Lecture7 (Chap 7)Updated

7/28/2019 Lecture7 (Chap 7)Updated

1/55

Light, Optic and Interference

The Reflection and Refraction of Light

Mirrors and Lenses

Optical Instrument

Principle of Linear Superposition


2/55

25.2The Ref lect ion of Lig ht

LAW OF REFLECTION

The incident ray, the reflected ray, and the normal

to the surface all lie in the same plane, and the angleof incidence equals the angle of reflection.


3/55

25.2The Ref lect ion of Lig ht

In specular reflection, the reflected rays are parallel to each

other.


4/55

25.3The Format ion o f Images by a Plane Mirror

The persons right hand becomes

the images left hand.

The image has three properties:

1. It is upright.

2. It is the same size as you are.

3. The image is as far behind the mirroras you are in front of it.


5/55


A ray of light from the top of the chess piece reflects from the mirror.

To the eye, the ray seems to come from behind the mirror.

Because none of the rays actually emanate from the image, it is

called a vir tual image.


6/55


The geometry used to show that the image distance is equal

to the object distance.


7/55


Conceptual Example 1Full-Length Versus Half-Length Mirrors

What is the minimum mirror height necessary for her to see her full

image?

(a) Equal to her height?

OR

(b) Equal to half her height?


8/55

25.4Spher ical Mirrors

If the inside surface of the spherical mirror is polished, it is a concave

mirror . If the outside surface is polished, is it a conv ex mir ror.

The law of reflection applies, just as it does for a plane mirror.

The pr inc ipal axisof the mirror is a straight line drawn through thecenter and the midpoint of the mirror.


9/55


A point on the tree lies on the principal axis of the concave mirror.

Rays from that point that are near the principal axis cross the axis

at the image point.


10/55


Light rays near and parallel to the principal axis are reflected

from the concave mirror and converge at the focal point.

The focal length is the distance between the focal point and

the mirror.


11/55


The focal point of a concave mirror is halfway between

the center of curvature of the mirror C and the mirror at B.

Rf21


12/55


Rays that lie close to the principal axis are called paraxial rays.

Rays that are far from the principal axis do not converge to a single

point. The fact that a spherical mirror does not bring all parallel

rays to a single point is known as spher ical abberat ion.


13/55


S


14/55


Rf21

When paraxial light rays that are parallel to the principal axis

strike a convex mirror, the rays appear to originate from the focal

point.

25 5 Th F t i f I b S h i l Mi


15/55

25.5The Format ion of Images by Spher ical Mirrors

CONCAVE MIRRORS

This ray is initially parallel to the principal axis

and passes through the focal point.

This ray initially passes through the focal point,

then emerges parallel to the principal axis.

This ray travels along a line that passes through

the center.



16/55


Image format ion and the pr incip le of revers ibi l i ty



17/55


When an object is located between the focal point and a concave mirror,and enlarged, upright, and virtual image is produced.



18/55


CONVEX MIRRORS

Ray 1 is initially parallel to the principal axis and appears to originate from

the focal point.

Ray 2 heads towards the focal point, emerging parallel to the principal axis.

Ray 3 travels toward the center of curvature and reflects back on itself.

25 5 The Format ion of Images by Spher ical Mirrors


19/55


The virtual image is diminished in size and upright.

The convex mirror always form a virtual image of the object no

matter where in front of the mirror the object is placed.


20/55

25 6 The Mirror Equat ion and Magni f icat ion


21/55

25.6The Mirror Equat ion and Magni f icat ion

These diagrams are used

to derive the mirror equation.

fdd io

111

o

i

o

i

d

d

h

hm

For concave mirrors,

(Usedi for virtual image)

Magnification Equation:



22/55


Example 5A Virtual Image Formed by a Convex Mirror

A convex mirror is used to reflect light from an object placed 66 cm infront of the mirror. The focal length of the mirror is -46 cm. Find the location

of the image and the magnification.

1

cm037.0cm66

1

cm46

1111

ii dfd

cm27id

41.0

cm66

cm27

o

i

d

dm



23/55


Summary of Sign Con vent ions for Spher ical Mir rors

mirror.concaveaforis f

mirror.convexaforis f

mirror.theoffrontinisobjecttheifis od

mirror.thebehindisobjecttheifis od

image).(realmirrortheoffrontinisobjecttheifis id

image).(virtualmirrorthebehindisobjecttheifis id

object.uprightanforis m

object.invertedanforis m


24/55

Exercise 1

A spherical Christmas tree ornament is 6.00 cm

in diameter. Determine the magnification of an

object placed 10.0 cm away from the ornament.

M = 0.130

26 1 The Index of Refract ion


25/55

26.1The Index of Refract ion

sm1000.3 8cLight travels through a vacuum at a speed

Light travels through materials at a speed less than its speedin a vacuum.

