Geometric Optics The Law of Reflection Optics The Study of Light.

Geometric Optics

The Law of Reflection

Optics

The Study of Light

Areas of Optics Geometric Optics

Light as a ray. Physical Optics

Light as a wave. Quantum Optics

Light as a particle.

Reflection Reflection occurs when light

bounces off a surface. There are two types of reflection

Specular reflection Off a shiny surface

Diffuse reflection Off a rough surface

Mirrors are great reflectors

Plane Mirror

shiny

+dark

-

shiny

+shiny

+dark

-dark

-

Spherical Mirrors

convex concave

Light Rays

Mathematical rays never bend

But light rays can, if they interact with materials!

Let’s take a closer look at a plane mirror

Plane Mirror

+ -Incident ray

Reflected ray

normal

A normal is a line that is

perpendicular to the mirror.

Ray tracing Ray tracing is a method of

constructing an image using the model of light as a ray.

We use ray tracing to construct optical images produced by mirrors and lenses.

Ray tracing lets us describe what happens to the light as it interacts with a medium.

Law of Reflection Lab

a) Hold a plane mirror upright on a sheet of graph paper that is on top of cardboard. Stick three pins through the graph paper on the same side of the mirror such that they appear to your eye to be in a straight line. (See the whiteboard for details)

b) Draw a normal to the surface of the mirror, as well as the incident and reflected rays.

c) Measure the angles of incidence and reflection.d) Repeat two or three times with different angles.e) Tabulate your angles of incidence and reflection.What can you say about the angles of incidence and

reflection?TURN IN ONE DRAWING PER GROUP. Include each

person’s name and period number.

Law of Reflection

The angle of incidence of reflected light equals the angle of reflection.

r = I

Note that angles are measured relative to a normal to the mirror surface.

shiny (+) dark (-)

plane mirrorlight source

incidentray

normal

reflectedray

r

i

Sample Problem A ray of light reflects from a plane mirror with

an angle of incidence of 37o. If the mirror is rotated by an angle of 5o, through what angle is the reflected ray rotated?

Solution

i = 370

r = 37050

50

i = 420

r = 420

42o + 5o = 47o relative to horizontal47o - 37o = 10o rotation of reflected ray

Sample Problem Standing 2.0 m in front of a small

vertical mirror, you see the reflection of your belt buckle, which is 0.70 m below your eyes What is the vertical location of the mirror

relative to the level of your eyes? If you move backward until you are 6.0 m

from the mirror, will you still see the buckle, or will you see a point on your body that is above or below the buckle?

Solution

buckle

mirror

eye

The mirror must be 0.35 m below the level of your eye.If you move backward, you’ll still see the belt buckle, since the triangles will still be similar.

0.7

0 m

0.3

5 m

0.3

5 m

2.0 m r

i

Wednesday, August 30, 2006

The Plane Mirror and Optical Images

Optical images Nature

real (converging rays) virtual (diverging rays)

Orientation upright inverted

Size true enlarged reduced

Ray tracing: plane mirror Construct the image using two rays.

+ -

object5 cm

Image-5 cm

Reflected rays are diverging.

Extend reflected

rays behind mirror.

Name the image:Virtual, upright, true size

Spherical mirrors There are two types of spherical mirrors

shiny shiny

concave convex

+ + --(where reflected rays go) (where reflected rays go) (dark side)(dark side)

Focal length, f, is positive Focal length, f, is negative

Parts of aSpherical Concave Mirror

Principle axis

+ - These are the

main parts of a spherical concave mirror.

The focal length is half of the radius of curvature.

The focal length is positive for this type of mirror.

R = 2f

Focus fCenter

R

Identification of the focus of a spherical concave mirror

+ - Rays parallel to the principle axis all pass through the focus for a spherical concave mirror.

Ray tracing: spherical concave mirror

The three “principle rays” to construct an image for a spherical concave mirror are the p-ray, which travels parallel to the

principle axis, then reflects through focus. the f-ray, which travels through focus, then

reflects back parallel to the principle axis. the c-ray, which travels through center, then

reflects back through center. You must draw two of the three principle

rays to construct an image.


Construct the image for an object located outside the center of curvature.

It is only necessary to draw 2 of the three principle rays!

C F

Real, Inverted, Reduced

Image

p

f

c

C F

Real, Inverted,

True Image


Construct the image for an object located at the center of curvature.

Name the image.

C F

Real, Inverted, Enlarged

Image


Construct the image for an object located between the center of curvature and the focus.

