Mirrors Physics 202 Professor Lee Carkner Lecture 20.

Mirrors

Physics 202Professor Lee

CarknerLecture 20

PAL #19 EM Waves Laser from air to glass to water to glass to

air

airn=1

airn=1

glassn=1.52

glassn=1.52

watern=1.33

1 2

34

5

PAL #19 EM Waves To find each angle, n1 sin 1 = n2 sin 2

sin-1 [(n1 / n2 ) sin 1]= 2

Input angle for each eqn is output angle from previous egn

2 = sin-1 [(1 / 1.52 ) sin ] =

3 = sin-1 [(1.52 / 1.33 ) sin ] =

4 = sin-1 [(1.33 / 1.52 ) sin ] =

5 = sin-1 [(1.52 / 1 ) sin ] =

Mirrors

Trace back the reflections of light from object O and it will converge to make image I I is distance i from mirror

Images in the “real world” are called real

Plane Mirror

has curvature of infinity Plane mirror images:

Are the same size as

the object

Are the same distance behind the mirror as the object is in front of it

i = -p Note that distances “in the

mirror” are negative

Spherical Mirrors

The center of curvature (C) is the center of the sphere that the mirror is a section of A distance r from the mirror

The distance to the center of the mirror from the focal point is the focal length (f)

f = ½ r

Concave and Convex Concave mirror:

The center of curvature is in front of

the mirror

The image is larger than the object

Convex mirror: The center of curvature is behind

mirror

The image is smaller than the object

Ray Drawing Can find image properties by drawing

rays from object to image First draw (to scale):

mirror (including curvature) center of curvature (at r)

Follow ray drawing rules for two rays from top of object Where they intersect is top of image

1) A ray that that is initially parallel to the central axis reflects through the focal point

2) A ray that passes through the center of curvature reflects back along itself

3) A ray that reflects from the mirror at the intersection with the central axis reflects symmetrically about the central axis

Mirror Equation Where are the images and how large are they?

When measuring from the center of the mirror:

i is the distance to the image

when I and F are on the back side of the mirror i and f are negative

1/p + 1/i = 1/f

Magnification If h is the height of the object and h’ is

the height of the image, then the magnification is given by:

|m| = h’/h

m = -i/p For m = 1, image and object are same size For |m| < 1, image is smaller If m is negative, image is inverted

Concave

Produces a upright, virtual image

Concave mirrors are used to provide magnification (e.g. a shaving or make-up mirror)

Produces an inverted, real image

A real image is projected onto something, it is not behind the mirror

Convex

For a convex mirror: The center of

curvature is behind the mirror

The image is closer to the mirror (|i| < p)

The image is smaller than the object

Next Time

Read: 34.7-34.8

What is the proper equation for I1?

A) ½ I0B) I0 cos2

C) I0 cos2 6

D) ½ I0 cos2

E) ½ I0 cos2 6

What is the proper equation for I2?

A) ½ I1B) I0 cos2

C) I0 cos2 6

D) ½ I0 cos2

E) ½ I0 cos2 6

What is the proper equation for I3?

A) ½ I2B) ½ I0 cos2

C) I0 cos2 6cos2 6

D) ½ I0 cos2 cos2 3

E) ½ I0 cos2 6cos2 6

If you removed the middle polarizer, what would be the

intensity I3?

A) 0B) ½ I0 cos2

C) I0 cos2 6cos2 6

D) ½ I0 cos2 cos2 3

E) ½ I0 cos2 6cos2 6

What is the direction of polarization of polarized sunglasses?

A) Left - rightB) Up - downC) At a 45 degree angle to the verticalD) One lens is up-down and one is left

rightE) They could be polarized in any

direction

Do polarized sunglasses work on light that is not glare (not reflected)?

A) No, the goggles do nothingB) Yes, but only if the light is polarized in

some other wayC) Yes, but they only reduce the light by

a small amountD) Yes, they make unpolarized light half

as brightE) No, but they still look cool