optics f

The IB Physics Compendium 2005: Optics 1

9. OPTICS

9.1. Light as an electromagnetic wave

The dual nature of light and the EM spectrum

Light can be described in two complementary ways: as particles, "photons", with the energy E = hf

(see Atomic physics) or as electromagnetic waves which can travel i vacuum with "the speed of light", c = f



. Different frequency or wavelength intervals represent different types of EM waves, such as radio waves, microwaves, infrared (heat) radiation, visible light (colours red -> violet), ultraviolet radiation, X-rays and gamma rays (see Waves, section 4.1.). The light as a wave motion can be described as oscillations of a magnetic field B and electric field E perpendicular to the direction of travel. For unpolarised light, these oscillations occur in all directions perpendicular to the direction of travel.

Fig o01a: E- and B-oscillations, unpolarised light

Experimental measurements of the speed of light c

 Römer's method: relative motion of Earth and Jupiter

.

In the 1600s a rather "good" value for the speed of light was obtained by studying the time period for a moon of Jupiter to revolve around its planet. This time was slightly shorter when the earth was getting closer to Jupiter and longer when earth was receding from Jupiter. The relative motion of the planets -would of the order of magntiude of planetary orbital speeds (e.g. 30 kms

-1

for earth) which is a small part - about 1/10000 - of the speed of light (c = 300 000 kms

-1

) but a change of

1/10000 in the time for a moon to revolve Jupiters (e.g. about 40 h) makes several dozens of seconds which was measurable.



Michelson's experiment with rotating mirrors (not to be confused with Michelson-

Morley's experiment in relativity, see section 8.2!)

In the 1800s, a more precise measurement of c was made using the equipment below:



Thomas Illman and Vasa övningsskola


Fig. o01b: Michelson's rotating mirror experiment

The eight-sided rotating mirror M1 reflects light from a source S towards a stationary mirror M2 back to another side of M1 and into the telescope of an observer O. From the difference in rotating speed of M1 between ones which make the ray of light visible for O one can find the time for M1 to turn 1/8 of a revolution; this is then the time needed for the light to travel twice the distance between M1 and M2. Michelson placed M2 on a mountain top about 35 km from M1.

Ex. If the distance from M1 to M2 was 30 km then twice the distance is 60 km, which light travels in s = vt = ct =>t = s/c = 60km/300 000 kms

-1

= 0.0002 s. The mirror must turn 1/8 of a revolution faster or slower in that time or once in 0.0016s; that is the difference in rotating speed is 1/0.0016 =

625 revolutions per second.

9.2. Refraction of light

Refractive index

Recall from Waves, section 4.10. Snell's law n

1

sin



1

= n

2

sin



2

[DB p. 6] and the refractive index n = c / v [DB p. 6] where the angles



1

and



2

are angles of incidence and of refraction from the normal to a plane boundary surface (e.g. between air and water), v the speed of a wave (here light) in the materials, and n the refractive index.





Fig. o02a = w10a

Parallel shift

A ray entering a plane boundary between media 1 and 2 and proceeding to the same medium as 1 at another boundary parallel to the first one (in plain English: light going through a sheet of glass) will at the second boundary be be refracted back to a direction parallel to the original one, but somewhat shifted to the side. This is a consequence of Snell's law:



n

1

sin



1

= n

2

sin



2

= n

3

sin



3

where sin



3

= sin



1

and then



1

=



3

if n

1

= n

3

.

Fig o02b: Parallel-shifting refraction

The bent stick

Since objects appear to nearer the surface than they are, a straight object sticking into the water (e.g. an oar) seems to be bent at the surface.




The IB Physics Compendium 2005: Optics

Apparent depth d' and real depth d

4

Fig o02c : Apparent and real depth

Let the light enter the water at the angle of incidence



This angle of refraction



2

1

and then be refracted to

 as the angle of incidence and is refracted to an angle of refraction =



1

2

by the water.

is its angle of incidence when hitting the bottom where it is reflected; for reflection the angle of incidence and of reflection are the same. It then hits the water surface with



back into the air.

2

Now let the real depth of the water be d and the apparent depth d' = the distance from the water surface down to a point where the extension of a ray entering the water and one re-emerging from it would cross. If the distance between the points of entry and reappearing for the ray is called 2x then:



tan



1

= x/d' => d' = x/tan



1

and tan





we have n

1



sin



1







= n

2 sin

 then sin



1

 

sin



d' = x/tan

tan



2



d' = x/n

2



2

 sin



1

1

1

= n

2 sin



2

and if



, sin



1

= n

2 sin



2

2

 

= x/d becomes



2

2

becomes



2

= x/d => d = x/tan



but if n

1

= n air

= 1 (about) then

2

1

is very small (we look at the bottom almost straight from above)

, and tan



1

 

2

= x/d which inserted into the previous gives

= x/(n

2

(x/d)) = d/n

2

; thus

1

1

= n

2



2

and

so

becomes d' = x/



1

= x/n

2



2

and d' = d/n water

(not in DB)

Total internal reflection

At any boundary, part of the light is transmitted into the other medium and refracted, part of it is reflected at the boundary. If the ray moves from a medium 1 with a higher to a medium 2 with a lower n-value, then total internal reflection may occur. This means that all the light is reflected, and none is refracted into medium 2. Medium two is then said to be optically more dense than medium





1. In practice this may happen when light is to go from water to air or glass to air, not air to glass or water.

[Notice however, that for sound which travels faster in water than in air, total internal reflection may happen when a sound wave is to go from air to water. This may explain why sounds are effectively reflected from the surface of a lake if the surface is undisturbed; e.g. sounds from people across a lake can be heard well in the evening].

