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Let the same things be supposed as before; that is to say, let CO, perpendicular to CR, represent a portion of a wave the continuation of which in the crystal is IK, so that the piece C will be continued on along the straight line CI, while O comes to K. Now if one takes a second period of time equal to the first, the piece K of the wave IK will, in this second period, have advanced along the straight line KB, equal and parallel to CI, because every piece of the wave CO, on arriving at the surface CK, ought to go on in the crystal the same as the piece C; and in this same time there will be formed in the air from the point I a partial spherical wave having a semi-diameter IA equal to KO, since KO has been traversed in an equal time.
Since the sums of the times along AK, KB, and along AC, CB are equal, if from the former sum one deducts the time along KB, and if from the other one deducts the time along RB, there will remain the time along AK as equal to the time along the two parts AC, CR. Consequently in the time that the light has come along AK it will also have come along AC and will in addition have made, in the medium from the centre C, a partial spherical wave, having a semi-diameter equal to CR. And this wave will necessarily touch the circumference KS at R, since CB cuts this circumference at right angles.
I must try something short and sweet: KB, urgent signal, 'Heave all aback'; or LM, urgent, 'The berth you're now in is not safe'; or what do you say to PQH? 'Tell my owners the ship answers remarkably well." "It's premature," I replied; "but it seems calculated to give pain to Trent. PQH for me."
Therefore the whole line AH will represent the time along AD, DB. Similarly the line AC or AF will represent the time along AC; and FH being by construction equal to 3/2 of CB, it will represent the time along CB in the medium; and in consequence the whole line AH will represent also the time along AC, CB. Whence it appears that the time along AC, CB, is equal to the time along AD, DB. And similarly it can be shown if L and K are other points in the curve CDE, that the times along AL, LB, and along AK, KB, are always represented by the line AH, and therefore equal to the said time along AD, DB.
I must try something short and sweet: KB, urgent signal, 'Heave all aback'; or LM, urgent, 'The berth you're now in is not safe'; or what do you say to PQH? 'Tell my owners the ship answers remarkably well." "It's premature," I replied; "but it seems calculated to give pain to Trent. PQH for me."
Then AK being an incident ray, and KB its refraction within the medium, it needs must be, according to the law of refraction which was known to Mr.
Supposing the line AB to be protected along the outer half of it, AK, by no natural obstacle the state of affairs which I have represented by a dotted line αγ; but suppose the second half of it, KB, should be protected by a natural obstacle, though not a very formidable one such as I have represented by the continuous line γβ.
And by prolonging the arc KQ till it meets AD at Y, the sum of the consequents is DY. Then KX ought to be to DY as 3 to 2. Whence it would appear that the curve KDE was of such a nature that having drawn from some point which had been assumed, such as K, the straight lines KA, KB, the excess by which AK surpasses AD should be to the excess of DB over KB, as 3 to 2.
Similarly, if one considers some other point of the wave IK, such as h, it will go along hm, parallel to CI, to meet the surface IB, while the point K traverses Kl equal to hm; and while this accomplishes the remainder lB, there will start from the point m a partial wave the semi-diameter of which, mn, will have the same ratio to lB as IA to KB. Whence it is evident that this wave of semi-diameter mn, and the other of semi-diameter IA will have the same tangent BA. And similarly for all the partial spherical waves which will be formed outside the crystal by the impact of all the points of the wave IK against the surface of the Ether IB. It is then precisely the tangent BA which will be the continuation of the wave IK, outside the crystal, when the piece K has reached B. And in consequence IA, which is perpendicular to BA, will be the refraction of the ray CI on emerging from the crystal.
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