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For if the angle DAQ or CBA is such that in the triangle ACB, CB is equal to 2/3 of AB, or is greater, then AN cannot form one side of the triangle ANB, since it becomes equal to or greater than AB: so that the portion of wave BN cannot be found anywhere, neither consequently can AN, which ought to be perpendicular to it. And thus the incident ray DA does not then pierce the surface AB.
"Then why don't you try harder?" "I do, but I am so stupid." "You are not, Dexter. You always learn easily enough with me." "Yes, with you," said the boy quickly, "but you don't want me to say angle ABC is equal to the angle CBA, and all such stuff as that." "Don't call it stuff," said Helen, smiling in spite of herself; "it is Geometry." "But it is rum stuff all the same.
When the angle DAQ is still large enough to enable the ray DA to pass, it is evident that the light from the portion AC of the wave is collected in a minimum space when it reaches BN. It appears also that the wave BN becomes so much the smaller as the angle CBA or DAQ is made less; until when the latter is diminished to the limit indicated a little previously, this wave BN is collected together always at one point.
For the triangles ACB, BNA being rectangular and having the side AB common, and the side CB equal to NA, it follows that the angles opposite to these sides will be equal, and therefore also the angles CBA, NAB. But as CB, perpendicular to CA, marks the direction of the incident ray, so AN, perpendicular to the wave BN, marks the direction of the reflected ray; hence these rays are equally inclined to the plane AB.
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