Flat Mirrors Consider an object placed in front of a flat mirror –Object O is placed a distance p in front of the mirror ( p = object distance) –Light.

Flat Mirrors• Consider an object placed in front

of a flat mirror– Object O is placed a distance p in front

of the mirror (p = object distance)– Light reflecting from mirror appears to

originate at point I• An image of object O is formed at a distance q behind the mirror

(q = image distance)• It is a virtual image (light does not really converge there)• When light actually converges at an image point, it is a real image

(can be projected on a screen)

– We can use (at least 2) rays to determine the image orientation and position

• Notice p = q for a flat mirror• Image is upright and virtual

Flat Mirrors• The height h of the object equals the image height h’• Lateral magnification M:

– General definition for any type of mirror– For flat mirrors, M = 1

– “Magnification” can mean enlargement or reduction in size in optics

• A flat mirror also produces an apparent left – right reversal– A waving left hand appears to be a waving right hand in

the mirror

h

hM

heightobject

heightimage

Example Problem #23.1

Solution:

Class interactive – Solution will be determined in class

Does your bathroom mirror show you older or younger than your actual age? Compute an order-of-magnitude estimate for the age difference, based on data that you specify.

Concave Spherical Mirrors• Consider light from an object O

striking a spherical concave mirror– If rays diverge at small angles,

they all reflect through same image point

– Large diverging angles mean rays intersect principle axis at different points, resulting in a blurred image

• Geometry of concave mirrors

p

q

h

hM

(from yellow and blue

similar triangles)

Concave Spherical Mirrors• Additional triangle geometry yields the mirror

equation:

– If an object is very far from the mirror, then 1 / p ≈ 0 and q ≈ R / 2

– In this case, the image point is called the focal point F and the image distance q is called the focal length f = R / 2

– Note that focal point focus point– The mirror equation can be written in

terms of focal length:

Rqp

211

fqp

111

Convex Spherical Mirrors• Geometry of a convex mirror

• We can use the same mirror equation as for concave mirrors, but we just need to use a particular sign convention for each quantity in the equation

Sign Conventions for Mirrors

Also, R > 0 (R < 0) for concave (convex) mirrors

Ray Diagrams for Concave Mirrors

Ray Diagrams for Convex Mirrors

• Note that the sideview mirror on the passenger side of a car is a convex mirror (hence the warning given on the mirror)

Example Problem #23.16

Solution (details given in class):

10.0 cm in front of the mirror

A convex spherical mirror with a radius of curvature of 10.0 cm produces a virtual image one-third the size of the real object. Where is the object?

Example Problem #23.19

Solution (details given in class):

(a) Concave

(b) 1.0 m in front of the mirror

A spherical mirror is to be used to form an image, five times as tall as an object, on a screen positioned 5.0 m from the mirror.

(a) Describe the type of mirror required.

(b) Where should the mirror be positioned relative to the object?

Images Formed by Refraction• Consider light from object O

refracting at a spherical surface between 2 transparent media– From Snell’s law and

geometry:

– Note that real images are formed on the side opposite that of the incident light

R

nn

q

n

p

n 1221

pn

qn

h

hM

2

1

Images Formed by Refraction• If the refracting surface is flat, R and we get:

• Now the (virtual) image is on the same side of the surface as the object

pn

nq

1

2

Interactive Example Problem: Lens Design 101

Animation and solution details given in class.

(PHYSLET Physics Exploration 34.3, copyright Pearson Prentice Hall, 2004)

Example Problem #23.25

Solution (details given in class):

2.00

A transparent sphere of unknown composition is observed to form an image of the Sun on its surface opposite the Sun. What is the refractive index of the sphere material?

Thin Lenses• A thin lens forms an image by

refraction of light– Has 2 refracting surfaces– Examples of thin lenses shown at right

• Lenses in group (a) converge parallel rays to a focal point on the opposite side of the lens

• Lenses in group (b) diverge parallel rays so they appear to originate from a focal point on the same side of the lens

(converging lens)(diverging lens)

Geometry of Thin Lenses

• Magnification by a lens

• Thin–lens equation

• Lens maker’s equation

p

q

h

hM

(same as for a mirror)

fqp

111 (same as for a mirror)

21

111

1

RRn

f( f = focal length in air)

(R1 (R2) = radius of curvature of front (back) surface)

( n = index of refraction of the lens material)

Sign Conventions for Thin Lenses

Ray Diagrams for Thin Lenses• Converging lenses

Ray Diagrams for Thin Lenses• Diverging lenses

Example Problem #23.35

Solution (details given in class):

(b) 39.5 mm

(a) 39.0 mm

A certain LCD projector contains a single thin lens. An object 24.0 mm high is to be projected so that its image fills a screen 1.80 m high. The object-to-screen distance is 3.00 m. (a) Determine the focal length of the projection lens. (b) How far from the object should the lens of the projector be placed in order to form the image on the screen?

Interactive Example Problem: Building a Converging Lens

Animation and solution details given in class.

(PHYSLET Physics Problem 35.11, copyright Pearson Prentice Hall, 2004)

Combinations of Thin Lenses• If 2 thin lenses are used to form an image, use the

following procedure:1. The image produced by the first lens is calculated as

though the second lens were not present

2. The image formed by the first lens is treated as the object for the second lens

3. The image formed by the second lens is the final image of the system

• The overall magnification is the product of the magnifications of the separate lenses

• This procedure can be extended to 3 or more lenses

Example Problem #23.40

Solution (details given in class):

Position = 9.26 cm in front of the 2nd lens

Magnification = +0.370

An object is placed 20.0 cm to the left of a converging lens of focal length 25.0 cm. A diverging lens of focal length 10.0 cm is 25.0 cm to the right of the converging lens. Find the position and magnification of the final image.

Example Problem #23.50

Solution (details given in class):The image is virtual, upright, located 25.3 cm behind the mirror, with an

overall magnification of +8.05.

Front of mirror Back of mirrorFront of lensBack of lens

The object shown above is midway between the lens and the mirror. The mirror’s radius of curvature is 20.0 cm, and the lens has a focal length of –16.7 cm. Considering only the light that leaves the object and travels first towards the mirror, locate the final image formed by this system. Is the image real or virtual? Is it upright or inverted? What is the overall magnification of the image?

HW Problems #23.51, 23.59#23.51:

#23.59:

Aberrations• Spherical aberration

– Light passing through lens at different distances from the principal axis is focused at different points

– Apertures are used to help narrow the incoming beam of light

• Chromatic aberration– Different wavelengths of

light refracted by a lens focus at different points

– Can be reduced by using combination of converging and diverging lenses