Fig o02d: Total internal reflection at the critical angle

Above light is to leave e.g. glass (1) and enter air (2) at the angle of incidence



1

. Since n

1

>n

2

:



n

1 sin



1

= n

2 sin



2

so sin



2

= (n

1

/n

2

)sin



1

> sin



1

and



1

>



2

Therefore



2

will become 90 o

when



1

is smaller than that; any further increase in

 total internal reflection. The angle of incidence



1

which gives



2

= 90 o

1

will cause

is the critical angle,



C

.



n

1



n

1 sin

 sin



1

= n

2 sin



2

becomes n

1 sin



C

= n

2

gives sin



C

= n

2 sin 90 o

= n

2

since sin 90 o

= 1

C

= n

2

/n

1

and if medium 2 is air where n

2



1 sin



C

= 1/n [not in DB] where n = the refractive index of the optically more dense medium that the light cannot leave. Note that for water with n = 1.33 we get



C

= arcsin(1/1.33)



49 o

.

The "underwater bright circle" and reversible rays

This was for light "attempting" to leave a medium with n > 1. An example of this would be a light source at the bottom of a pool, from which light can be refracted into the air only for angles of incidence smaller than the critical angle as they hit the water surface from below. If light was sent in the opposite direction, it should follow the same path back (it can be shown by swapping



1

and



2

that rays are reversible) and therefore only light coming from air and leaving surface at an angle of refraction smaller than



C

can reach an observer at the bottom. A diver looking up from the bottom will see a "bright circle" at the surface above him through which all light from the world above the surface must come to the diver. Since this leaves all angles of incidence from 0 o

to 90 o available, the diver on the bottom could if the surface is smooth enough see as much as one at the surface, but the view would be distorted.





Fig o02e: Underwater circle, reversed rays

9.3. Technical applications of refraction

Prismatic reflectors

In a prism with two sides at a 45 o

angle to a third one rays entering this third side at a zero angle of incidence will pass into it unrefracted and "attempt" to leave it into air in a situation where the angle to the surface and therefore also the angle to the normal = the angle of incidence is 45 o

. If the glass is made of a material with a critical angle of less than this (n = 1.5 will give the critical angle 42 o ), then the ray will be totally internally reflected towards another side where the same occurs. (Recall that for reflection the angles of incidence and reflection are always the same). The result of this is that the ray is reflected back towards where it came from, as is it had hit a mirror. This is useful in optical instruments like binoculars since 100% of the light is reflected and none lost, which always happens to some extent even in very good mirrors.

Fig o03a: Rays in a prismatic reflector

[Note I: One material, diamond, has the exceptional n-value of 2.42. Diamonds can be given a shape such that light entering from above will be totally internally reflected back up only if the material really is diamond, and not fake materials with n-values of around 1.5 or 1.6. This makes it possible for a jeweller to quickly determine if a diamond is genuine.

Note II: The n-values for water and ethanol are about 1.33 and 1.36. For solutions of ethanol in water the value will be something in between, and an unknown ethanol content can be inferred from small changes in the direction of rays from e.g. an educational He-Ne laser after passing a transparent sample container with parallel walls. Unlike density measurements, this is not affected by sugar content.





Note III: The unauthorized home production of distilled alcohol is illegal in Finland. Do not accuse your physics teacher of encouraging anyone to break this law.]

Optical fibres

In an optical fibre light is conducted, if necessary along a curved path, by repeated total internal reflection inside a transparent material. Often the material is in two layers, the inner "core" and the outer "cladding", such that n core

> n cladding

. o03b: Optical fibre

Optical fibres are used for



telecommunications, where (laser) light carries information. Since the frequency of light is much higher than that of radio waves, more information can be carried.



in medicine to access inner organs without major surgery, either to observe (endoscopy) and diagnose or to treat with stronger laser light

9.4. Dispersion of light (n depends on wavelength = colour)

Earlier we noted that the optical refractive index n depends on the material, e.g. n = 1.33 for water and close to 1 for air. But in a given medium, it also depends on the wavelength (or the frequency) of the light. The n-values for certain wavelengths of light in water are:

Wavelength (nm)

761

656

Colour red orange n

1.329

1.331

589

527

431

397 yellow green blue violet

1.333

1.335

1.341

1.344

The same phenomenon occurs in glass, and leads to white light with all colours present being split up in a prism:





Fig. o04a White light dispersed in a prism

The prisms in spectroscopes split up the light from a given source (sunlight, light from special lamps containing heated vapour of chemical elements to be studied) so that "spectral lines" (caused by a narrow slit that the light has to pass before the prism) can be viewed in a microscopelike device. These spectral lines were important in developing the atomic model, with light of different frequencies being emitted as electrons fall from a higher shell to a lower one, or absorbed in the opposite process giving dark lines in the spectrum.

9.5. Lenses

How do lenses work?

Lenses are glass (or plastic) objects (see later for specific lens types) with curved surfaces where refraction occurs when light enters and leaves the lens. The angle of incidence



1

is to the normal to the tangent of the lens surface where it the ray enters the lens. The angle of refraction



2

is given by

Snell's law. In a curved lens, the angle of incidence for leaving the lens and going on into air is not the same as



2

but some other angle



3

Snell's law gives the angle of refraction



depending on the geometry of the situation. After that,

4

into air. Since the surface of the lens is curved there is no simple relation between the angles.

Fig o05a: Refraction at the surfaces of a lens,





Lens types and concepts

In a convex or converging lens, parallel incident rays will converge to a focus or focal point after passing it, in a concave or diverging they will diverge (and appear to originate in a focal point

"before" the lens). Since light can enter the lens from either side, it will have two focal points F .

Fig o05b : Convex and concave lens.



The principal axis (PA) is a line through the focal points.



The focal length (f) is the distance from the optical center (O) of the lens to the focal point

Image construction for lenses:" paraxial"," focus" and "center" rays

When constructing the image produced by a lens, the object is often represented by an arrow and 3 key rays of light from the tip of the arrow followed.

