Refraction

Snell’s Law: n1 sin(1) = n2 sin(2)

where n1 = c/v1 and v1 = [1/]1/2

and = Ko where K = K() .

This means that n = n(), and this means that -transmitted depends on the frequency.

Thus different colors will be bent slightly differently. This is called DISPERSION.

Refraction and Dispersion

DISPERSION• rainbows & prisms break white light into

component colors

Refraction and Dispersion

DISPERSION• rainbows & prisms break white light into

component colors• signal degeneration (except thru vacuum):

signal pulses are Fourier series of frequencies

and each frequency travels at different speed

so pulse shape disperses

Refraction and Thin Lenses

Can use refraction to try to control rays of light to go where we want them to go.

Let’s see if we can FOCUS light.

Refraction and Thin Lenses

What kind of shape do we need to focus light from a point source to a point?

lens with some shape for front & back

screen

pointsourceof light

s = object distances’ = image distance

Refraction and Thin Lenses

Let’s try a simple (easy to make) shape: SPHERICAL.

Play with the lens that is handed outDoes it act like a magnifying glass?

Refraction and Thin Lenses

Let’s try a simple (easy to make) shape: SPHERICAL.

Play with the lens that is handed outDoes it act like a magnifying glass?

Does it focus light from the night light?

Refraction and Thin Lenses

Let’s try a simple (easy to make) shape: SPHERICAL

Play with the lens that is handed outDoes it act like a magnifying glass?

Does it focus light from the night light?

Does the image distance depend on the shape of the lens? (trade with your neighbor to get a different shaped lens)

Refraction and Thin Lenses

Spherical shape is specified by a radius.

The smaller the sphere (smaller the radius),

the more curved is the surface!

RR

R2R1

Refraction and Thin LensesIn the computer assignment Thin Lenses (Vol. 5,

#2), in the Introduction, the following relation is derived if we make a couple of assumptions:

Assumptions:THIN lensesSMALL ANGLES

{ }n n

n} * {

1

R R =

1

s s'glass air

air 1 2

1 1

Refraction and Thin Lenses

This is called the THIN LENS equation:

NOTE:

– all the parameters on the left (n,R1,R2) refer to how the lens is MADE;

– all the parameters on the right (s,s’) refer to how the lens is USED.

{ }n n

n} * {

1

R R =

1

s s'glass air

air 1 2

1 1

Refraction and Thin LensesTherefore, we break the THIN LENS

equation:

(nglass – nair) 1 1 1 1

nair R1 R2 s s’Into the LENS MAKERS equation and the LENS USERS equation:

(nglass – nair) 1 1 1 1 1 1

nair R1 R2 f f s s’where f is a distance called the focal length.

* { =+ +

* {

}

+ } = & = +

Refraction and the Lensmakers Eq.

Note that the order of R1 and R2 does NOT matter - which means that it doesn’t matter which side of the lens is front and which is back.

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

Refraction and the Lensmakers Eq.

QUESTION: Can the focal length, f, be negative?

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

Refraction and the Lensmakers Eq.

Can the focal length, f, be negative?

Can the radii, R1 & R2, be negative?

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

The NORMAL (convex) lens above is considered to have R1 and R2 both positive, and so f must be positive.

R1>0

R2>0

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

This (concave) lens is considered to have R1 and R2 both negative, and so f must be negative!

R1<0

R2<0

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

Refraction and the Lensmakers Eq.

What about these two shapes? Is f positive or negative?

HINT: which radius in each case is bigger?

• LEFT: |R1| < |R2|, R1 wins, R1>0 so f>0

(lens is thicker in middle)

• RIGHT: |R1| > |R2|, R2 wins, R2<0 so f<0 (lens is thinner in middle)

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

Negative focal lengths

Positive focal length lenses focus light – that is, they bend the rays so that the rays tend to converge.

Negative focal length lenses tend to make the light diverge rather than converge.

Often a negative focal length lens is used when one lens converges the light too much – the negative lens then makes the light converge less quickly.

The same formulas work for negative lenses as for positive lenses. We just have to be careful with the signs of s and s’.

The computer homework on Thin Lenses talks about negative focal length lenses and one of their uses: in eye glasses.

