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Refraction

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Snell’s Law: n1 sin(1) = n2sin(2)

where n1 = c/v1and 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.

- DISPERSION
- rainbows & prisms break white light into component colors

- 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

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.

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

lens with some shape for front & back

point

source

of light

screen

s’ = image distance

s = object distance

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

Play with the lens that is handed out

Does it act like a magnifying glass?

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

Play with the lens that is handed out

Does it act like a magnifying glass?

Does it focus light from the night light?

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

Play with the lens that is handed out

Does 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)

Spherical shape is specified by a radius.

The smaller the sphere (smaller the radius),

the more curved is the surface!

R1

R2

R

R

In 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 lenses

SMALL ANGLES

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.

Therefore, 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 11 1

nair R1 R2ffs s’

where f is a distance called the focal length.

* {

}

+

=

+

* {

} =

&

=

+

+

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.

QUESTION: Can the focal length, f, benegative?

Can the focal length, f, be negative?

Can the radii, R1 & R2, be negative?

R1>0

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

R2>0

R1<0

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

R2<0

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)

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.

If 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.

For 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 fand 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.

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.

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.

f > 0

s > 0 AND s > f

s’> 0 AND s’> f

f

f

s

s’

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

as s gets bigger,

s’gets smaller

(but still need s’ > f)

f

f

s

s’

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

as s approaches infinity

s’approaches f

f

f

s

s’

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

f > 0

s > 0 AND s > f

s’> 0 AND s’> f

f

f

s

s’

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

as s gets smaller,

s’gets larger

f

f

s

s’

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

as s approaches f,

s’approaches infinity

f

f

s

s’

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

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.

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

f

f

ray 1

- 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)

f

f

ray 1

ray 2

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

object

image

ray 1

f

f

ray 3

ray 2

- Note that a real image is formed.
- Note that the image is up-side-down.

object

image

ray 1

f

f

ray 3

ray 2

- By looking at ray 3 alone, we can see
by similar triangles that M = h’/h = -s’/s.

object

h

s’

image

h’<0

f

s

f

ray 3

note h’ isup-side-down

and so h’ < 0

This is the situation when the lens is used

in a camera or a projector. Image is REAL.

object

image

ray 1

f

f

ray 3

ray 2

What happens when the object distance, s, changes?

object

h

s’

image

ray 1

s

h’

f

f

ray 3

ray 2

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

object

s’

h

image

h’

ray 1

f

s

f

ray 3

ray 2

To 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?

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!

Let’s now look at the situation where

s< f (but s is still positive):

s

f

f

Example: 1 / 5 cm = 1 / 4 cm + 1 / s’

We can still use our three rays. Ray one goes

through the focal point on the left side.

ray 1

s

f

f

Ray two goes through the focal point on the

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

ray 1

s

f

f

ray 2

Ray three goes through the center of the lens

essentially undeflected.

ray 1

h’

s

f

f

ray 2

s’

ray 3

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

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

ray 1

h’

s

f

f

ray 2

s’

ray 3

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

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

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

h’

s

f

f

s’

ray 3

This is the situation when the lens is used

as a magnifying glass! Image is VIRTUAL.

ray 1

h’

s

f

f

ray 2

s’

ray 3

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

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.

Are there any limits to the magnifying power

we can get from a magnifying glass?

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.

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!

- 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)

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.

A 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.

objectimage lens eye

In 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

objectimage lens eye

object distance for eye

A 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.

imageobject lens eye

In 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 distanceobject distance

imageobject lens eye

object distance for eye

- Use first lens (objective lens) to form
an image (use as a projector lens)

objective lens

Image1

Object 1

- Use second lens (eyepiece)
as a magnifying glass to magnify Image 1

objective

lens

eyepiece

Image 1

Object 1

Image 2

Object 2

M1 = -s1’/s1 M2 = -s2’/s2

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

L = s1’ + s2

s1

s1’

s2

objective

lens

1/s1 + 1/s1’ = 1/f1

eyepiece

s2’

1/s2 + 1/s2’ = 1/f2

NOTE: s2’ = -25 cm

so Mtotal < 0 !

Image 1

Object 1

Image 2

Object 2

- 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

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.

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.

M1 = -s1’/s1 M2 = -s2’/s2

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

L = s1’ + s2

s1

s1’

s2

objective

lens

1/s1 + 1/s1’ = 1/f1

eyepiece

s2’

1/s2 + 1/s2’ = 1/f2

NOTE: s2’ = -25 cm

so Mtotal < 0 !

Image 1

Object 1

Image 2

Object 2

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.

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.

To 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.

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 cms1’ = 27.5 cm.

We now put these values into our remaining two equations.

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 .

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

- Camera:focus by changes s’
- Projector:focus by changing s
- eye:focus by changing f (shape)
- magnifying lensfocus by changing s’
- what about seeing underwater?