negative focal length−| f | Use the

mirror equation 1/p + 1/i = −1/| f | to show that the condition M = |−i/p| = 1cannot not be satisfied

600 17 Light Waves and Optics(25) Six objects are located at the following positions from a spherical mirror:

(i) p = ∞, (ii) p = 15 cm, (iii) p = 10 cm, (iv) p = 7.5 cm, (v) p = 5 cm, and (vi) p = 2.5 cm Locate and describe the image for each object when the

spherical mirror is: (a) concave, with a focal length of 5 cm, (b) convex, with afocal length of 5 cm

(26) Repeat Exercise 25, this time sketching the lateral magnification M for each object’s location p.

Section 17.6 Formation of Images by Refraction

(27) (a) A cylindrical glass rod (n2 = 1.6) has a hemispherical end of radius

R = 2 cm An object of height h = 0.2 cm is placed in air (n1= 1) on the axis

of the rod at a distance p= 6 cm from the spherical vertex, see Fig.17.34

(a) Locate and describe the image (b) Repeat part (a) when p= 2 cm

Fig 17.34 See Exercise (27)

I C

(28) A spherical fish bowl filled with water (n1 = 1.33) has a radius of 15 cm.

A small fish is located at a horizontal distance p= 20 cm from the left side

of the bowl, see Fig.17.35 Neglecting the effect of the glass walls of the bowl,where does an observer see the fish’s image? What is the lateral magnification

Back Front

(29) (a) An object is placed 30 cm from a converging lens with a 10 cm focal length.Find the position of the image and its lateral magnification Is the image real orvirtual? Is it upright or inverted? (b) Repeat part (a) for an object placed 5 cmaway.

(30) Repeat Exercise 29 with a diverting lens

(31) Use a ray diagram to explain the results of Exercises 29 and 30

(32) An object is placed 20 cm from a symmetrical lens that has an index of

refrac-tion n = 1.6 The lateral magnification of the object produced by the lens is

M = 1/4 (a) Determine the type of the lens used and describe the image.

(b) What are the values of the two radii of curvature of the lens?

(33) A thin converging lens of focal lens f1= +15 cm is placed in contact with a

thin diverging lens of unknown focal length f2 Find f2when incident Sunrays

on the converging lens are focused by this combination at a point 25 cm behindthe diverging lens

(34) A converging lens of focal lens f1 = +2 cm is placed at a distance L = 4 cm

in front of a diverging lens of focal lens f2= −14 cm An object is placed atinfinite distance from the converging lens Where will the object be focused?

(35) Repeat Exercise 34 with f1 = +10 cm and f2= −16 cm

(36) Two lenses with focal lenses f1 = +16 cm and f2= +20 cm are at a distance

L= 64 cm apart An object is placed 48 cm in front of the first lens Locate anddescribe the image formed by the system

(37) Repeat Exercise 36 with L= 19 cm

(38) A converging lens of f2= +17 cm is placed behind a diverging lens of unknown

focal lens f1 by a distance L = 12 cm Find f1when parallel light rays strike thediverging lens and leave the converging lens parallel

(39) Repeat Exercise 38 when the two lenses exchange positions

(40) An object is moving with velocityv = −dp/dt toward a converging lens of focal length f such that p > f Find the image velocity di/dt as a function of p Find p when v = di/dt.

Interference, Diffraction 18

and Polarization of Light

In this chapter we treat light as waves to study interference, diffraction, and

polar-ization This study is known as wave optics or physical optics.

We found inChap 16that the superposition of two sound waves could be structive or destructive The same phenomena can be observed with light waves.When a resultant wave at a given position or time has an amplitude larger than

con-the individual waves, we refer to con-their superposition as constructive interference.

However, when a resultant wave has an amplitude smaller than the individual waves,

we refer to their superposition as destructive interference.

18.1 Interference of Light Waves

If you have two ordinary light bulbs, incandescent or fluorescent, the light waves theyemit have random phases These phases change in time intervals that are less than ananosecond apart; thus, the conditions for constructive and destructive interferenceare maintained for only very short time intervals, too short for our eyes to notice

Such light sources are said to be incoherent.