Designing LensesIf we specify the focal length for a lens, then we have

three parameters to specify: nglass, R1, and R2 .Our design will entail choosing one of the two possible

designs: for positive lenses either or .For the first case, , we could make the choice of

R1 = R2= R. This simplifies the above equation to become: . For the second case, ,

R1>0 and R2<0, we need to be careful. We can’t choose |R1| = |R2 | since this will give 1/f = 0. We also have to make sure that |R1| < |R2 | to keep our shape.

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

f

1 = }

R

2{*}

n

nn{

air

airglass

Designing LensesFor negative lenses, we again have two choices for

shapes similar to the choices for positive lenses.

For the first case, R1<0 and R2<0, we could make the choice of R1 = R2= R. This simplifies the above equation to become: , where both f

and R are negative. For the second case, R1>0 and R2<0, we again need to be careful. We can’t choose |R1| = |R2 | since this will give 1/f = 0. In this case, have to make sure that |R1| > |R2 | to keep our shape.

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

f

1 = }

R

2{*}

n

nn{

air

airglass

Designing Lenses: Examples

If we are to design a lens to have a positive focal length of 30 mm, then we can use the symmetrical design so that R1=R2=R. We need to choose one more parameter, so we can solve for the remaining one. Let’s choose nglass = 1.67.

Our equation now becomes:{(1.67 – 1)/1}*(2/R) = 1/(30 mm).

We can solve this for R:

R = (30 mm)*(0.67)*2 = 40.2 mm.

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

Designing Lenses: Examples

Let’s now choose a harder example: that of a negative focal length contact lens. Here we want R1>0 and R2<0, and we want |R1| > |R2 | . Let’s say we want a focal length of -55 mm, and we choose a clear plastic of index of refraction 1.67. We need to choose either R1 or R2. Let’s choose R2 = -100 mm.

Our equation becomes: (1.67 – 1)*{1/R1 + 1/(-100mm)} = 1/(-55mm) , or

(.67)/R1 – (.0067/mm) = -.0182/mm, or (.67)/R1 = -.0115/mm, or R1/(.67) = - 1mm/.0115 , or R1 = - .67mm/.0115 = -58.35 mm.

But this can’t work since we wanted R1>0 for a contact lens design.

We need to choose |R2| to be a smaller number (< .67* |f|) so that R1 can be positive. If we choose R2 = -30mm, then we get R1 = 161.4 mm.

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

Refraction and the Lens-users Eq.

1

f =

1

s +

1

s'

f f

ss’

s > 0 AND s > f

s’ > 0 AND s’ > f

f > 0

Example: 1 / 5 cm = 1 / 10 cm + 1 / 10 cm

Refraction and the Lens-users Eq.

1

f =

1

s +

1

s'

f f

ss’

as s gets bigger,s’ gets smaller(but still need s’ > f)

Example: 1 / 5 cm = 1 / 30 cm + 1 / 6 cm

Refraction and the Lens-users Eq.

1

f =

1

s +

1

s'

f f

s s’

as s approaches infinitys’ approaches f

Example: 1 / 5 cm = 1 / 200 cm + 1 / 5.128 cm

Refraction and the Lens-users Eq.

1

f =

1

s +

1

s'

f f

ss’

s > 0 AND s > f

s’ > 0 AND s’ > f

f > 0

Example: 1 / 5 cm = 1 / 10 cm + 1 / 10 cm

Refraction and the Lens-users Eq.

1

f =

1

s +

1

s'

f f

s

s’

as s gets smaller,s’ gets larger

Example: 1 / 5 cm = 1 / 6 cm + 1 / 30 cm

Refraction and the Lens-users Eq.

1

f =

1

s +

1

s'

f f

ss’

as s approaches f,s’ approaches infinity

Example: 1 / 5 cm = 1 / 5.1 cm + 1 / 255 cm

Refraction and the Lens-users Eq.

Before we see what happens when s gets smaller than f, let’s use what we already know to see how the lens will work.