Fig o05c: Image construction in a convex lens





I.

An incident ray parallel to the principal axis: the refracted ray or its extension backwards goes through the focal point

II.

An incident ray through the optical centre continues in same direction

III.

Convex lens : An incident ray through the focal point: the refracted ray will be parallel to the principal axis Concave lens : a ray towards the focal point on the transmission side will be parallel to the principal axis.

Fig o05d: Image construction in a concave lens



Where any two of the rays I-III or their extensions backwards intersect, the image of the tip of the arrow will be found.

The distance from object to lens (optical center) is u and from the lens to the image is v.



For rays from the base of the arrow, when that is placed on the PA, I-III will be identical

and the image at the PA at the same distance from O as the image of the tip

Real and virtual image points

If the image of the arrow (study the image of the arrow tip) is found with extended rays, not physically intersecting rays (this is always the case for concave lenses and sometimes for convex ones), the image is virtual . It can then not be "focused" on a screen, even if the image can be seen through the lens. If the image is found with intersecting rays, the image is real .





"Erect" and "inverted images

The image point of the arrow base is on the PA, but the image point of the tip may be on the same side as the tip of the object, in which case the image is called erect ; in the opposite case it is

"inverted" .

Magnification (linear) of images

The height of the image h i

may be smaller, greater or equal to that of the depicted object, h o

. The magnification of the image m = h i

/h o

, that is m > 1 if the image is larger and m < 1 if it is smaller than the object.

Characteristic images for convex and concave lenses

By drawing and constructing the various cases you can verify that the following is true:

Convex lens u v m orientation real/virtual

1

2 u < f u = f f > v > u >1 erect no image (v =



) - -

3 2f > u > f v > 2f

4 u = 2f v = 2f

5 u > 2f 2f > v > f

6 u =

 v = f

>1 inverted

=1 inverted

<1 inverted

<1 inverted virtual

- real real real real

Note that rays from infinity that are parallel to each other, though not to the principal axis are refracted to points in the focal plane.

Concave lens u v m orientation real/virtual

Any u v < u <1 erect virtual

Note: For all lenses, real images are inverted and virtual images are erect.

The lens equation (= the mirror equation)

The image of the arrow, representing any object, will follow this law (for " thin " lenses, for reasons of spherical aberration to be presented later).

1/f = 1/u + 1/v [DB p. 13] where u = distance from lens to object, v = distance to image, f = focal length or mirror with the following sign rules :

 u is positive for real objects, negative for "virtual object points" (that is: if what the lens depicts is not a physical object, but the image produced by another lens, which may be the




The IB Physics Compendium 2005: Optics 12 case in a system of two or more lenses, then u is negative if it is not on the "correct" side of the lens).

 v is positive for real images, negative for virtual images

 f is positive for convex and negative for concave lenses

General manipulation of the lens equation:

1/f = 1/u + 1/v gives 1/v = 1/f - 1/u with the common denominator fu so 1/v = u/uf - f/uf so 1/v = (uf)/uf or v = uf / (u - f)

Convex lens (f positive)

1 . u < f makes (u - f) negative and v negative. Absolute value of (u-f) is smaller than f, so v = ux where x = f/(u-f) >1 and then v > u

2 . u = f makes (u - f) = 0 and v = f / 0 =



3 . 2f > u > f makes (u - f) positive and v positive. (u - f) is still smaller than f so f / (u - f) is smaller than 1 and therefore v < u.

4 . u = 2f makes (u - f) = 2f - f = f and v = uf/f = u; u is positive if v is (for all real objects depicted).

5 . u > 2f makes (u - f) > f and v = ux where x = f / (u - f) < 1 so v < u.

6 . u =



makes 1/u = 0 and from the original 1/f = 1/u + 1/v we get 1/f = 1/v so v = f

Concave lens (f negative)

If a real object is depicted then u > 0 and then (u - f) with a negative f is always positive and > f. So with v = ux where x = f / (u - f), x is always positive but smaller than 1, so v < u.

Linear magnification

The linear magnification m, that is the relation between image and object size can be found as: m = h i

/ h o

= v / u [DB p.13]

It follows from this that whenever the image is virtual and erect, u is positive but v negative and therefore m negative while a real and inverted image is obtained when v and u both are positive and m then positive:

Positive m => inverted, real image

Negative m => erect, virtual image





Near point

In the human eye, a convex lens is used to focus the incoming light on the retina (Sw. näthinna, Fi. verkkokalvo). The shape of this lens can be changed by the ciliary muscles around it. The smalles distance from the from which the image of an object can be focused is called the near point , and it is ca 25 cm for young adults. The largest distance is the far point . The far point is often taken to be infinity.

Angular magnification M in simple magnifying glass

When the angle an object subtends at the eye viewing without aid at the near point (d

N

= 0.25 m) and with a (usually convex, "simple magnifying glass") lens are compared for a distance of the image to the lens v = (-)d

N

, the angular magnification (or magnifying power) M is defined as:

M =

 i

/

 o

[DB p. 13]

Fig o05e : object at 25 cm vs virtual image at 25 cm (object at u < f)

Here where strictly get



tan

 i

= h i

/v = h i

/d

N

and tan

 o

= h o

/d

N

,

Note that

 o

= the angle we would have if we placed the object at the near point, at the distance d

N from out eye not using any lens at all. With the lens we can place the object much nearer the eye, at a distance u < f where f is clearly smaller than d

N

. With the lens we can still focus on the virtual image at the ideal distance d

N

.



but for small angles tan x



x so



approximately

 i

/

 o

= (h i

/d

N

)/(h o

/d

N

) = h o

/h i

but from before we have:



h o

/h i

= u/v so







m =

 i

/

 o

= v/u

When the image is at the near point, v = (-)d

N

, the lens formula gives:



