The IB Physics Compendium 2005: Optics

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The IB Physics Compendium 2005: Optics

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:

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

n1 sin 1 = n2 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:

n1 sin 1= n2 sin 2 = n3 sin 3 where sin 3 = sin 1 and then 1 = 3 if n1 = n3.

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.

Apparent depth d' and real depth d

Fig o02c : Apparent and real depth

Let the light enter the water at the angle of incidence 1 and then be refracted to 2 by the water. This angle of refraction 2 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 2 as the angle of incidence and is refracted to an angle of refraction = 1 back into the air.

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2 = x/d => d = x/tan2

 we have n1sin1 = n2sin2 but if n1 = nair = 1 (about) then

sin1 = n2sin2 and if 1 is very small (we look at the bottom almost straight from above) then sin11, sin22 , and tan11so

 sin1 = n2sin2 becomes 1 = n22 and

 d' = x/tan 1 becomes d' = x/1 = x/n22 and

 tan2 = x/d becomes 2 = x/d which inserted into the previous gives

 d' = x/n22 = x/(n2(x/d)) = d/n2 ; thus

d' = d/nwater(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 n1 >n2 :

n1sin1 = n2sin2 so sin2 = (n1/n2)sin 1 > sin 1 and 12

Therefore 2 will become 90o when 1 is smaller than that; any further increase in 1 will cause total internal reflection. The angle of incidence 1 which gives 2 = 90o is the critical angle, C .

n1sin1 = n2sin2 becomes n1sinC = n2sin 90o = n2 since sin 90o = 1

n1sinC = n2 gives sinC = n2/n1 and if medium 2 is air where n2 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)  49o.

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 0o to 90o 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 45o 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 45o. If the glass is made of a material with a critical angle of less than this (n = 1.5 will give the critical angle 42o), 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 ncore > ncladding.

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) / Colour / n
761 / red / 1.329
656 / orange / 1.331
589 / yellow / 1.333
527 / green / 1.335
431 / blue / 1.341
397 / violet / 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 spectroscopessplit 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 depending on the geometry of the situation. After that, Snell's law gives the angle of refraction 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 axisConcave 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.

 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

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

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 hi may be smaller, greater or equal to that of the depicted object, ho. The magnification of the image m = hi/ho , 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 / u < f / f > v > u / >1 / erect / virtual
2 / u = f / no image (v =) / - / - / -
3 / 2f > u > f / v > 2f / >1 / inverted / real
4 / u = 2f / v = 2f / =1 / inverted / real
5 / u > 2f / 2f > v > f / <1 / inverted / real
6 / u =  / v = f / <1 / inverted / 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 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 = (u-f)/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 = hi / ho = 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.