Refraction and the Lens-users Eq.

f f

– Any ray that goes through the focal point on its way to the lens, will come out parallel to the optical axis. (ray 1)

ray 1

Refraction and the Lens-users Eq.

f f

– Any ray that goes through the focal point on its way from the lens, must go into the lens parallel to the optical axis. (ray 2)

ray 1

ray 2

Refraction and the Lens-users Eq.

f f

– Any ray that goes through the center of the lens must go essentially undeflected. (ray 3)

ray 1

ray 2

ray 3

object

image

Refraction and the Lens-users Eq.

f f

– Note that a real image is formed.

– Note that the image is up-side-down.

ray 1

ray 2

ray 3

object

image

Refraction and the Lens-users Eq.

f f

– By looking at ray 3 alone, we can see

by similar triangles that M = h’/h = -s’/s.

ray 3

object

image

s

h s’

h’<0

note h’ is up-side-downand so h’ < 0

Refraction and the Lens-users Eq.

f f

This is the situation when the lens is used

in a camera or a projector. Image is REAL.

ray 1

ray 2

ray 3

object

image

Refraction and the Lens-users Eq.

f f

What happens when the object distance, s, changes?

ray 1

ray 2

ray 3

object

image

s

s’

h’

h

Refraction and the Lens-users Eq.

f f

Notice that as s gets bigger, s’ gets closer to f and |h’| gets smaller.

ray 1

ray 2

ray 3

object

image

sh’

s’h

FocusingTo focus a camera, we need to change s’ as s

changes. To focus a projector, we need to change s as s’ changes. We do this by screwing the lens closer or further from the film (image) or slide (object).

But what about the eye? How do we focus on objects that are close and then further away with our eyes? Do we screw our eyes in and out like the lens on a camera or projector?

Focusing

But what about the eye? How do we focus on objects that are close and then further away with our eyes? Do we screw our eyes in and out like the lens on a camera or projector? - NO, instead our eyes CHANGE SHAPE and hence change f as s changes, keeping s’ the same!

Refraction and the Lens-users Eq.

f f

Let’s now look at the situation where

s < f (but s is still positive):

s

Example: 1 / 5 cm = 1 / 4 cm + 1 / s’

Refraction and the Lens-users Eq.

f f

We can still use our three rays. Ray one goes

through the focal point on the left side.

s

ray 1

Refraction and the Lens-users Eq.

f f

Ray two goes through the focal point on the

right side (and parallel to the axis on the left).

s

ray 1

ray 2

Refraction and the Lens-users Eq.

f f

Ray three goes through the center of the lens

essentially undeflected.

s

ray 1

ray 2

ray 3

s’

h’

Refraction and the Lens-users Eq.

f f

Notice that: s’ is on the “wrong” side, which

means that s’ < 0 , and that |s’| > |s| so f > 0.

s

ray 1

ray 2

ray 3

s’

h’

Example: 1 / 5 cm = 1 / 4 cm + 1 / -20 cm

Refraction and the Lens-users Eq.

f f

Notice that: h’ right-side-up and so h’ > 0,

M = h’/h = -s’/s . M > 0 (s’ < 0 but -s’ > 0).

s

ray 3

s’

h’

Refraction and the Lens-users Eq.

f f

This is the situation when the lens is used

as a magnifying glass! Image is VIRTUAL.

s

ray 1

ray 2

ray 3

s’

h’

Refraction and the Lens-users Eq.

The same lens can be used as:

• a camera lens: s >> f, s > s’,

M < 0, |M| < 1

• a projector lens: s > f, s’ > s,

M < 0, |M| > 1

• a magnifying glass: s < f, s’ < 0,

M > 0, M > 1

Refraction and the Lens-users Eq.

Notes on using a lens as a magnifying glass:

• hold lens very near your eye

• want IMAGE at best viewing distance

which has the nominal value of 25 cm

so that s’ = -25 cm.

Refraction and the Lens-users Eq.

Are there any limits to the magnifying power

we can get from a magnifying glass?

Refraction and the Lens-users Eq.

Are there any limits to the magnifying power

we can get from a magnifying glass?

– Recall: 1/s + 1/s’ = 1/f and M = -s’/s

where s’ = -25 cm.

– For biggest M, need smallest s which means smallest f.

Refraction and the Lens-users Eq.

For biggest M, need smallest s

which means smallest f.

Recall lensmakers eq:

• For smallest f, need smallest R’s.

