Geometrical Optics – Part II

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Geometrical Optics – Part II Chapter 24 1

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Geometrical Optics – Part II. Chapter 24. Going Backwards. Stuff. We continue with mirrors and lenses and even refractive surfaces. Quiz on Friday - PowerPoint PPT Presentation

Transcript of Geometrical Optics – Part II

Page 1: Geometrical Optics – Part II

Geometrical Optics – Part II

Chapter 24

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Page 2: Geometrical Optics – Part II

Going Backwards

2

'

0'

1

)2

(1/s 2

2

'

11

ss

R

Rs

Rss

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Stuff We continue with mirrors and lenses and even

refractive surfaces. Quiz on Friday For a while, office hours will be in, of all

places, my office. We really don’t need MAP-318 except before exams. And the hours are too confusing.

Next Exam is on Wednesday, December 2nd. I give up on the remaining evil clickers.

Clicker grade=0. Let’s move on.

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Page 4: Geometrical Optics – Part II

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When the Center of Curvatureis on the same side of theoutgoing ray, RR is positive.

Otherwise, if the center of curvature is not on the same sideas the outgoing ray, RR is negative.

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Concave Mirror/Paraxial Approximation

Consequently

Rss

R

h

s

h

s

h

2

'

11

2

'

MIRROREQUATION

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Image Formation

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0'

0

0

s

R

s

y’<0(from the diagram) so image is inverted.

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The geometry……

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s

s'-m

and '

'

s

y

so Triangles,Similar

diagram)in image inverted fromsign (- '

s

y

y

ym

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Let’s try an example

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Page 9: Geometrical Optics – Part II

A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 15 cm from the mirror.

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10 cm 5 cm

Normal to mirrorand bounces backalong incomingpath.

mmys

sm

s

fRss

4

5.'

5.7'

12

'

11

Page 10: Geometrical Optics – Part II

A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 10 cm from the mirror.

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10 cm 5 cm

mmys

sm

cms

fRss

8

0.1'

10'

12

'

11

Page 11: Geometrical Optics – Part II

A concave spherical mirror has a radius of 10 cm. Calculate the location and size of an 8mm object a distance 2.5 cm from the mirror.

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10 cm 5 cm

eyemmys

sm

cms

fRss

8

0.2'

5'

12

'

11

virtualimage

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The Concave Mirror

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More Convex Mirror

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Page 14: Geometrical Optics – Part II

Graphical Methods are very useful to check your work.

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Moving on to refractive surfaces

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Spherical Refractive Surfaces

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air glass

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A closer look atthe Math ….

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bbaa

b

a

nn

)(

'

s

h

b

a

b

aab n

n

n

nR

h

s

h

Ignoring

R

nn

s

n

s

nR

hnn

s

hn

s

hn

bnnn

nnnn

n

n

abba

abba

baab

aabb

b

a

)(

'

)('

)(

)(

Page 18: Geometrical Optics – Part II

No for the height of the image

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sn

sn

y

ym

s

ynn

s

ynn

s

ys

y

s

a

bbbaaa

b

aa

'''

''

'

tan

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Check this out – how big is R?

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Page 20: Geometrical Optics – Part II

From the math:

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0'

)(

'

s

n

s

nR

nn

s

n

s

n

ba

abba

sn

sn

y

ym

s

a ''

1

1'

''

0'

m

sn

sn

snsns

n

s

ns

n

s

n

a

b

ba

ba

ba

Page 21: Geometrical Optics – Part II

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Page 22: Geometrical Optics – Part II

The Thin Lens We ignore the

thickness of the lens.

We will use mostly geometrical methods.

Any ray that bends is assumed to bend only once at the center of the lens.

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From whence it came

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Surface 1

Surface 2

n=1 n=1.5 n=1

Surface 2n>1

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The thin lens - geometry

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parallel

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More Geometry

Lens is thin Actual thickness of the lens is ignored.

Image from first surface provides the object for the second surface.

Paraxial Ray Approximation sin(x)=tan(x)=x cos(x)=1 x is in RADIANS

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Page 26: Geometrical Optics – Part II

More Geometry

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Triangle PQO andtriangle P’Q’O aresimilar.

s

s

y

y ''

We will show that fora very thin lens:

F1=F2=f

fss

s

sm

f

fs

y

y

or

fs

y

f

y

1

'

11

'''

'

'

AOP'at Looking

The Thin Lens Equation

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This, of coursedepends on where the

object is placed with respect to f.

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Thin Lens (con’t)

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Image thatwould form

if material “a”was all on this

side of the lens.

Object for secondsurface.

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Procedure for equation

•Solve for image position for first surface•Use image as object for the second surface.•Use the refraction equation in both cases.

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b

111

a

's

n

:surface secondFor

's

n

:#1 SurfaceConsider

R

nn

s

n

R

nn

s

n

bcc

abb

For a lens. na=nc=1So we can call the middle one just n

12 ' : ssNote

Mess with the algebra and you will get:

221

111

1

'

1

s'

n-

1

's

1

R

n

s

R

n

s

n

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FINALLY – with some algebra and obvious substitutions, we get:

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fRRn

ss

111)1(

'

11

21

The Lensmaker’s Equation

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Two Ways to do this STUFF Algebraically using the lens equation (with the

1/f if you know it) Using graphical Methods

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Graphical Methods:

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Graphical Methods:

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Most important case: converging lensObject to left of F1

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Most important case: converging lens

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Most important case: converging lens

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Most important case: converging lens

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Most important case: converging lens

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Most important case: converging lens

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