DEFINITION OF THE INDEX OF REFRACTION

The index of refraction of a material is the ratio of the speed

of light in a vacuum to the speed of light in the material:

v

cn

materialin thelightofSpeed

in vacuumlightofSpeed

Refraction is the incidence of change in speed as a ray of light goes

from one material to another causing the ray to deviate from its incident

direction

26 1 The Index of Refract ion


26/55

26.1The Index of Refract ion

26 2 Snells Law and the Refraction of Light


27/55

26.2Snell s Law and the Refraction of Light

SNELLS LAW OF REFRACTION

When light travels from a material withone index of refraction to a material with

a different index of refraction, the angle

of incidence is related to the angle of

refraction by

2211 sinsin nn

SNELLS LAW

26.2 Snells Law and the Refraction of Light


28/55


Example 1Determining the Angle of Refraction

A light ray strikes an air/water surface at an

angle of 46 degrees with respect to thenormal. Find the angle of refraction when

the direction of the ray is (a) from air to

water and (b) from water to air.

26.2 Snells Law and the Refraction of Light


29/55


54.0

33.1

46sin00.1sinsin

2

112

n

n (a)

(b)

332

96.0

00.1

46sin33.1sinsin

2

112

n

n

742

26.2Snells Law and the Refraction of Light


30/55

g

APPARENT DEPTH

Example 2Finding 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?



31/55

g

69.000.1

31sin33.1sin

sin 1

22

1

n

n

441

313.30.2tan 12



32/55

g

1

2

n

ndd

Apparent d epth,

ob server direct ly

above object



33/55

g

Conceptual Examp le 4On 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 swimmers eyes. To theswimmer, 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?

Answer: Greater

26.3Total Internal Reflect ion


34/55

When light passes from a medium of larger refractive index into one

of smaller refractive index, the refracted ray bends away from the

normal.

Crit ical ang le 21

1

2 sin nnnnc



35/55

Example 5Total 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 beamenter the air or will there be total internal reflection? (b) Repeat part (a)

assuming that the diamond is surrounded by water.



36/55

4.2442.2

00.1sinsin 1

1

21

n

nc(a)

(b)

3.3342.2

33.1sinsin 1

1

21

n

nc



37/55

Conceptual Examp le 6The 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 diamondexhibit such brilliance? Why does it lose much of its brilliance when

placed under water?

26.4Polar izat ion and the Reflect ion and Refract ion o f Ligh t


38/55

1

2tann

nB Brewsters law

26.6Lenses


39/55

Lenses refract light in such a way that an image of the light source is

formed.

With a converging lens, paraxial rays that are parallel to the principal

axis converge to the focal point.

26.6Lenses


40/55

With a diverging lens, paraxial rays that are parallel to the principal

axis appear to originate from the focal point.

26.6Lenses


41/55

26.7The Format ion of Images by Lenses


42/55

RAY DIAGRAMS



43/55

IMAGE FORMATION BY A CONVERGING LENS

In this example, when the object is placed further than

twice the focal length from the lens, the real image is

inverted and smaller than the object.



44/55

When the object is placed between F and 2F, the real image is

inverted and larger than the object.



45/55

When the object is placed between F and the lens, the virtual image is

upright and larger than the object.



46/55

IMAGE FORMATION BY A DIVERGING LENS

A diverging lens always forms an upright, virtual, diminished image.

26.8The Thin-Lens Equat ion and the Magni f ication Equat ion


47/55

fdd io

111

o

i

o

i

d

d

h

hm

Thin lens equation: Magnification equation:



48/55

Summary of Sign Convent ions for Lenses

lens.convergingaforis f

lens.divergingaforis f

lens.theofleftthetoisobjecttheifis od

lens.theofrightthetoisobjecttheifis od

image).(reallenstheofrightthetoformedimageanforis id

image).(virtuallenstheofleftthetoformedimageanforis id

image.uprightanforis m

image.invertedanforis m



49/55

Examp le 9The 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

m6.19m50.2

1

m0500.0

1111

oi dfd(a)

m0510.0id real image

(b) 0204.0m50.2m0510.0

o

i

ddm

m0347.0m70.10204.0 oi mhh

26.9Lenses in Combinat ion


50/55

The image produced by one lens serves as the object for

the next lens.

26.12The Compoun d Microscope


51/55

To increase the angular magnification

beyond that possible with a magnifying

glass, an additional converging lenscan be included to premagnify the

object.

Ang ular magni f icat ion o fa compound m ic roscope

eo

e

ff

NfL

M

26.13The Telescope


52/55

Angular magni f icat ion of

an astronom ical telescopee

o

f

fM

27.1The Princip le of Linear Superposi t ion


53/55

When two or more light waves pass through a given point, their electric

fields combine according to the principle of superposition.

The waves emitted by the sources start out in phase and arrive atpoint P in phase, leading to con struc t ive interference.

,3,2,1,012 mm



54/55

The waves emitted by the sources start out in phase and arrive at

point P out of phase, leading to destruct ive inter ference.

,3,2,1,021

12 mm



55/55

If constructive or destructive interference is to continue ocurring at

a point, the sources of the waves must be coherent sourc es.

Two sources are coherent if the waves they emit maintain a constant

phase relation.

Lecture7 (Chap 7)Updated

Documents

Transcript of Lecture7 (Chap 7)Updated