• But smallest R’s mean very small size!

{n n

n R R} =

1

fglass air

air 1 2

} *{

1 1

Refraction and the Lens-users Eq.

• Magnifying glass has limits due to size

• As we will see in a little bit, magnifying

glass has limits due to resolving ability

• NEED MICROSCOPE (two lens system)

Systems of Two Lenses

Before we go to the microscope, let’s consider putting two lenses in a system. The basic idea is that the image of the first lens becomes the object for the second lens.

We (will see / have seen) in lab how this works with two lenses.

This also works if we consider the eye to have a lens and eyeglasses to be a second lens.

Near SightedA near sighted person can see things clearly if they are close, but

has trouble focusing on things far away.

When things are close, the eye has to bend the light a lot in order to focus, and the near sighted person’s eyes can do that. When things are far away, the eye doesn’t have to bend the light as much, but the near sighted person’s eyes have trouble with this.

What we want is a lens that will take a real object that is far away and make an image that is closer (but still on the same side of the lens).

The eye can then focus on this (closer) image.

object image lens eye

Near SightedIn this case, the object distance is positive and somewhat large,

while the image distance is negative (on the opposite side of where it normally would be) and smaller than the object distance. This image of the lens becomes the object for the eye, and it is close enough for the near sighted person to focus on.

So when we use the thin lens equation, 1/f = 1/s + 1/s’, we find that we need a negative focal length lens to correct for nearsightedness.

object distance image distance

object image lens eye

object distance for eye

Far SightedA far sighted person can see things clearly if they are far away,

but has trouble focusing on things close.

When things are close, the eye has to bend the light a lot in order to focus, and the far sighted person’s eyes have trouble with that. When things are far away, the eye doesn’t have to bend the light as much, and the far sighted person’s eyes can do this.

What we want is a lens that will take a real object that is near and make an image that is further away (but still on the same side of the lens).

The eye can then focus on this (further away) image.

image object lens eye

Far SightedIn this case, the object distance is positive and somewhat small,

while the image distance is negative (on the opposite side of where it normally would be) and larger than the object distance. This image of the lens becomes the object for the eye, and it is far enough away for the far sighted person to focus on.

So when we use the thin lens equation, 1/f = 1/s + 1/s’, we find that we need a positive focal length lens to correct for farsightedness.

image distance object distance

image object lens eye

object distance for eye

Microscope

• Use first lens (objective lens) to form

an image (use as a projector lens)

Object 1

objective lens

Image 1

Microscope

• Use second lens (eyepiece)

as a magnifying glass to magnify Image 1

Object 1

objective lens

Image 1Object 2

eyepiece

Image 2

Microscope

M1 = -s1’/s1 M2 = -s2’/s2

Mtotal = M1 * M2 = (s2’*s1’) / (s2*s1)

Object 1

objective lens

Image 1Object 2

eyepiece

Image 2

s1 s1’ s2

s2’

L = s1’ + s2

1/s1 + 1/s1’ = 1/f1

1/s2 + 1/s2’ = 1/f2

NOTE: s2’ = -25 cmso Mtotal < 0 !

Microscope

• DESIGN: 8 quantities:– s1, s1’, s2, s2’, f1, f2, L, M

• DESIGN: 4 equations:– two thin lens equations– equation for L– equation for M

• Usually know s2’ = -25 cm and M

• Get to choose two others

Example of designing a microscope

Design a microscope with a magnification of 100.

We have four equations: 1/fobj = 1/s1 + 1/s1’ , 1/feye = 1/s2 + 1/s2’, L = s1’ + s2 , and M = (s1’/s1)*(s2’/s2).

We have eight parameters: s1, s1’, s2, s2’, fobj, feye, L, and M.

We are given (or know) two of these: M = -100, and s2’ = -25 cm.

Example of designing a microscope

Note that the diagram for the microscope is on a previous slide that we repeat on the next slide.

Since we have four equations and six unknowns, we need to choose two of the six unknowns. However, we must make sure that only the Magnification, M, and the final image distance, s2’, are negative, and that the remaining six parameters are all positive.

