Optics pH2310 02/03 Solutions from GAG A 1 (a) Chromatic Resolving Power is defined as where Sh is the wavelength

separation of two spectral lines of wavelengths h and h + 6h just resolved on the Rayleigh Criterion as shown below. Because many spectrometers produce fringes of several orders we have to restrict our definition to a comparison of lines of the same order (the same p). (May get answers based on the Taylor criterion, which is better for considering multi-beam fringes; adjacent lines overlap at half-max height.) lmark (112 mark awarded for sketch below or Taylor equivalent)

{resultant intensity

position

(Some sort of sketch of the following would be helpful, but most of the marks would be awarded for a

qualitative description alone)

* l Reflectivity = 90%

Intensity

I I 1 I I I I I 1 I 1

5 6 - 1 1 2 3 4

4 1

Transmitted Intensity

Intensity 'distribution of fringes due to 2-beam (above) and multiple beam interference

b

Reflectivity = 5%

We see that as reflectivity increases we get multi-beam transmission and the sharpness of the fringes

increases. We will then have sharp bright fringes on a dark background, which is an ideal spectrometer.

(2.5 marks)

PH2310A Optics 02/3 Solutions GAG

1 (b) The arrangement shown below is for producing Newton's rings by reflection

(Only the small inset is needed).

4 beam divider collimated beam of monochromatic light reflecting

p u r f a c e s see inset

dark background

Newton's Rings are formed due to interference in the film between an optical flat and lower surface of a ' convex lens. At the centre t = 0, for perfect contact.

1 Now 2t = (p + 7 )A for a bright fringe, so the central fringe (a spot) is black, the zero order fringe.

Consider the positions of the fringes. Let ap = the radius of the pth dark fringe, tp = depth of film at the p th dark fringe and R the radius of curvature of the lens, we see that ~2 = (R - tp )2 + ap2 and since tp << R, ap2 - 2Rtp.

2 Now tp = for the dark fringe and therefore F = ph

(1 C) (i) The X- and Y- motions are antiphase so motion is like a backslash i.e along the line X=-y

(ii) Again a straight line but X displacement is twice as large as the y displacement i.e along the

line y= x/2. So (i) and (ii) are both plane polarised.

(iii) X motion is 900 ahead of the y-motion, so a circle is described in an anticlockwise sense. So I

light is Left handed circularly polarised.

PH2310A Optics 02/3 Solutions GAG

(l d) The incident ray, reflected ray and normal to the surface are coplanar.

The angle of incidence is equal to the angle of reflection. Using the fact that the angles of incidence and reflection are equal

X = ( i+8) - ( i - 8) = 28 (see figure) we have,

50 mm illuminated and viewed from above

20 fringes correspond to a wedge increase in thickness of 10h, as light go there and back and we get a

fiinge for each wavelength increase in path difference.

Exercise is one of comparing sides of similar triangles; let t be the thickness of the wire so that

thickness1 length = t l 50mm = 10hI lmm. So t = 50.10. 600. 10-~.= 0.3mm.

(If) Interference by Division of Amplitude; by dividing a single beam by means of partial reflection

and transmission (each is a replica-phase-map of original), and then recombined to produce interference.

Interference by Division of Wavefront; in which the neighbouring parts of a wavefront are divided and

then recombined to produce interference.

(Examples quoted are those included in the course)

Interference by Division of Amplitude; Newton's Rings, Michelson Interferometer, Fabry-Perot plates.

Inteiference by Division of Wavefront; Young's slits, Lloyd's mirror, Fresnel's Biprism

PH2310A Optics 02/3 Sollctiotls GAG 15:03 February 6, 2003

Q2b When a = 0 the first factor in the intensity distribution (the sinc squared function) tends to 1. sin 2P = 2 sin p cos p so squaring this, the sine squared terms cancel leaving us with 4 cos2 P, where P is related to the phase so we have the 2-beam interference pattern for a double-slit which is observed in Young's double slit experiment (for example). (I am not requiring students to relate P to the geometry of the slits at this stage-- that comes in the last part). Q2c Shown below is a sinc2 function which is the first factor in the intensity distribution. It provides an envelope curve (upper bound) for the function plotted above producing the type of curve shown (bottom). The local features are generally unchanged but there is a very rapid fall-off in intensity from the straight through position, followed by weaker bands of lower intensity.

PH2310A Optics 02/3 Solutions GAG 4 15:03 February 6, 2003

and (2) = d sin 8 S

Slit

It is necessary to turn to the diffraction at the obstacle, where 8 is the angle with the forward normal and a the slit width and then the maximum phase difference over the slit leads to the expression a = d h . a sine. Similarly for the slit spacing we arrive at P = d h d sine.

Missing oders occur ( as shown in the plot above) when principal maxima of the diffraction envelope (first factor) coincide with the zeros of the (second factor).

For an interference maximum, p = d h .d sine = pn, where p = 0, c l , k 2 . . .and p, an integer, is the order of the interference fringe.