1/f = 1/u + 1/v which becomes 1/u = 1/f - 1/v



the image is virtual, so v is negative; therefore d

N

= - v so



1/u = 1/f - (1/-d

N

) = 1/f + 1/d

N

The (tangent of) the angle of the image

 i

= h i

/v but also



N

(1/f + 1/d

N

) or i

= h o

/u so



M =

 i

/

 o



gives m = d

= (h o

/u)/( h o

/d

N

) = d

N

/u = d

N

(1/u) which with 1/u = 1/f + 1/d

N

Near point: M = d

N

/f + 1 [not in DB]

For the image at infinity we can use from above



m = d

N

/u and when v =



we have 1/v = 0 and then



1/f = 1/u + 1/v becomes 1/f = 1/u or f = u so m = d

N

/u becomes

Far point: M = d

N

/f [not in DB]

The far point is used when the eye is relaxed .

Lens systems

In more complicated optical instruments two or more lenses may be used. The image produced by one lens will then act as an "object" for the next. If this "object" is at the wrong side of the lens, then it is a "virtual object" and its distance to the second lens given a negative sign:

Let the two lenses 1 and 2 have the focal lengths f

1

and f

2

and be at a very small distance from each other. The object distance to the first lens is u

1

and the image distance v

1

. We then get:



1/f

1

= 1/u

1

+ 1/v

1

but now approximately v

1

= - u

2

so



1/v

1

= -1/u

2

so 1/u

2

= -1/v

1

= -(1/f

1

- 1/u

1

) = 1/u

1

- 1/f

1



Now 1/f

2

= 1/u

2

+ 1/v

2

which with the above leads to



1/f

2

= 1/u

1

- 1/f

1

+ 1/v

2

, where we move "-1/f

1

" to get



1/f

1

+ 1/f

2

= 1/u

1

+ 1/v

2

This means that the lens system takes the light from an object at the distance "u" (here u

1

) to the lens system and produces an image at the distance "v" (here v

2

) from the lens system such that the same would have been done by a single lens with the focal length f where

1/f = 1/f

1

+ 1/f

2

[not in DB]

This was derived for a case where the two lenses are very close to each other. If the distance between them cannot be ignored, the calculations would be somewhat different.

The astronomical telescope

The most modern telescopes are reflectors, meaning that they are based on mirrors rather than lenses. During the history of astronomy, especially from the 1600s to the 1800s, refractors or astronomical telescopes based on a lens system were important.





In an astronomical telescope we have two convex lenses, an objective lens with the focal length f o towards the distant object to be observed, and an eyepiece lens with the focal length f e

towards the observer.

Fig o05f: Ray diagram for astronomical telescope

The distant star can reasonably be assumed to be at a distance u o

=



from the objective lens which has a long focal length fo and produces an image at vo given by



1/f o

= 1/u o

+ 1/v o

where 1/u o

= 0 so v o

= f o

This image is diminished, inverted and real. It is formed in the focal point on the transmission side of the objective lens. Although it is diminished compared to the original image, the eyepiece lens with as short f e

is placed close to the image from the objective lens so the image subtends a large angle for the eyepiece lens. By relaxing the eye and adjusting the eyepiece so that the image of it is at infinity, v e

=



, we have



1/f e

= 1/u e

+ 1/v e

or when 1/v e

= 0 we have 1/f e

= 1/u e

so u e

= f e

.



this means that the "object" for the eyepiece lens = the image of the objective lens is placed in the focal point if the eyepiece lens, so



the distance between the lenses = the length of the telescope = f o

+ f e

.





The image of the eyepiece lens is greatly magnified, inverted and virtual. (A virtual image of an erect real object is erect, but this is a virtual image of an inverted real set of object points - real since the image formed by the objective lens is on the incoming side of the eyepiece lens.

The angle subtended by the object for a naked eye is

 o

; this angle is small and approximately equal to its tangent. Look at an angle between the principal axis and a ray through the optical center of the objective lens: this angle is the same on both sides so



tan

 o

=

 o

= h io

/f o

where h io

is the height of the image formed by the objective lens



h io

is then the "object height" h oe

for the magnification in the eyepiece lens



now for the angle eyepiece lens h oe

/f e

= h io

/f e

= tan

 e

=

 e

is the angle viewed by the observer



the total magnification of the telescope is then m =





m = (h io

/f e

)/(h io

/f o

) = f o

/f e

or finally e

/

 o

where m telescope

= (-)f objective

/f eyepiece

[not in DB] where a minus sign may be inserted to indicate that the image is inverted. (But the rule earlier was

"positive m => inverted, real image, negative m => erect, virtual image", so this inverted virtual image does not quite fit that rule for single lenses).

It follows from the formula that the f o

should be large and f e

small to maximise the magnification.

This leads to long telescopes. The inverted image is not a problem in astronomical observations where a star looks about the same upside down; when observing the moon or features of the planets

(e.g. Mars) this must be remembered.

For terrestrial telescopes, the inverted image can be avoided with a third convex lens (terrestrial telescopes, used by admirals to observe enemy ships), or a different lens system with a concave eyepiece lens (theater binoculars, where the objective lens produces an image on the "wrong", transmission side of the concave eyepiece, this image acts as a virtual object for the eyepiece lens).

Compound microscope

The compound microscope is also constructed with two convex lenses, the objective lens O and the eyepiece lens E. Recall that a convex lens produces a magnified image when 2f > u > f or when u < f; that is when the object distance is either slightly larger or smaller than the focal distance.

The first alternative is used in the objective lens O, where the object is placed a bit outside the f o

on the emission side. A virtual, inverted and magnified lens is produced at a distance v o

larger than f o on its transmission side. The eyepiece lens E is placed so that the image produced by O is immediately inside f e

on its emission side, and acts as a virtual object for E. If we denote the distance between O and E, the length of the microscope L, then the distance from O to its image = v o



L - f e

.





