INF-GEO Solutions, Geometrical Optics, Part 1

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1 INF-GEO Solutios, Geometrical Optics, Part Reflectio by a symmetric triagular prism Let be the agle betwee the two faces of a symmetric triagular prism. Let the edge A where the two faces meet be perpedicular to the plae which cotais the icidet ad emerget rays. wo parallel beams of light are reflected off the two symmetric faces of the prism. a. Show that the agle betwee the two reflected beams is twice the agle betwee the two reflectig surfaces. Fritz Albregtse 200

2 First: the simple case where the icidet beams are parallel to the symmetry plae of the prism: Let the plae of symmetry halve the top agle of the prism. Let half the agle betwee the two reflected beams be called β/2. /2 /2 θ i β/2 θ r he agle of icidece θ i equals the agle of reflectio θ r - ad the same is true for their 90 complemets. Sice the two parallel beams are parallel to the symmetry axis, 90 -θ r =90 -θ i =/2. Now we have a triagle with agles /2, /2 ad 80 β/2. he sum of agles withi the triagle is 80, so 80 = /2 + / β/2 => β = 2. 2 Fritz Albregtse 200

3 Secod: he geeral case of two parallel beams: Oe beam strikes the left had face of the prism uder a agle of icidece θ, givig the relatio γ = θ. he other beam strikes the right had face uder a agle of icidece θ 2, givig the relatio γ = β + θ θ β γ θ 2 he two expressios for γ give: β + θ 2 90 = θ => β + θ + θ 2 = 80 + Ad sice θ + θ 2 = 80, 90-θ 80- we get β = θ 2 => β = 2 Fritz Albregtse 200 3

4 2 Refractio i plae parallel slab of glass a. Verify the expressios for the displacemets d ad l i sectio β -β s d Let the agle of icidece be, while β is the agle of refractio as the beam eters the plae parallel slab. he idex of refractio is, ad the thickess of the slab is. he displacemet, d, give relative to the thickess of the slab, is Ad (/s) = si (90 β). So we get d Sell s law gives: d = ( β ) s si ( β ) s si si cos β cos si β = = s si(90 β ) cos β 2 2 si = si β si β = si, cos β = si d si cos β cos si β = = cos si cos β 2 2 si 4 Fritz Albregtse 200

5 he twice reflected beam: β s 2k l k he twice reflected beam will be displaced relative to the first by a amout l, give i uits of the thickess of the slab: We have established that So that l 2k π 2 s si β = si = cos 2 2si β = cos cos β si = si β si β = si, cos β = l = 2 si cos 2si cos = si si 2 si 2 Fritz Albregtse 200 5

6 3 Dispersio i a plae parallel slab of glass Assume that a thi beam is icidet o a plae parallel slab of glass i air, as i sectio But ow the beam is ot moochromatic; it is white light, so the beam is spread out ito a spectrum as it passes through the slab. a. Will the emergig rays of differet colors be parallel or ot? For each color there will be a differet value of the idex of refractio,, givig differet displacemets d for differet wavelegths. But all the displaced beams will be parallel to the iput beam, Ad therefore also parallel to each other. b. What determies the thickess of the beam as it exits the slab? - he dispersio of the glass slab used, i.e. the variatio of the idex of refractio with wavelegth, - he thickess,, of the slab. - he agle of icidece. 0,5 displacemet of beam 0,4 0,3 0,2 0, d/ l/ (d/) (l/) -0, agle of icidece, degrees 6 Fritz Albregtse 200

7 4 Critical agle ad total iteral reflectio Assume that we have a a semi-circular bowl of water at 25 C. A light-ray from a m laser eters perpedicular to the surface 4/0 of the radius from the bowl cetre. We wat to obtai grazig refractio ad total iteral reflectio of the light beam that is reflected towards the water / air iterface. a. Does the material of the bowl play ay role i this? No. he material i the bowl oly reflects the beam towards the water / air iterface. b. How much do we have to raise the refractive idex of the water by icreasig the saliity? he refractive idex of water ca be foud at refractio idex of water at 20 C 30 refractive idex,34,335, saliity, percet It icreases approximately liearly from for pure water. he slope of the curve is , ad we make the assumptio that this may be extrapolated liearly. Fritz Albregtse 200 7

8 At the poit where the reflected beam hits the water surface, we have si(θ i /2) = 0.4 => θ i =2 arcsi(0.4) Grazig refractio occurs whe w( s)si( θ i ) = a a w( s) = = = si[ 2arcsi( 0.4) ] s = s = θ i /2 θ i Fritz Albregtse 200

9 5 Atmospheric refractio Make the simplifyig assumptio that the Earth s atmosphere is uiform (thus havig a uiform idex of refractio), ad that it exteds to a height h. Beyod that, we assume that there is vacuum.he Earth s radius is R. a. Verify that as we observe a object settig o the horizo, uder these assumptios it is actually a agle δ below the horizo, give by δ = R arcsi R + h arcsi R R + h h φ δ R A object o the horizo is lifted by refractio by a agle δ. Assumig vacuum outside a uiform atmosphere, Sell s law gives: si(δ+φ) = si φ δ+φ = arcsi( si φ ) But si φ is give by si φ = R/(R+h) => φ = arcsi[r/(r+h)] so δ + arcsi[r/(r+h)] = arcsi[ R/(R+h)] => δ = arcsi[ R/(R+h)] - arcsi[r/(r+h)] Fritz Albregtse 200 9