Spotlight

To observe detailed interference effects in light waves, the following conditions

must be fulfilled:

• The sources must be coherent, i.e., they must maintain a constant phase

with respect to each other.

• The sources must be monochromatic, i.e., of a single wavelength.

H A Radi and J O Rasmussen, Principles of Physics, 603 Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_18,

© Springer-Verlag Berlin Heidelberg 2013

A common method for observing interference is to allow a single monochromaticlight source to split to form two coherent light sources and then allow the light waves

from the two sources to overlap This can be achieved by using the diffraction of

light waves from a small opening as introduced in Fig.17.2 Figure18.1shows theoverlap of monochromatic coherent light waves after being diffracted from two slitswhenλ ≈ a, where a is the width of each slit.

Fig 18.1 Spreading of light

waves from each slit (which is

known as diffraction) ensures

overlapping of waves, and

hence interference effects can

be observed when the light

from the two slits arrive at a

viewing screen (which is not

shown in this figure)

18.2 Young’s Double Slit Experiment

Figure 18.2shows a schematic diagram of the apparatus used by Thomas Young

in 1801 to demonstrate the interference of light waves Plane monochromatic light

waves arrive at a barrier that has two parallel slits S1 and S2 These two slits serve as

a pair of coherent light sources because the emerging waves originate from the samewave front and hence have the same phase relationship

Diffraction of the light by the two slits sends overlapping waves into the regionbetween the barrier and the viewing screen When light waves from the two slits

combine constructively at any location on the screen, they produce a bright band.

On the other hand, when light from the two slits combine destructively at any location

on the screen they produce a dark band These bands are called fringes, and the pattern of bright and dark fringes is called an interference pattern Figure18.2bshows a representation of the interference pattern observed on the screen

18.2 Young’s Double Slit Experiment 605

Screen Barrier

Wave fronts from

Wave fronts from S1

Fig 18.2 (a) When light waves are diffracted from the two slits S1and S2, waves overlap and undergo

interference Constructive interference in the region between the barrier and screen is represented by red circles on the red lines Destructive interference is represented by the yellow lines (b) Representation of the photograph that we get for the interference pattern in Young’s double slit experiment

With the help of Fig.18.3a, we can quantitatively specify the positions of brightand dark fringes in Young’s experiment In this figure, we show the following:

• Light waves of wavelengthλ illuminating a barrier having two narrow slits

• The two slits S1 and S2 are separated by a distance d

• Point Q is half way between the two slits

• The central line QO between the barrier and the screen has a distance D (where

D  d)

• Point P on the screen makes an angle θ above the central line QO The central line

QO will be taken as a reference line for measuring positive angles above or below

the line

• The distance from P to S1 is r1, and the distance from P to S2 is r2.

The waves from S2 must travel a longer distance to reach point P than the waves starting at S1 This difference L in distance is called the path difference When

Wave Fronts

(b) (a)

Screen Barrier

Fig 18.3 (a) Locating the fringes for Young’s double slit experiment geometrically (the figure is not

to scale) (b) For the condition D  d, we can approximate rays r1and r2as being parallel, making an angleθ to the central line QO, and the path difference between the two rays is r2− r1= d sin θ

D  d, the rays r1 and r2 are approximately parallel Using Fig.18.3b the pathdifference will be:

Constructive interference (maximum light intensity) occurs at P when the two waves

are in phase (φ = 0, ±2π, ±4π, rad), or when the path difference d sin θ is an

integer multiple of the wavelength λ That is, when:

d sin θm = m λ (m = 0, 1, 2, )

MaximaBright fringes



(18.2)

The number m for a bright fringe is called the fringe order number The central

bright fringe atθm = 0, where m = 0, is called the zeroth-order maximum The first maximum on either sides of point O, where m = 1, is called the first-order maximum,

and so forth

Destructive interference occurs at P when the path difference d sin θ is an odd

multiple of half the wavelength That is when:

d sin θm = (m −1

2) λ (m = 1, 2, 3, )

MinimaDark fringes