Fig. o05g : Ray diagram for the compound microscope.

Then:



m o

= v o

/u o

= (L - f e

)/u o

but for small angles (the objects viewed with a microscope are not far away, but they are like bacteria very small so this can be justified) where the angles and their sines and tangents are about equal, we can assume that the angular magnification of the objective lens is about the same as the linear magnification:



M o

= m o



the eyepiece lens E then magnifies the image of O in which it acts like a simple

 magnifying glass. If the eye is relaxed then we have from before

M e

= d

N

/f e

, where d

N

= the near point of the eye = 0.25 m



then M tot

= M o

M e

or



M



if f



M tot e

= (d

N

/f e

)(L - f e

)/u o

and f o

are small compared to L then (L - f e

)



L and u tot



(d

N

/f e

)L/f o

= Ld

N

/f e f o o



f o

which gives

M microscope

= Ld

N

/f e f o

[not in DB]

The total magnification is limited by the difficulty in manufacturing lenses with very short focal distances without excessive aberrations, see below. There is also the additional difficulty with the wavelength of light which due to the Rayleigh criterion and the necessarily limited diameter of the lens (see later) puts restrictions on the resolution. This can be improved in electronic microscopes where the de Broglie wavelength of electrons facilitates more detailed images (see Atomic physics).

Aberrations : spherical and chromatic

These equations for lenses are only approximately true. The errors which occur in reality are:

 spherical aberration : only rays close to and parallel with the principal axes (paraxial rays) follow these equations with high precision. For more exact result in other cases parabolic instead of circular/ spherical surfaces shouls be used. Another method is to use additional corrective lenses to counteract the phenomenon.





 chromatic aberration : for lenses there is an additional error if light of different wavelengths is used, since the refractive index then is different (see the dispersion phenomenon). This can be fixed for two colours with a double lens, one convex and one concave ("color-corrected lens" or "achromatic doublet"). For other colours more complex lenses made of several elements are needed.

These phenomena can be investigated graphically by drawing large circular cross-sections of lenses, a few parallel incident rays, the tangent and normal at the point of entry, measuring the angles of incidence with an angle ruler, calculating the angle of refraction with Snell's law (using slightly varying n-values for chromatic aberration), and repeating the process where the ray leaves the lens material.

[Lens-makers equation (not needed in the IB)

If the lens surfaces are spherical (intersecting for convex, non-intersecting for concave lenses), with spheres of radius values r1 and r2 then the focal length f is given by the lens-maker's equation:

1/f = (n - 1)(1/r

1

+ 1/r

2

) [not in DB] where n = the refractive index of glass or other lens material.]

9.6. Planar mirrors

Mirror and diffuse reflection

In a mirror all or most of the light is reflected from one planar (or curved but simple) surface, whereas in diffuse reflection the reflecting surface is made up of many pieces which reflect the light in different directions. Often the light is only partially reflected, and partially absorbed into the material.

Fig o06a: Mirror and diffuse reflection

For a mirror, the direction of the incident ray i is defined as the angle to the normal to the mirror surface, the angle of incidence. The angle of reflection is that between the normal and the reflected ray .

Law of reflection: angle of incidence = angle of reflection





The image of an object in front of a planar mirror can be described as in the figure below. Since the image is formed by the extensions of rays, not by rays intersecting in the image point, the image in a planar mirror is always virtual .

Fig o06b : Reflection in a plane mirror



Note:



distance from image to mirror = distance from object to mirror



line from object point to image point at 90 o angle to mirror



the image is laterally inverted image (right and left switched)

A real image is formed where rays of light cross (can be put on a screen)

A virtual image is "formed" where extension of rays cross (cannot)

The same properties are true for extended objects (not small as a point)

9.7.* Curved mirrors

In addition to planar mirrors, we can have concave and convex spherical mirrors. These can be described as spherical surfaces, here represented by circles, with a center of curvature C and a focal point F; and a principal axis PA through these.

Fig o07a : Convex and concave spherical mirror with C, F,





Image construction for spherical mirrors



I: incident ray parallel to principal axis: the reflected ray or its extension through focus



 incident ray throuh centre of curvature: the reflected ray returns in the opposite direction

III: incident ray or its extension through focus : the reflected ray parallel to principal axis

20

Fig o07b : Curved mirrror construction rules

As for the images produced by lenses, important properties are:

'



Real or virtual



Erect or inverted



Diminished, same size or enlarged



Distance from mirror

The result will for a concave mirror be different at different distances, for a convex mirror always the same. This can be checked using a spoon.

The mirror equation= the lens equation

1/f = 1/u + 1/v [DB p. 13]

The equation can be used with u and v in the same meaning as for lenses, and with the same sign rules (positive for real object/image, negative for virtual). The sign rules for the focal length f is, however reverse compared to lenses:

Lens Mirror

Convex f > 0 f < 0

Concave f < 0 f > 0





The lens-maker's equation is for mirrors replaced by the relation between the focal length f and the radius of curvature R (the radius of the sphere with center in C that the mirror surface is a part of): f = ½R [not in DB] which means that the lens/mirror equation for mirrors can be written

1/f = 1/u + 1/v = 2/R [not in DB]

9.8. Lasers

Characteristics of laser light

One importan type of light used since around 1960 is laser light, which is



monochromatic = only one colour (wavelength, frequency) is present



coherent = the wavecrests are in pace with each other and interfere constructively

("amplify" each other)

Spontaneous and stimulated emission

LASER means "light amplification by stimulated emission of radiation". If a photon of light hits an atom it may be absorbed an its energy cause an electron to rise to a higher shell = energy level.

From this this or another electron will soon fall down to fill the empty place in the lower shell, emitting a photon. This is "spontaneous emission".