Name the image.

C F

No image is formed.

Construct the image for an object located at the focus.


C F

Virtual, Upright,

Enlarged Image

Construct the image for an object located inside the focus.

Name the image.


Problema) Construct 2 ray diagrams to illustrate what

happens to the size of the image as an object is brought nearer to a spherical concave mirror when the object outside the focus.

b) Repeat part a) for an object which is brought nearer to the mirror but is inside the focus.

Solution a)

The image becomes larger when you move the object closer.

Solution b)

The image becomes smaller when you move the object closer.

Mirror equation #1 1/si + 1/so = 1/f

si: image distance so: object distance f: focal length

Mirror equation # 2 M = hi/ho = -si/so

si: image distance so: object distancem hi: image height ho: object height M: magnification

Sample Problem A spherical concave mirror, focal length 20 cm,

has a 5-cm high object placed 30 cm from it.a) Draw a ray diagram and construct the image.

c) Name the image

Sample Problem A spherical concave mirror, focal length 20 cm,

has a 5-cm high object placed 30 cm from it.b) Use the mirror equations to calculate

i. the position of image

ii. the magnification

iii. the size of image

Parts of aSpherical Convex Mirror

These are the main parts of a spherical convex mirror.

The focal length is half of the radius of curvature, and both are on the dark side of the mirror.

The focal length is negative for this type of mirror.

Principle axis

CenterFocus

+ -

Ray tracing: spherical convex mirror

Construct the image for an object located outside a spherical convex mirror.

Name the image.

F C

Virtual, Upright

, Reduce

d Image

Problem Construct 2 ray diagrams to illustrate what

happens to the size of the image as an object is brought nearer to a spherical convex mirror.

Problem A spherical concave mirror, focal length 10 cm,


Problem A spherical concave mirror, focal length 10 cm,


i. the position of imageii. the magnificationiii. the size of image

c) Name the image

Problem A spherical convex mirror, focal length 15 cm,


Problem A spherical convex mirror, focal length 15 cm,


i. the position of imageii. the magnificationiii. the size of image

c) Name the image

SummaryConcave vs convex mirrors Concave

Image is real when object is outside focus

Image is virtual when object is inside focus

Focal length f is positive

Convex Image is always

virtual

Focal length f is negative

Definition: Refraction Refraction is the movement of light

from one medium into another medium. Refraction cause a change in speed of

light as it moves from one medium to another.

Refraction can cause bending of the light at the interface between media.

Index of Refraction

speed of light in vacuumspeed of light in medium

n = c/v

n =

When light slows down… …it bends. Let’s take a look at a

simulation. URL:

http://www.walter-fendt.de/ph14e/huygenspr.htm

http://www.walter-fendt.de/ph14e/huygenspr.htm

Snell’s Lawn1sin 1 = n2sin 2

n1

n2

1

angle of incidence

2

angle of refraction

n1 < n2

When the index of refraction increases, light bends toward the normal.

n1

n2

1

2When n1 < n2

1 > 2

n1 > n2

n1

n2

1

2

When the index of refraction decreases, light bends away from the normal.

When n1 > n2

1 < 2

Problem Light enters an oil from the air at an angle of 50o with

the normal, and the refracted beam makes an angle of 33o with the normal.

a) Draw this situation.b) Calculate the index of refraction of the oil.c) Calculate the speed of light in the oil

Problem Light enters water from a layer of oil at an angle of

50o with the normal. The oil has a refractive index of 1.65, and the water has a refractive index of 1.33.

a) Draw this situation.b) Calculate the angle of refraction.c) Calculate the speed of light in the oil, and in the water

ProblemLight enters a prism as shown, and passes through the prism.

a) Complete the path of the light through the prism, and show the angle it will make when it leaves the prism.

b) If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism.

c) How would the answer to b) change if the prism were immersed in water?

30o

60oglass

air

Problem

Light enters a prism made of air from glass.a) Complete the path of the light through the prism, and show

the angle it will make when it leaves the prism.b) If the refractive index of the glass is 1.55, calculate the angle

of refraction when it leaves the prism.

30o

60o

glass

air

Dispersion The separation of white light into

colors due to different refractive indices for different wavelengths is called dispersion.

Dispersion is often called the prism effect.

Dispersion

Which color of light has the greatest refractive index?

Critical Angle of Incidence The smallest angle of incidence for

which light cannot leave a medium is called the critical angle of incidence.

If light passes into a medium with a greater refractive index than the original medium, it bends away from the normal and the angle of refraction is greater than the angle of incidence.

