Download Check focal lengths of board ray optics set Equipment o Graph

Phys 233
Day 13, O3: Ray Optics & Lenses
Thurs. 10/25 O3: Lenses
RE-O3
Mon., 10/29
Tues. 10/30
HW8: O1: 4; O2: 7, 8; O3: 4, 8
RE-O4
PL7; Quiz 6: O1, O2, O3
RE-O5; Lab Abstract & Conclusions 2nd Draft
Thurs. 11/1
O4 Opt Systems
L7: Mirrors & analysis, (O3)
O5 Instruments
1
Check focal lengths of board ray optics set
Equipment
o Graph paper, rules, protractors
o Board Laser Optics set with extension cord
Reflection
o In water: Wave tank simulation http://www.falstad.com/ripple/ set for
setup:refraction.
o In light: Arbor Scientific optics kit & power supply
Plane & Spherical mirrors – big concave Mirror
o Laser & smoke in a can
o Asymmetric object, like my tea-pot.
o
O3 Introduction to Lenses
O3.1
Introduction to Lenses
O3.2
Ray Tracing and Numerical Approaches
Circle back and apply to mirrors too.
O3.3
Diverging Lenses – Virtual Images
O3.4
Thick Lenses
Last Time
Chapter 26 The Refraction of Light: Lenses and Optical Instruments
Introduction
26.1 The Index of Refraction
1
 vlight .vacuum c
o
o
 v light . c
26.2 Snell’s Law
o What repercussion does this have for light’s transmission?
 Refracts
 n1 sin 1 n 2 sin 2
where the n’s are called indices of refraction.
Spherical Mirrors
Last time we began looking at spherical mirrors. The fundamental principle involved is
simply that, r = - I . These angles are measured off the normal to the reflecting surface.
But if we curve our surface into a segment of a sphere, then rays hit at different locations
are reflected around the different local normals,
Phys 233
Day 13, O3: Ray Optics & Lenses
2
25.1
The Formation of Images by Spherical Mirrors
We’ve only considered two very special cases – light originates parallel and is reflected
into the focal point, and light originating at the focal point ends up parallel. But what if
the light originates somewhere else, not parallel?
Demo: Wave tank – create circular waves somewhere away from the focus point. See it
converge elsewhere & then continue on its way. It’s as good as if there were a source at this
point of convergence – there is an image of a source.
Real Image
o Let’s trace a few light beams coming from an object just out from the focal point.
o
o The four easiest to trace are the ones that have simple rules:
 1) Inbound ray parallel to axis will reflect through focal point
 2) Inbound ray through focal point will reflect parallel to axis
 3) Inbound ray along radial (line from center point) will reflect on itself.
 4) Inbound ray toward midpoint of mirror is flipped
o Okay, but what’s all this ray tracing good for? Where we mentally perceive a
point of an object to be is where the light rays entering our eyes appear to be
diverging from. For example, if you hold your finger up infront of you, how you
mentally judge the distance to your finger is how cross-eyed you have to go to
point both eyes at it – the closer the finger, the more cross-eyed. Now, laying all
three traced images on the same sketch shows us something:

o As far as the eye is concerned the light might as well have originated at this
second point. In fact, our minds expect light to just travel in straight lines, so we
are fooled into imagining the object to sit there. This is an image.
o Notice that we aren’t getting completely made the fool; the light does really
converge at image point, just like it does at the real object. Therefore, we
distinguish this as a real image.
o Magnification, how much bigger or smaller the image is than the object.
image.height hi
 m
object .height ho


looking at the picture, we can see that we have two similar right triangles
– same set of angles, just different lengths of sides. It may sound vaguely
familiar from trigonometry that the legs scale, i.e.
Phys 233
Day 13, O3: Ray Optics & Lenses
hi
Si '
(negative sign comes from the fact one height is
ho
So
measured up and the other down)
So magnification can be expressed in terms of either heights or
distances form the mirror.
hi
Si '
m
ho
So
o Mirror Equation
 Now, how does the image location depend upon the object location?
Drawing a few pictures, you can see that they are strictly related. Here’s
the easiest picture to use.


ho
f
hi’
Si’






So
The Red and Blue triangles are similar, that is, Again we have similar
triangles, so we have
( hi ) ho ( hi )
f
So
( hi ) ho ( hi )
f
So
So
h
Si '
S
but, i
, so we can replace ho with ho ( hi ) o so
ho
So
Si '
( hi ) ( hi ) ( hi )
f
Si
So
since all terms have the same factor of –hi, we can cancel that off and
simply have
1
f
1
Si '
1
So
3
Phys 233
Day 13, O3: Ray Optics & Lenses
4
o Example3: A concave mirror has a focal length of 42 cm. The image formed by
this mirror is 97 cm in front of the mirror. What is the magnification?