Fig o08a: Spontaneous and stimulated emission

If an electron is at a higher level when a photon hits it, the photon may cause it to rise to an even higher level, but also to disturb it so it drops down to a lower level. If the incoming photon had the same energy as the energy difference between the shells the electron falls between, then the emitted photon will have the same energy E = hf and then also the same frequency and wavelength.

It turns out that it also is "coherent" with the first photon, that is the wavecrests are in phase and they move in exactly the same direction.

This is stimulated emission.

Population inversion - impossible with two levels

An incoming photon is equally probable to cause absorption as stimulated emission. Which of these occurs the most depends on how the electrons in the atoms or molecules hit by the photon are distributed. Stimulated emission will dominate if there are more electrons in a higher of two shells than in the lower of these two. This is called "population inversion", since it is common for atoms to




The IB Physics Compendium 2005: Optics 22 have their atoms as low as possible in the shells, following the chemical rules for how many fit in a certain shell. Heating the material ("thermal pumping") may cause some to rise to a higher shell, but it can be shown that one can never get more than 50% in the higher of the two shells (and in practice much less).

Three-level laser

All lasering materials involve at least three (or more) energy levels. They must be such that the following happens:

Fig o08b: Three-level lasers

Electrons are raised from E

1

to E

3

by absorbing ordinary light. They quickly fall down to E

2

(causing spontaneous emission of photons with an energy equal to the difference between E

3

and

E

2

, but these are irrelevant here. Since the falling down is spontaneous, no electrons are lifted from

E

2

t E

3

). The "decay" (not radioactive) from level E

2

back to E

1

is much slower than the other processes, causing a population inversion between E

2

and E

1

(that is, many more electrons in E

2 than in E

1

). If even one incoming photon happens to have the right energy for stimulated emission we will now get two (coherent) photons with the right energy. They will produce 2, 4, 8 etc. With 4 or more levels other similar possibilities can be found. All this is true assuming that the suitable energy levels with required rate of "decay" are found in the laser material, which is why only some materials can be used.

These photons are moving back and forward between two mirrors at the ends of a "laser cavity" (the place where the material with the desired sort of energy levels is). This is actually a case of standing or stationary wave , like on a string attached at both ends, which limits the possible wavelength or frequencies. To get the laser light out of the cavity, one of the mirrors is such that it does not reflect the light perfectly but lets some of it "leak" out.

Laser applications

Holograms: A virtual three-dimension can be produced be produced by letting the light from two laser beams be recorded on a film. One is allowed to reach the film "directly" (via mirrors, though); the other is reflected from the object. These two beams will produce an interference patterns, where all interference is caused by the shape of the object, if the the beams are split from the same laser beam, with all their waves coherent, so that no interference is caused by the light itself (other than perfectly constructive interference making it stronger). The image can then be reconstructed with a similar beam.





Laser surgery: Laser light can be used to change the shape of parts of the eye to correct defects of vision, and (possibly via optical fibers) for other types of surgery

Telecommunications: Signals sent in optical fibers are laser light, since it has a well-defined direction and only one wavelength, which reduces the problem of dispersion in the fibre material

Military use: Strong laser pulses can be used to shoot down missiles and artillery shells.

9.9. Diffraction

The Huygens principle and diffraction

Recall from Waves that a wavefront travels by every part of it emitting half-circular wavelets which together form the new wavefront:

Fig o09a = w07b

If the wave front has to pass an obstacle, some wavelets will be suppressed, and the resulting new wavefront may alter its appearance. Some common cases are:

A. Single edge: The waves will bend around the edge of e.g. a solid disk. For ordinary light the bending is not usually noted, but if a precise point source S of monochromatic light is used with a circular disk, waves bending around its edges will interfere constructively and produce a bright spot

P in the center of the shadow of the disk.

[In particle physics, particles accelerated towards a target, e.g. alpha particles on a nucleus, will also be diffracted around it, since the alphas can be assigned a de Broglie wavelength.]

B. Narrow (single) slit: The waves after a slit with width b will produce a set of intensity maxima in directions following sin



=



/b or approximately



=



/b [DB p13] , see more below.

C. Circular aperture: The waves will produce a bright spot with bright rings around it, and intensity minima around it. The radius of the central bright spot (= the distance from the center to the first minimum follows



= 1.22



/b [DB p. 13] .





Fig o09b: Diffraction from circular disk

Single slit diffraction (B.)

Consider a set of parallel rays entering a narrow single slit with the width b. One intensity maximum will obviously be found in the "straight ahead" direction, or at the angle



= 0 o

from their original direction. The first intensity minimum should be found from the following graph:

Fig o09c: Parallel rays entering slit (b wide), diffracted to



(see text)

The incident rays are diffracted so those at one edge have a path difference (head start) of

 compared to those from the other edge of the slit. A ray at one edge will then have the path difference ½ 

compared to one at the center of the slit and will interfere destructively with it.

Another ray just next to the same edge will do the same with one just next to the center, and thus pairs of rays will all extinguish each other in this direction. From the geometry of the situation it follows directly that we have an intensity minimum when:



sin



=



/b or for a small angle



approximately that



=



/b [DB p. 13]

We could also have found corresponding results with 2



, 3



, ... This means that if we let the diffracted rays hit a screen rather far away we will get an intensity maximum at



= 0 o

and minima




The IB Physics Compendium 2005: Optics 25 to both sides at the distances d =



/b, 2



/b, 3



/b, ... from the central maximum, and smaller intensity maxima in between:

Fig o09d : graph of relat intensity vs angle for diff at single slit

It can be shown that the intensity minima will be found in directions given by b sin



= m



[not in DB] where m = 1, 2, 3, ....

The effect of this on double slit interference will be returned to later.

