Today’s Lecture
Lenses – Image Formation
Mirror and Lens Review
Imaging
f
l
Three principal rays emanating from every point of the object:
1. A ray parallel to the axis of the lens on the front side passes
through the focal point on the back side.
2. A ray that passes through the center of the lens and does not
get refracted.
3. A ray that passes through the focal point on the front side
emerges parallel to the axis of the lens on the back side.
Ray Diagram for Converging Lens
l
f
If the object is further away from the lens than the front focal point,
l > f , => the image is real, inverted, and located behind the lens.
If the object is located at more than 2f, l > 22f, => the image is
reduced. If the object is located at f < l < 2f, the image is
magnified. WHY?
What is different between a concave mirror and a converging lens?
Ray Diagram for Converging Lens
f
l
If the object is closer to the lens than the front focal point,
l < f, => the image is virtual, upright, magnified, and in front of
the lens
The image can only be viewed from behind the lens!
Since the image is larger than the object, this is
configuration for a magnifying glass!
Ray Diagram for Diverging Lens.
The situation is pretty much the same no matter where the
object is located.
l
f
The image is virtual, and can only be seen through the lens.
The image is upright and is always reduced in size compared with
the object.
Note the analogy with convex mirrors!
Lens Magnification
l
The two shaded triangles, blue and gold, are
right-angled and similar. Therefore, for the
lens magnification, M, we have:
In an exact analogy with mirrors, the “−” sign
is taken to signify the fact that the image is
inverted.
l’
Lens Equation
l
l’
Consider the similar right triangles defined by the angle θ. From
the magnification we know that –h’/h = l’/l. Since these triangles
are similar:
The “Lens equation” is the exact form as the mirror equation. l and l’ are
both considered positive for the case shown. Whereas l’ is positive in
the mirror equation when the image is in front of the mirror.
http://www.mtholyoke.edu/~mpeterso/classes/phys301/geomopti/lenses.html
Summary for
Converging Lens
Image real, inverted, reduced
(magnified for l < 22f )
Virtual, upright, magnified
Lens Magnification – Diverging lens
1 1 1
+ =
l l' f
f  0, ℓ − anything, ℓ ′  0 and |ℓ ′ |  ℓ;
M  −ℓ ′ /ℓ, M  1
Virtual, upright, reduced
To Summarize for both types of lenses:
What happens to the image if we put in an aperture?
If we remove the lens?
Top and bottom of image
are less bright.
Top half of image
is less bright
Who has seen the lens?!
I
O
I
O
Who has seen the lens?!
O
I
O
I
Lens aberrations, spherical:
Ideally, the surface of a lens should be parabolic…
Those are very difficult to make, though!
Spherical shape is rather close to a parabola for paraxial rays, which
are close to the axis of the lens.
The rays, which come at greater angles (further away from the axis)
converge in a different, closer point.
The image of the point O gets blurry…
Lens Aberrations, Chromatic
Regular glass materials are dispersive: indices of refraction are
different for different colors.
Rays of different colors are refracted through different angles and
converge in different points – the image gets blurry!
Remedies – try to use a single color illumination or a composite
lens compensating for the dispersion.
High end microscope objective collect light from wide angles
and compensate for all possible aberrations…
Imagine a camera with a single lens with a focal distance f = 35mm.
By how much and in what direction should the lens be moved to
move its focus from an object, which is far-far away to an object at a
distance of 1.5m? Far-far away means l = ∞, and l’= f = 35mm.
New distance l = 1.5m; l’= ?
A movie projector.
Where is the film with respect to the focal point?
The diameter of an eyeball is about 22mm in diameter.
It focuses light emanating from different objects onto the retina to produce
sharp images.
Most of the focusing job is done by the cornea, which acts as a fixed lens
with a focal distance of 22mm.
The lens is only doing some fine tuning to move the focusing from objects
which are far-far away to objects which are close.
A healthy human eye can
clearly see (focus) objects at
distances from infinity to about
25cm. How is that achieved?
By changing the focal
distance of the lens! We
always have l’≈ 22mm.
Far-far away means l = ∞, and f = l’ ≈ 22mm.