Quantities
f = 42 cm
Si’ = 97 cm
So = ?
m=?
Relations
hi
di
1
1
1
m
,
ho
do
di do
f
Algebra (I’ll pause along the way to find the object distance).
1
1
1
1
1 1 di f
di do
f
do
f di
di f
do
m
di f
97cm 42cm
d i f 97cm 42cm
di
97 cm
1 .3
do
74 cm
74cm
Virtual Image
o Now let’s trace some light rays coming from an object that is just in from the
focal point.
o Have them do this on their paper – trace four key rays.
o
o Notice that these rays do not cross paths again after leaving the image. All the
same, the eye expects them to a) be converged on the source and b) follow strictly
straight paths. Thus the only way the eye can make sense of the light it receives is
to imagine the source to be back behind the mirror where the rays received would
intersect if they went straight.
o Here we’re getting doubly fooled; not only is the object not where we imagine it,
neither do the light rays even pass through that point. We call what we see a
virtual image.
Comparing Image and Object
o You’ve probably noticed three basic differences between the image and the
object.
 location
 size
 orientation (real is upside down, virtual is right side up.)
o Given an object’s size, orientation, and location and a mirror’s radius of
curvature, you can trace a few representative rays, obeying r = - I, and learn the
image’s location, size, and orientation.
o You’ll be asked in one of your homework problems to do this.
Phys 233
Day 13, O3: Ray Optics & Lenses
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Concave Mirror
o Convex Mirror
 Consider this arrangement: (have them sketch along)
ho
f
C
Even though the light rays from the tip of the object can’t actually
penetrate the mirror to go through the focal point or the center, we can still
say that
1) A ray headed for the focal point will reflect parallel to the axis
2) A ray running parallel to the axis will reflect along the path out
from the focal point
3) A ray headed toward the center will reflect back on itself.
4) A ray that hits the mid point of the mirror will just flip
ho
Location of
image
f
C
Now, the three reflect rays don’t actually criss-cross at some point but they radiate out as if
originating at the location behind the mirror. This is a “virtual” image since the light rays don’t
really go there.