Diffraction from a circular aperture (C.):

The above treatment of diffraction in a single slit was based on a long, narrow slit such that its length l >>b so that the same diffraction phenomenon in a dimension at a 90 o

angle to the one previously studied can be ignored. For a circular aperture this is obviously not true, and furthermore the "width" of the slit is not constant.

Fig o09e: long narrow slit and circular aperture as seen by approaching rays





It will turn out to give a diffraction pattern with a central circular maximum ("Airy disk") and circular minima and maxima around it, such that the directions to them (and therefore the distance between them on a screen comparatively far away; set this distance to one length unit at note that sin x



tan x



x for a small x) will follow



= 1.22



/b [DB p. 13]

Optical resolution: the Rayleigh criterion

Two sets of parallel rays (coming from, say, two different distant objects) can be optically resolved or distinguished from each other if they are so far from each other on the screen (or retina of the eye, or photographic film, or electronic image recording device) that the central maximum of one is coinciding with the the first minimum of the other; that is, that the angle between them is precisely the



= 1.22



/b mentioned above.

Fig : o09f: Rays from distant objects diffracted through an aperture ("just resolved")

For the optical resolution of rays passing a long narrow slit one would use the



=



/b.

The result of this is that the resolution of objects in for example telescopes is limited not only by how well various lens and/or mirror aberrations have been remedied, but also by the relation between the wavelength of the light and the diameter of the telescope. The resolution can be improved by increasing this aperture using a larger telescopes (or possibly several telescopes acting together as

"pieces" of a very large telescope). Especially for radio telescopes, using waves with a higher wavelength, a large aperture is needed. Shorter wavelengths (e.g. ultraviolet or X-rays) are better than visible light for finding the precise direction to a star, which may be useful in the parallax method of finding the distance to a star (see the Astrophysics option).





"Well resolved", "just resolved" and "not resolved" images

27

Fig o09g



Well resolved : the intensity peaks have a minimum in between them, the two objects (e.g. stars) appear as separated



Just resolved: the Rayleigh criterion is barely fulfilled, one intensity curve has a minimum where the other has a maximum and vice versa, but in between the intensity is nowhere near zero. The objects appear to "touch", but the shape of two circles can be seen



Not resolved: the intensity peaks do not coincide, but in between them constructive interference produces an even higher resulting peak; the two objects appear as one elliptic object

Multiple slit diffraction and diffraction gratings

Recall from Waves that when light passes two or more slits such that the distance between them (or between their centers) is equal and = d, then we have d sin



= n



[DB p.6 and 13]

If many slits are used we have a diffraction "grating". The difference between the intensity maxima from a grating compared to a double slit is that they are much narrower, since a small change in direction that does not significantly alter the constructive interference between rays from two slits next to each other will mean a considerable path difference for rays from slits further apart, e.g. with 1000d in between.

Diffraction gratings are useful for investigating spectra and especially for experimental determination of an unknown wavelength.

Single - double slit diffraction envelope and missing orders

We have so far noticed that



a double slit (or many slits as in a diffraction grating) will produce several bright spots on a screen; that is peaks in the light intensity



a single slit will produce a similar set of peaks





Since every slit in a double/multiple slit arrangement also produces the single slit peaks by itself in addition to the peaks produced in "cooperation" with the others, the real pattern will be like the one in:

Fig o09h : Two sets of intensity peaks, envelope sketch

We have that

 

double/multiple slit maxima follow d sin



= n



, n = 0,1,2,3,...

 

single slit minima follow b sin



= m



, m = 1,2,3,...

This means that some double slit maxima will be missing (the "missing orders") for n-values such that (dividing the two equations): d sin



/ b sin



= n



/ m



or n/m = d/b

9.10. Thin film interference

Optical path and reflections at liquid boundaries

When a ray of light hits the boundary between two transparent media, it will be partially reflected and partially refracted into the new medium. For reflections we hade the rule in the Waves chapter that when e.g. a wave on a rope hits an end where the rope is attached to a wall it is inverted when reflected; that is, it is phase-shifted by the "path difference"



. When the wave reaches an open end of the rope, it is also reflected, but without any inversion or phase shift. If the wave reaches a heavier rope, some of it will be reflected, and then it will be phase shifted like from a wall; while when it reaches a thinner rope, it will be reflected without phase shift. Optically, this means that:



the partial reflection occurs with a phase shift



when going from an optically less to more dense medium (from lower to higher refractive index n )



the partial reflection occurs without phase shift when going from an optically more to a less dense medium (from higher to lower n )

The phase shifts in different media are complicated by the fact that waves move at different speeds in them; to account for that one can find out how they interfere by comparing the optical path (or the difference in it); which is optical path = refractive index * distance travelled





Recall that for light n = c/v => v = c/n (v = the velocity in the medium); the distance travelled in a given time s = vt = ct/n which differs from the distance travelled in vacuum by the factor 1/n; this is neutralised by using instead the optical path.

Interference in a thin oil film on water

Assume that monochromatic light is incident on the surface of an oil slick on water, with the angle of incidence =



1

. Some of it will then be reflected from the air-oil boundary and some will be refracted into the oil at the angle of refraction



2

which also is the angle of incidence to the oilwater boundary. Of this light, some will be refracted into the water (but this part we no longer include in our study) and some will be reflected back at the angle of reflection



2

and subsequently some of it will be refracted back into the air, with an angle of refraction equal to



1

.

Fig o10a: Ray 1 hits air-oil at A, reflected; ray 2 hits at A, refracted to B at oil-water, reflected to C at oil-air where refracted back into air.





In the figure above, d = the thickness of the oil film. Let the distance AB = x; it follows that the distance BC = x. We can extend the distance d from A vertically down to the oil-water boundary and equal distance d down into the water. Call the point then reached D and note that the distance

AD = 2d and for geometrical reasons the angle ADC =



2

.

