Apparent Depth. B' l'

Transcription

1 REFRACTION by PLANE SURFACES Apparet Depth Suppose we have a object B i a medium of idex which is viewed from a medium of idex '. If '<, the diagram of this would be like that below. ' θ' '< B θ θ B' l' θ' a O l What does the observer O, who is lookig alog the ormal to the iterface, see whe he looks at object B? The diagram shows a typical ray leavig B. It makes a agle θ with the ormal whe it his the iterface. It leaves the iterface at agle θ' where from Sell's law, siθ'=(/')siθ. (1) If the object is a distace l from the iterface ad the ray itersects the iterface at a poit a distace a from the lie of sight, by trigoometry taθ=a/l. (2) If the extesio of the refracted ray itersects the optical axis at l', the from trigoometry taθ'=a/l'. (3) plae surfaces, page 1 W. F. Log, 1992

2 Now let's assume that the rays that actually eter the observer's eye are very early parallel to the optic axis, the so-called paraxial approximatio. The from the small agle approximatio θ siθ taθ ad θ' siθ' taθ' so equatios (1), (2), ad (3) become θ'=(/')θ. θ=a/l. θ'=a/l'. Substitutig (2') ad (3') ito (1') gives '/l'= /l. Note that the result is idepedet of a, so all rays from B itersect the axis at the same poit B'. That would look like the followig. (1') (2') (3') (4) ' '< B B' O To the observer all the light seems to be comig from poit B' a distace l' beeath the iterface, ot from poit B a distace l below the iterface. B' is the image of B ad equatio (4) is the fudametal paraxial equatio for a plae surface relatig object ad image distaces. Note, this oly happes for paraxial rays. If rays make large eough agles with the iterface, the refracted rays do ot all itersect at poit B', ad the image is somewhat blurred or aberrated. plae surfaces, page 2 W. F. Log, 1992

3 Example: A pey is at the bottom of a swimmig pool three meters deep. How far below the surface of the water does the pey seem to be to a diver lookig dow o it. I this case l=3m, =1.33 (the idex of water), ad '=1 (the idex of air) so the apparet depth of the pey is l'=('/)l=(1/1.33)(3m)=2.26m. Note that the coi looks closer to the surface tha it really is. Total Iteral Reflectio θ c I goig from deser to rarer medium, part of the eergy of a ray is reflected, part refracted away from the ormal to the surface. At oe particular icidet agle θ c, the refracted ray makes a 90 agle with the ormal. For ay agle of icidece greater tha θ c, all the eergy of the ray is reflected. The agle θ c, is called the critical agle. It satisfies Sell's law siθ c ='si90 =' siθ c ='/. Critical agle has a umber of practical applicatios i producig a mirror surface without silverig. It is used i fiber optic probes, diamods, a variety of reflectig prisms, etc. plae surfaces, page 3 W. F. Log, 1992

4 Example: What is the critical agle for light goig from water to air? θ c =arc si('/)=arc si(1/1.33)=49. Example: A hero is hutig a frog hidig beeath a lily pad of 50cm diameter. What is the maximum distace the frog may go beeath the lily pad before the hero ca see him? lily pad 25cm x c The frog will remai ivisible to the hero so log as rays from the frog's body are totally reflected at the air-water iterface. The maximum depth for which this will be true is beeath the ceter of the lily pad. At that maximum depth, rays from the frog will strike the edge of the lily pad at the critical agle which we calculated to be 49 i the previous example. From the diagram that distace, x, is give by 25cm/x=ta49 so x=25cm/ta49 =21.9cm. Prisms Let a ray of light strike a wedge of refractive material, a prism. The wedge has apical agle α ad idex ' ad is surrouded by a medium of idex. If, as is always the case i realistic problems, the prism is more optically dese tha the surroudig mirror, the ray is bet toward the prism base as it passes through each prism face. The questio of iterest is, what is the agle through which the icomig ray is tured? plae surfaces, page 4 W. F. Log, 1992

5 α copy of icomig ray ' Calculatig is, i geeral, a complicated geometric problem. But i ophthalmic optics, oly thi prisms, those of small apical agle, are importat ad oly rays icidet early ormally are cosidered. That simplifies the problem a lot. The diagram below shows such a prism. α α θ' copy of icomig ray ' Let's simplify the calculatio by havig the ray icidet ormally o the first face of the prism. It is ot, the deviated by the first face ad strikes the secod face such that it makes a agle α with the ormal. It leaves the prism at a agle θ' with the ormal. From Sell's law, 'siα=siθ' or 'α θ'. But the agle of deviatio is related to θ' by θ'=α+. (5) (6) plae surfaces, page 5 W. F. Log, 1992

6 Elimiatig θ' from (5) ad (6), =[('/)-1]α. (7) Eve though (7) was derived for a early ormally icidet ray, it is true for ay ray icidet early ormally. That meas that all such rays are deviated through exactly the same agle. If the rays come from a object B, as i the diagram below, a observer sees them as if they came from a image B', displaced vertically from B but ot displaced logitudially. B' B Ophthalmic prisms are thi prisms. They are measured i prism diopters of prism power. A prism which causes a deviatio of a ray of light of 1cm per meter travelled by the ray has oe prism diopter of prism power. The diagram below shows how a oe diopter prism deviates a ray of light. 1m 1cm From the diagram, the agle of deviatio is (radias)=1cm/1m=0.01. If the prism has prism diopters of power, it moves the beam cetimeters for every meter travelled ad = cm/1m=0.01 or sice =1 for prisms i air. =100(radias)=100('-1)α(radias) (8) plae surfaces, page 6 W. F. Log, 1992

7 Example: What is the power of a crow glass prism of apical agle 4? α=4 =(π/180 )x4 =0.070 radias ad '= Pluggig ito (8) gives =100( )(0.070)=3.65prism diopters 4 p.d. Superpositio of Thi Prisms As show i the diagram above, a ophthalmic prism may be represeted as a vector with the magitude of the prism ad directio from apex-tobase. It would have bee icer if the opposite directio sice the image moves toward the apex ad the prism sort of poits that way, but that just is't how it's doe. Superimposed prisms may be combied by just addig the correspodig vectors ad they may be resolved ito compoets, just like ay other vectors. plae surfaces, page 7 W. F. Log, 1992

8 Example: The apex-to-base lie of a 6p.d. prism makes a agle of 60 with the x-axis. What is the prism power i the horizotal ad vertical meridias? Draw the vector represetatio of the prism. From trigoometry x =6cos60 =3p.d., y =6si60 =5.2p.d. =6 y The whole vector could be writte i compoet form as =(3.0, 5.2). 60 x Example: What prism is equivalet to a 5p.d. 45 prism superimposed o a 3p.d. base dow prism? Resolve each prism ito compoets. The vector correspodig to the first prism is 1 =(5cos45,5si45 )=(3.5,3.5). The vector correspodig to the secod prism is 2 =(0,-3). The total prism is the vector sum, = 1+ 2 =(3.5+0,3.5-3)=(3.5,0.5). The magitude of the resultat prism is = [(3.5)2+(0.5)2]=3.54p.d. ad the agle is θ=arc ta(0.5/3.5)=8 8' ' plae surfaces, page 8 W. F. Log, 1992