An object at l = 25cm means
The adaptive lens driven by the eye muscles changes the focal
distance of the eye by “only” 9%. But this is quite a lot!..
Myopic (shortsighted) eye – the lens is always “too strong”, that is
too much converging and if the object is far away it creates its
image in front of the retina.
So, the light pattern on the retina becomes blurry, out of focus.
The eye can be helped by a negative, diverging lens, which
creates virtual images of far away objects closer to the eye.
Rephrasing it: the diverging lens + the “too much” converging lens
in the eye make a composite lens of the right converging power.
The power of the corrective
lens is P=1/flens (diopters, m-1)
Hyperopic (farsighted eye) – the lens is sometimes not quite strong
enough. If an object is close it focuses behind the retina.
So, the light pattern on the retina becomes blurry, out of focus.
The eye can be helped by a positive, converging lens, which
creates magnified virtual images of close objects further away from
the eye.
Rephrasing it: the additional converging lens + the “not strong
enough” converging lens in the eye make a composite lens of the
right converging power.
Optics in Review
Electromagnetic waves (light) all propagate in vacuum at c=3x108 m/sec
In a transparent medium with refractive index n, the speed is v=c/n. The
frequency of the wave remains constant, hence λ=nλn.
With an angle of incidence between two different transparent mediums, θ1,
Snell’s Law
Mirror’s and Lenses
What is the difference between the lens and mirror equations?
Meter Stick
If you look into the corner of the cube
at 45o (as shown) what mark on the
meter stick do you see if
(a) The tank is empty?
(b) The tank is half full?
(c) The tank is full?
40cm doh!
Critical Angle
When light is propagating in glass with n=1.52, what is the critical angle
when the glass is immersed in (a) water (n=1.33)?
(b) Benzene (n=1.50)?
(c) diiodomethane (n=1.738)?
Displaced Laser Beam
l
Find the displacement of a laser beam
that is approaching a 5cm thick layer
of ice with an angle of incidence of
θ1 = 30o when ice has an index of
refraction of n=1.3.
Defining the diagonal through the slab
as l we find:
From Snell’s law:
Hence the displacement is:
Image Formation with a Mirror
A candle is 36cm from a concave mirror with a focal length 15cm.
(a) Where is the image?
(b) What is the magnification?
(c) Is the image real or virtual, upright or inverted, enlarged or reduced?
The image is in front of the mirror, hence real.
The magnification is negative, hence inverted.
The object is located > 2f, hence image is reduced.
Image Formation with a Mirror
The image of an object in a 27cm focal length concave mirror
is upright and magnified by a factor of 3. Where is the object?
From the magnification factor of 3 we know that:
From the lens equation:
The image is 18cm front of the mirror which is inside the focal point!
Image Formation with a Mirror
At what two distances could you place an object from a
45cm focal length concave mirror in order to obtain an
image that is 1.5 times the object’s size?
First from the magnification
we know that:
From the lens equation:
First
The image is behind the mirror and virtual. Is it upright or inverted?
M is positive so the image is upright!
Image Formation with a Mirror
At what two distances could you place an object from a
45cm focal length concave mirror in order to obtain an
image that is 1.5 times the object’s size?
First from the magnification
we know that:
From the lens equation:
Second
The image is in front of the mirror and real. Is it upright or inverted?
M is negative so the image is inverted!
Image Formation with a Lens
A lens has a focal length of 35cm.
(a) Find the type and height of the image when a 2.2cm high object is
placed f+10cm from the lens.
The image is real, behind the lens, and inverted.
Image Formation with a Lens
A lens has a focal length of 35cm.
(b) Find the type and height of the image when a 2.2cm high object is
placed f−10cm from the lens.
The image is virtual, in front of the lens, and upright.
Image Formation with a Lens – Fine Print
You have a magnifying lens has a focal length of 30cm. How far from the
page should you hold the lens in order to see the print enlarged 3x?
First note that the image is on the
same side of the lens as the object,
the page! This means that the image
distance is negative. Hence:
From the lens equation:
This result means that the lens is 20cm above
the page and the image is 60cm below the page,
and the image is upright.