The diffraction envelope zeros occur at a = d h a sine. = qn, where q = + 1, + 2 .q is an integer

If d:a is an integer ratio greater than or equal to 2 then have missing orders when the p and q values coincide. In the plot shown d:a as 3: 1 so the 3d ,6"', 9' orders of the interference pattern are missing. (We can't have d:a as 1: 1 as that wouldn't be a grating)

PH2310A Optics 02/3 Solutions GAG 15:03 February 6,2003

Q3a 1) A ray, which is parallel to the axis, emerges from the lens in a direction such that it passes through the secondary focal point F2 of a converging lens, or appears to come from the secondary focal point of a diverging lens. 2) A ray through the centre of the lens is (essentially) undeviated. 3 A ray through (or heading towards) the first focal point F1, emerges parallel to the axis. 3b) (candidates are free to choose to do the this part either as a scale drawing and measure from it or draw diagrams approximately to scale and calculate magnification, tube length)

Astronomical Telescope

e From similar triangles, we see, angular magnification M = -

U 0

When the astronomical telescope is in normal adjustment, i,e. with its final image at infinity, ue W f 2 e - -- f 0

uo - h21fo (geometrically, paraxial rays) , numerically M = - f e

For use in normal adjustment, the image (like the object) is at infinity. Note that the magnification is again the ration of the focal lengths of the objective

f 0 and eyepiece M = -

f e

The tube length is equal to fo-fe rather than fo+fe which can be considerably shorter

than the astronomical telescope for low magnifications 3c) Angular magnification is ratio of the focal lengths of the objective to eye-piece (Convex for astronomical, concave for Galilean) So magnification is 0.210.05 = 4 for each instrument. Instrument length is for astronomical telescope equal to the sum of the focal lengths of the objective and eye-piece (= 0.25m) and for the Galilean telescope equal to the difference of the focal lengths of the objective and eye-piece (= 0.15m). 3d) The terrestrial telescope is used when an erect image is required and a shorter instrument length (the tube length is equal to fo-f, , rather than fo+fe which can be considerably shorter than the astronomical telescope for low magnifications) So terrestrial telescope has advantages for bird watching or opera glasses, a day at the races etc. The disadvantage of terrestrial is that the eye-ring (best place for eye for viewing) is inside the telescope.

PH2310A Optics 0213 Solutions GAG 6 15:03 February 6, 2003

G3 t+Lz.rn a .----"--.- r! - - iy.;.o.."%+ 7.45';-

I O w C fc W' \ a ci;sa

Y QN. ; \5 S~L" 3 ~ ~ 5 ~ ~ ~ r ' ' ro e m s ~ h i s

.$ Q T- f m 5 ~ . . ~ ) ,,... . irror -.-. .&--&a -

l?, so ( \S 6 ~ - I Q [h S

--

(b) White light fringes are only observed at the zero-path-difference position, which would be midway between the positions described being mirror images of each other. So 13.5 mm is correct setting. The problem(s) with observing white light fringes is that they can only be observed when the path-difference setting is within 4 or 5 fringes of the z.p.d.p. and that assumes that the mirrors are almost perpendicular so that fat richly coloured fringes are available. Using a white light source alone it can be a very frustrating business to find them. With a monochromatic source the approximate position of the 2.p.d.p. can be found- using the straight line fringe position of the first part of this part of the question-but the w.1. fringes will not be observed until the w.1. source is substituted for monochromatic source. This again can be problematic. A good compromise is to use both a monochromatic source and a white light source simultaneously, so that fringe straightness and colours can both be observed simultaneously.

(C) For good visibility (and below for poor visibility); {Marks: 2.5 for each plot, l each for visibility and 2 for wavelenth separation)

2 2 = 3 COS X + 2 C O S ( X ) Visibility = Imax - Imin

= (5-0)/(5+0) =l Imax + Imin

y = 3 cos2, + 2 cos2 (X + i)

Imax - Irnin Visibility = I = (3-2)/(3+2)=0.2

max + Imin

PH2310A Optics 02/3 Solutions GAG

-'t If there are 1000 fringes between the positionsof good visibility the wavelengths differ by l pan in 1000 so the separation of lines centred on 590 nm is 0.59 nm.

PH2310A Optics 0213 Solutions GAG 15:03 February 6, 2003

This topic was approached in lectures by first considering a problem which students find relatively easy,

but also involves a cylindrical wavefront, namely "how to deduce the intensity distribution for

Fraunhofer diffraction using an amplitude-phase diagram", and then considering the Fresnel diffraction

as a modified version of this, so it is likely that this approach will be taken in the exam, but direct

approach is of course OK.

Consider the Fraunhofer diffraction at a single slit, which gives the sinc* function. Contributions from

equal elements over the wavefront are of equal magnitude so as the phase between successive elements

increases as we move further away from the straight through position, the arc of the circle which results

is bent round tighter and tighter.

The resultant amplitude is found by joining the ends of this arc and the resultant amplitude

squared to find the intensity.

(the problem is similar to taking a string of fixed length, length fixed by the width of the slit, and

stretching it out straight for the straight through position. For points further away from the straight

through position the string is arranged as the arc of a circle which eventually winds round into a full

circle (at the angle to first min position), then in a circle and a half, two circles etc For Fresnel diffraction the contributions from "each zone" successive contributions frmin-phase and anti-phase regions of are no longer equal, but are of decreasing strength, so we get a spiral instead of a circle, (Cornu's Spiral). The method of determining intensity distributions is the same however. The straight edge obstacle is dealt with as follows. The (amplitude) contribution from the total wavefront is obtained by joining the ends of the spiral. The amplitude at any point which is either in the geometric shadow or the geometrically illuminated section can be obtained by using a restricted part of the Cornu Spiral. The upper (inner) spiral point will always be one end point, but the other end point will be anywhere along the spiral depending on how much of the wavefront the point under consideration can "see". At the edge of the geometric shadow the origin is the other end point. Into the bright section we find a max corresponding to the most distant point on the curve followed by "wiggles" as more of the spiral is included. The rest of the question is about these contributions.

PH2310A Optics 0213 Sollitions GAG