To find out how rays 1 and to interfere, we need to find the optical path difference opd , which in the figure above is obtained from:



the geometric path difference = 2x - y where y = the distance travelled in air by a wavelet on the wavefront found on a line from A and hitting BC perpendicularly at E (that is, the distance it would have travelled in air if it like ray had been reflected at the air-oil

Now we will notice that since n air

< n oil

then a phase shift occurs for the ray reflected at the air-oil boundary, but if n oil

> n water

then there is none at that boundary. The condition for constructive interference must then include a phase shift of



or half a wavelength; therefore it becomes

(m + ½) 

= 2ndcos



2 boundary).



in air n =1 so y o

= ny = y



the optical path for a wavelet at the other end of the same wavefront must be the same, so y o

= z o

where z = EC



but EC is travelled in oil where n



1 so z o

= nz



so is 2x, so the corresponding optical path is 2nx



then opd = 2nx - nz = n(2x - z)



then from the figure we see that cos



2



(2x - z) = 2dcos



2

and then

 opd = 2ndcos



2

= (2x - z)/2d so

This means that in certain directions we will get constructive interference (bright bands or rings) and in between them destructive interference (dark ones). The relation between



Snell's law, n oil sin



2

= n air sin



1

.

1

or



2

is given by

The number of "fringes" (bright circles) that are formed by light entering the oil at angles of incidence from 0 o

(normal incidence) to 90 o

(grazing incidence) can be found by comparing:



normal incidence:



1

=



2

= 0 o . This gives (m + ½) 

= 2nd or m



= 2nd if we count the dark fringes, which gives m normal

.



grazing incidence: the relation between

 reflection (though with reversed rays) so the

 reflection:



1

=



2

is similar to what we have for total internal

2

is the critical angel for total internal

2

= arcsin(1/n). Using this value in (m + ½)



= 2ndcos



2

or m



= 2ndcos



2 gives us m grazing

.



the number of fringes that can be seen are then m grazing

- m normal

.

Dispersion and thin film interference

All this was done under the assumption that the light was monochromatic. Ordinary sunlight is a mixture of all colours of the spectrum; for these different wavelengths the refractive index is




The IB Physics Compendium 2005: Optics 31 different, and therefore the colours will have intensity maxima due to constructive interference in slightly different directions. This we observe as "rainbow colours" in oil slicks on water.

Thin air wedges: measuring small thicknesses

If two glass plates P

1

and P

2

are separated by a small object (e.g. a thin wire W) at one end, the air between then forms a wedge with a thickness d that varies depending on where on the wedge we are. For some d-values we will, using monochromatic light source L, have a constructive interference and for some destructive. If an observer using a microscope M which can be moved

("travel") horisontally observes light that has entered the wedge vertically, there will be alternating bright and dark fringes. (The light can be sent down to the wedge using a glass plate P

3

at a 45 o angle; some light from L will be reflected down to the wedge and some light returning upwards will pass through P

3

and enter the microscope M).

Fig. o10b: Interference in a wedge of air (P

1

above)

Note here that the ray entering P

1

vertically will be parallel-shifted a little but enter the air wedge vertically. Comparing to the oil film above, the air wedge corresponds to the oil and the lower P

2

to the water. The ray reflected from the surface of the oil corresponds to that reflected from the glassair boundary at the lower surface of P

1

. These are not phase-shifted (the glass in P

1

has a higher n than the air in the air wedge). The rays reflected from the air-glass boundary at the upper surface of

P

1

are, however, phase shifted by half a wavelength. As for the oil film, one ray is phase-shifted and the other not, so for constructive interference we can use

 

(m + ½)



= 2ndcos



2



2nd = 1 for a very small angle

 

then (m + ½)



= 2d with n wedge

= n air

= 1

Moving the microscope horisontally from one bright fringe to the next changes the value of m by 1 and therefore the thickness by half a wavelength. When we know how far to the side we have moved the microscope to change the thickness of the wedge by



/2, we can find the angle of the wedge and then the thickness of the wire W.





9.11* Polarisation

Light and other electromagnetic waves (photons) is a transverse oscillation of electric and magnetic fields; for unpolarised light this oscillation takes place in all directions perpendicular to the direction the wave (photon) moves in.

For polarised light, there are oscillations only in one direction perpendicular to the direction of wave travel. Polarisation can occur e.g. in these cases:



in certain long molecules used in Polaroid glasses



when light is scattered to angle perpendicular to the original



in reflection of light at the surface of another medium (e.g. water) when:the angle between reflected and refracted ray is 90 o = Brewster's law

Fig o11a : Brewster angle for polarisation

For refraction we have that sin

 since



2

 incidence

-



2

+ 90

=



1 o

+

 reflection we get that



) = n

2

or sin



1

/ cos



= 180 o

1

1

/ sin



2

= n

2

/n

1

= n

2

if medium 1 is air where n

1

is about 1. But

(to the the right of the normal) and for any reflection



2

= 90 o

-



1

= n

2

giving

and therefore sin



1

/ sin



2

= n

2

becomes sin



1 reflection

=

/ sin (90 o tan



1

= n

2 for the Brewster angle of polarisation.

Polarisation can also be accomplished in certain materials (as in some sunglasses) or when light is scattered by air molecules to a direction perpendicular to the original.

Malus' law

The intensity of light or any other wave is I = P / A where P = the power at which energy is transported by the wave and A = the area hit by the wave. The intensity I in the unit Wm

-2

is a measure of how bright the light is.

When light has to pass through two polarisers then the intensity of it is decreased to (Malus' law)





I = I

0 cos 2

 where I

0

= the original intensity of the light, I = the intensity left after the second polariser,



= the angle between the directions of polarisation for the two polarising filters. When this angle is 90 o , no light will pass.]




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