To derive the mathematical relationship between the focal distance, the
image distance, and the object distance, we could go through the same
work as we just did for the concave mirror; what we’d find is almost the
same equation except that the signs of the focal length and image distance
terms would be flipped. The same mirror / lens equations hold, but
different signage.
The reading gives some rules on page 13; the “virtual object” seems pretty
unphysical now, but they’ll make sense when we look at compound mirror
or lens systems in a later chapter – then the image produced by one mirror
can be thought of as the object that the next mirror images. Should the
second mirror itself be in the way, so the rays from the first mirror never
actually converge to form the first image, then that unformed image is a
‘virtual object’ for the second mirror.
Concave
Convex
Real Object / Image (rays really go through, in front of mirror)
Virtual Object / Image (rays don’t really go through, behind mirror)
f>0
f<0
S>0 / S’>0
S<0 / S’<0
Example. Say we have an object that’s 16cm from a convex lens of focal length 12 cm; where’s
the image?
1/16 + 1/si = 1/(-12cm) …
Phys 233
This Time
O3.1
Day 13, O3: Ray Optics & Lenses
6
Introduction to Lenses & Why they work
Why Lenses work.
o Last time, we thought about how light reflects from a surface, the law was
simply that of equal angles of incidence and reflection relative to the surface.
So, if you have a bent surface, rays could be reflected in whatever direction
you wanted. If you bent it specifically into a parabola, or it’s usually good
enough to use a circle, then rays that were coming in parallel to each other can
be aimed to converge to a single point or diverge as if from a single point, and
rays radiating from a point on an object can be reflected to radiate as from
another point; thus mirrors can be used to create images.
 We’ll pick back up with this in a little bit.
o We also met Snell’s law for refraction. So, when a light beam hits a surface,
the angle at which it passes into the material depends upon what we call the
two materials’ indices of refraction.
o Demo: laser shone into plexiglass block on board, see it bend on entry and
exit.
o Again, the surfaces can be sculpted into segments of paraboloids or spheres so
that parallel rays in to the material emerge converging down to a single point
or diverging as if from a single point. Thus lenses can be used to create
images.
o Demo: lasers rays shone through Plexiglas converging lens.
26.6The Formation of Images by Lenses
Now let’s say that we have an object, from which light radiates in all directions. Let’s
see what these two styles of lenses do with the light they intercept.
Ray Tracing
o Converging
 Object outside focal length
1) Parallel in, through focus out
2) Through focus in, parallel out
3) in through center, straight out
Putting these three representative traces together, we see that an image is reformed on the far
side of the lens from the object.
Four points are of note:
The Object is outside the focal point
The image is shrunk
Phys 233
Day 13, O3: Ray Optics & Lenses
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The image is on the far side of the lens from the object
The image is upside-down
The image is real: the light really radiates out from its location
ClassWork. An object is located 9.0 cm in front of a converging lense (f = 6.0 cm). Using
an accurately drawn ray diagram (graph paper with each grid representing 1cm), determine
where the image is located.
45.
SSM REASONING The ray diagram is constructed by drawing the paths of two rays
from a point on the object. For convenience, we choose the top of the object. The ray that is
parallel to the principal axis will be refracted by the lens and pass through the focal point on the
right side. The ray that passes through the center of the lens passes through undeflected. The
image is formed at the intersection of these two rays on the right side of the lens.
SOLUTION The following ray diagram (to scale) shows that di = 18 cm and reveals a
real, inverted, and enlarged image.
Scale:
3 cm
Ob ject
F
F
Image
Mirror / Lens Law
This is just geometry, and we can draw a mathematical relationship out of it. (if I just did this for
the mirrors, don’t do for lens)
f
ho +(-hi)
(-hi)
ho
(-hi)
Si
So
So
From the fist picture,
ho ( hi ) ( hi )
So
f
ho
So
( hi )
So
( hi )
f
Putting them together,
From the second picture
ho / S o
( hi ) / S i
ho
( hi )
So
Si
Phys 233
Day 13, O3: Ray Optics & Lenses
( hi ) S o
So Si
( hi )
So
8
( hi )
f
1
1
1
Si So
f
The exact same relationship as for concave mirrors!
Returning to the example, if we plug in the focal length of 6cm, the object distance of
9cm, lo and behold, we get an image distance of 18cm (actually work through to demonstrate
flipping).
Object inside Focal Length: Let’s bring the object in closer than the focal point
Have work, and then sketch through.
Example. Say the object is 25 cm from a lens with a 50 cm focal length. What’s the image
distance?
1 1 1
Si So f
1
1
1
Si
50cm
S i 25cm 50cm
Hm… a negative distance, what does that mean?
Demo/trace. Let’s set it up on the board and sketch it out on your graph paper.
Ah, the negative image distance just means that the image is on the same side as the object.
Of note
The object is closer than the focal point
The image is enlarged
The image is right-side up
The image is on the same side as the object
The image is virtual: the light does not actually radiate through the image
location.
Diverging: Now let’s look at the case for a diverging lens
1) Parallel in, as if from same focus out
2) toward other focus in, parallel out
3) in through center, straight out
Putting these all together we see
Phys 233
Day 13, O3: Ray Optics & Lenses
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The object is outside the focal point
The image is shrunk
The image is right-side up
The image is on the same side as the object
The image is virtual: the light does not actually radiate through the image
location.
Lens equation. Now, we could go through the same work as we did for the converging lens
relating similar triangles, and we’d derive almost the same relation can be derived for the
diverging lens as we did for the converging lens except we’d end up with a negative sign on the
focal length term. So, a handy way of handling that is to say you can use the same equation but
use a negative value for focal length.
Sign Convention
Real object / Image
Virtual object / Image
Converging Lens, Concave mirror
Diverging Lens, convex mirror
S o/ Si’ > 0
S o/ Si’ < 0
f>0
f<0
o Example 3. When a diverging lens is held 13 cm above a line of print as in the
figure, the image is 5.0 cm beneath the lens. What is the focal length of the lens?
46.
REASONING AND SOLUTION Using the thin lens equation, we find
1/f = 1/do + 1/di = 1/(13 cm) + 1/(–5.0 cm)
or
f = –8.1 cm
Ray tracing. Given the focal length of about “-8cm”, and an object 13 cm away, trace the
situation showing the three key rays to help locate the image.
Phys 233
Day 13, O3: Ray Optics & Lenses
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Thick Lens as two consecutive surfaces
The reading notes that it can be derived from Snell’s Law that, for a circularly curved, convex
interface between two materials, 1 & 2, then the object and image distances, indicies of
refraction, and radius are related by:
n1 n2 n2 n1
So Si '
R
One example is seeing where, within a spherical chunk of glass, an image would be formed,
Or taking the radius to be infinite, then you can find where someone in one medium (air) might
imagine something in another medium (a fish in water) is – the negative Si’ would just reflect
that the image is back on the opposite side from what’s drawn.
Another example is putting two of these back to back, then you’ve got a real thick lens.
So2
R2
So1
R1
Si2’
Si1’
This is the first example of essentially two optical elements in series, and also the first example
then of a ‘virtual object.’ The rays from the first surface head off to converge out at a location,
and that image that would get formed serves as a virtual object for the next curve.