Microscope Diagram

M1 = -s1’/s1 M2 = -s2’/s2

Mtotal = M1 * M2 = (s2’*s1’) / (s2*s1)

Object 1

objective lens

Image 1Object 2

eyepiece

Image 2

s1 s1’ s2

s2’

L = s1’ + s2

1/s1 + 1/s1’ = 1/f1

1/s2 + 1/s2’ = 1/f2

NOTE: s2’ = -25 cmso Mtotal < 0 !

Example of designing a microscope

At this point, let’s fill in the knowns into the four equations:1/fobj = 1/s1 + 1/s1’, 1/feye = 1/s2 + 1/s2’, or 1/feye = 1/s2 + 1/(-25cm) L = s1’ + s2 , andM = (s1’/s1)*(s2’/s2), or -100 = (s1’/s1)*(-25 cm/s2).

There are many ways to choose two unknowns. We could choose the two focal lengths, fobj and feye. We could choose the length and the focal length of the objective, L and fobj. For this example, I’m going to choose L and s1 since I want a reasonable length for the microscope and I want a small distance for the slide to be from the objective lens.

Example of designing a microscope

To be specific, let me specify: L = 30 cm, and s1 = 1 cm. The equations now

become:1/fobj = 1/1 cm + 1/s1’ 1/feye = 1/s2 + 1/(-25cm) 30 cm = s1’ + s2 , and-100 = (s1’/1 cm )*(-25 cm/s2).

Notice that this particular choice left all four equations with at least two unknowns in each equation. This is not a nice situation for doing the algebra, although it is certainly possible. Let’s try a different set of choices.

Example of designing a microscopeTo be specific, let me specify L and s2. There is an advantage

in specifying s2 in that this gives you a chance of evening out the magnifications due to the two lenses. If we split the magnification between the two lenses, each magnification should provide SQRT{M}. In this case, Meye should be about SQRT{M} = SQRT{100} = 10. With Meye = 10 =-s2’/s2 = --25 cm / s2 = 10, we would choose s2 to be 2.5 cm

L = 30 cm, and s2 = 2.5 cm. The equations now become:1/fobj = 1/s1+ 1/s1’ 1/feye = 1/(2.5 cm) + 1/(-25cm) 30 cm = s1’ + 2.5 cm , and-100 = (s1’/s1)*(-25 cm/2.5 cm).

Note that this choice gives us two of the equations with only one unknown: the second and third equations.

Example of designing a microscope

1/fobj = 1/s1+ 1/s1’ 1/feye = 1/(2.5 cm) + 1/(-25cm) 30 cm = s1’ + 2.5 cm , and-100 = (s1’/s1)*(-25 cm/2.5 cm).

From the second equation, 1/feye = 1/(2.5 cm) + 1/(-25cm)feye = 2.778 cm.

From the third equation, 30 cm = s1’ + 2.5 cm s1’ = 27.5 cm.

We now put these values into our remaining two equations.

Example of designing a microscope

1/fobj = 1/s1+ 1/(27.5 cm)1/(2.778 cm) = 1/(2.5 cm) + 1/(-25cm) 30 cm = 27.5 cm + 2.5 cm , and-100 = (27.5 cm/s1)*(-25 cm/2.5 cm).

Although the first equation has two unknowns, the fourth equation only has one, so we can solve for s1: -100 = (27.5 cm/s1)*(-25 cm/2.5 cm), or s1 = 2.75 cm.

Now we can substitute this value into the first equation and solve for our last unknown, fobj .

Example of designing a microscope

1/fobj = 1/(2.75 cm)+ 1/(27.5 cm)1/(2.778 cm) = 1/(2.5 cm) + 1/(-25cm) 30 cm = 27.5 cm + 2.5 cm , and-100 = (27.5 cm/2.75 cm)*(-25 cm/2.5 cm).

Solving the top equation for fobj gives fobj = 2.5 cm.

A solution to this design problem then is:

M = -100 L = 30 cm

fobj = 2.75 cm s1 = 2.75 cm s1’ = 27.5 cm

feye = 2.778 cm s2 = 2.5 cm s2’ = -25 cm

Using a lens:

• Camera: focus by changes s’

• Projector: focus by changing s

• eye: focus by changing f (shape)

• magnifying lens focus by changing s’

• what about seeing underwater?