If the angle of refraction is > 90o, the light cannot leave the medium.

Critical Angle of Incidence

This drawing reminds us that when light refracts from a medium with a larger n into one with a smaller n, it bends away from the normal.

n1

n2

n1 > n2


This shows light hitting a boundary at the critical angle of incidence, where the angle of refraction is 90o. No refraction occurs!

n1

n2

c

r = 90o

n1 > n2

Ray reflects instead of refracting.


Instead of refraction, total internal reflection occurs when the angle of incidence exceeds the critical angle.

n1

n2

c

r = 90o

n1 > n2

Ray reflects instead of refracting.

Calculating Critical Angle

n1sin(1) = n2sin(90o) n1sin(c) = n2sin(90o) sin(c) = n2/ n1 c = sin-1(n2/n1)

Problem (separate sheet)A. What is the critical angle of incidence for a gemstone with

refractive index 2.45 if it is in air?

B. If you immerse the gemstone in water (refractive index 1.33), what does this do to the critical angle of incidence?

Lenses There are two types of lenses.

converging

++-

-(where refracted rays go)

Focal length, f, is positive

Focal length, f, is negative

(where refracted rays go)

diverging

Thicker in middle

Thinner in middle

Lens ray tracing Ray tracing is also used for lenses. We use the same principle rays we

used for mirrors. the p-ray, which travels parallel to the

principle axis, then refracts through focus. the f-ray, which travels through focus, then

refracts parallel to the principle axis. the c-ray, which travels through center and

continues without bending. You must draw 2 of the 3 principle rays.

Lens Equations We use the same equations we

used for mirrors 1/si + 1/so = 1/f M = hi/ho = -si/so

Identification of the focus

All rays parallel to the principle axis refract through the focus of a converging lens.

F

Ray tracing: converging lens

Construct the image for an object located outside 2F.

It is only necessary to draw 2 of the three principle rays!

C F

Real, Inverted, Reduced

Image

F2F 2F

+-

p cf

Construct the image for an object located at 2F.

C F

Real, Inverted,

True Image

F2F 2F

+-


Construct the image for an object located between F and 2F.

C F

Real, Inverted, Enlarged

Image

F2F

+-


Construct the image for an object located at the focus.

C F

No image

F2F

+-


Construct the image for an object located inside the focus.

C F

Virtual, Upright,

Enlarged Image

F

+-


For converging lenses f is positive so is positive si is positive for real images and

negative for virtual images M is negative for real images and

positive for virtual images hi is negative for real images and

positive for virtual images

Diverging lens

Construct the image for an object located in front of a diverging lens.

C F

Virtual, Upright,

Reduced Image

F

+-

For diverging lenses f is negative so is positive si is negative M is positive and < 1 hi is positive and < ho

Problem A converging lens, focal length 20

cm, has a 5-cm high object placed 30 cm from it.

a) Draw a ray diagram and construct the image.

b) Use the lens equations to calculatei. the position of imageii. the magnificationiii. the size of image

c) Name the image

Solution a) and c)

C FF

+-a)

c) Real, Inverted, Enlarged

Solution b)

i. 1/si + 1/so = 1/f

1/si + 1/30 = 1/20

1/si = 1/20 - 1/30 = 3/60 –2/60 = 1/60

si = 60 cm

ii. M = - si/so = -60/30 = -2

iii. M = hi/ ho

hi = M ho= (-2)5 = -10 cm

Problem A converging lens, focal length 10




c) Name the image

Solution a) and c)

a)

c) Virtual, Upright, Enlarged

C FF

+-

Solution b)

i. 1/si + 1/so = 1/f

1/si + 1/5 = 1/10

1/si = 1/10 - 1/5 = 1/10 –2/10 = -1/10

si = -10 cm

ii. M = - si/so = -(-10)/5 = 2

iii. M = hi/ ho

hi = M ho= (2)2 = 4 cm

Problem A diverging lens, focal length -15




c) Name the image

Solution a) and c)

a)

c) Virtual, Upright, Reduced

C FF

+-

Solution b)

i. 1/si + 1/so = 1/f

1/si + 1/10 = 1/(-15)

1/si = -1/10 - 1/15 = -3/30 – 2/30 = -5/30 = -1/6

si = -6 cm

ii. M = - si/so = -(-6)/10 = 0.6

iii. M = hi/ ho

hi = M ho= (0.6)(4) = 2.4 cm

Geometric Optics The Law of Reflection Optics The Study of Light.

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