10 b. Calculate δ for R = 6378 km ad h = 20 km. Assume that = δ = arcsi[ R/(R+h)] - arcsi[r/(r+h)] = arcsi[.0003 * 6378 / 6398 ] - arcsi [ 6378 / 6398 ] = 85, = 0.22 = 0.22 *60 = 3.2 c. How does this compare to the statemets about atmospheric refractio i sectio ? O the horizo itself refractio is about 34', but oly 29' half a degree (oe solar diameter) above it. Our simple model where refractio oly occurs at the top of a uiform atmosphere is clearly too simple, as it uderestimates the horizotal refractio. Multiple choice geometrical optics. What do we mea by critical agle at a boudary betwee two optical media? he agle of icidece where equal parts of refractio ad reflectio occurs he largest agle of icidece where all light is reflected he smallest agle of icidece where o light is reflected he smallest agle of icidece where all light is refracted he agle of icidece where refracted light is taget to the boudary 0 Fritz Albregtse 200

11 2. Geometrical optics (20 poits). We ca use a umber of optical prisms to alter the directio of a light beam. A equilateral right agle prism will chage the directio by 90, as show i the sketch to the right. P γ β a) Below we give you two figures from the curriculum text showig the reflectio coefficiet of p-polarized light (polarized i the plae of the sketch) at the trasitio from air to glass (left) ad glass to air (right). Rp Rp β Reflectio coefficiet Agle of icidece Reflectio coefficiet 0 γ Agle of icidece Mark which part of the figures that describe the situatio at poits, β, ad γ i the first sketch. What do we call the pheomeo that occurs at the poit β? At, the light beam goes from air to glass at a icidece agle of zero, ad a small fractio of the icidet light is reflected (R = 0.04), as idicated by the circled poit to the left i the left had figure above. At β, the light beam is reflected at the glass/air iterface at a icidece agle of 45. Whe movig from a more dese medium ito a less dese oe (i.e. > 2 ), above a icidece agle kow as the critical agle, all light is reflected ad R =, as illustrated i the right had figure. his is kow as total iteral reflectio. he critical agle is approximately 4 for glass i air. hus, the reflectio coefficiet is exactly.0 at β, as idicated by the circled poit i the right had figure above. he reflectio agle is equal to the icidece agle at β (45 ). herefore, the beam strikes the glass/air iterface orthogoaly at γ, so the reflectio coefficiet (R = 0.04) here is foud i the left had circle of the right had figure above. b) We substitute the prism above by a right agle Brewster prism, where oe agle is give by θ B = arctg( 2 / ), where 2 is the refractive idex of glass, ad the refractive idex of air. We place the prism i the light path from P, as show i the figure to the right, so that the icidece agle is θ i = θ B 56. Fritz Albregtse 200

12 Now the refractio agle θ r is give by θ i + θ r = π/2. Draw ad explai the path of the light beam through the prism. γ ( θ B P β At a icidece agle equal to the Brewster agle, p-polarized light goig from air to glass is ot reflected, so there is purely refractio at. he icidece agle at β is give by π/2 θ r = π/2 (π/2 - θ i ) = θ i = θ B which is larger tha the critical agle (4 ). So there is total reflectio at β. he reflectio agle at β is equal to the icidece agle. So the agle betwee the icidet ray ad the glass/air iterface at γ is π - θ B θ r = π - θ B (π/2 - θ B ) = π /2. Which meas that the icidece agle at γ is 0, ad a small fractio R is reflected while (-R) is trasmitted, orthogoal to the iterface. c) How much light is reflected back to P i exercise b, compared to the equilateral prism i exercise a, if = ad 2 =.5? At ormal icidece (θ i = 0), the reflectio coefficiet i the two figures is give by R =[( - 2 )/( + 2 )] 2. For = ad 2 =.5 we get R = 0.25/6.25 = I exercise b, o light is lost from through γ. At γ, 4% (R) is reflected back to β. At β there is oly reflectio to, ad at there is o reflectio (see right had figure for icidece agle = 34 ), oly refractio to P. So 4% (R) is reflected ad refracted back to P. I exercise a, R is reflected at. At γ, R(-R) is reflected via β to. R(-R) 2 goes to P while R 2 (-R) is reflected back to γ via β. From γ, R 3 (-R) is reflected via β to. Now R 3 (-R) 2 goes to P ad R 4 (-R) 2 goes to γ. So R + R(-R) 2 + R 3 (-R) 2 + R 5 (-R) should be summed at P, givig R + R -2R 2 + R 3 + R 3 2R 4 + R 5 + R 5-2R 6 + R = 2R( R + R 2 R 3 + R 4 R 5 + R ) = 2R/(+R). So the ratio of the reflected light i b) to the reflected light i a) is R / (2R/(+R)) = ( + R)/2, or If we just cosider the first two cotributios i exercise a, R + R(-R) 2, the ratio becomes /(2-2R + R 2 ) = 0,5204, which is a little more tha 0.5, ad very close to the fial sum. 2 Fritz Albregtse 200

REFLECTION AND REFRACTION

REFLECTION AND REFRACTION RFLCTON AND RFRACTON We ext ivestigate what happes whe a light ray movig i oe medium ecouters aother medium, i.e. the pheomea of reflectio ad refractio. We cosider a plae M wave strikig a plae iterface