tinue in mutually perpendicular directions to the mirrors, where they are reflected back to B. On arriving back at B, the two beams are recombined as two superposed beams, D and F. If the time taken for the light to go from b to e and back is the phase and will reinforce each other, but if the two times differ slightly, the beap" same as the time from b to C and back, the emerging beams d and F will be will be slightly out of phase and interference will result. If the apparatus is"at rest"in the ether, the times should be precisely equal, but if it is moving toward the right with a velocity u, there should be a difference in the times. Let us see why First, let us calculate the time required for the light to go from b to e and pack. Let us say that the time for light to go from plate b to mirror E is 11, and the time for the return is t2. Now, while the light is on its way from b to the mirror the apparatus moves a distance ut1, so the light must traverse a distance L ut at the speed c. We can also express this distance as cti, so we have cII=L+ utI t1=L/(c-l) (This result is also obvious from the point of view that the velocity of light relative to the apparatus is c-u, so the time is the length L divided by c -u )In a like manner, the time t2 can be calculated. During this time the plate B advances a distance ut2, so the return distance of the light is L-ut2. Then we have ct2EL-ut2, or I2 =L/(c+ u Then the total time is 1+t2=2Lc/(2-u2) For convenience in later comparison of times we write this as t1+t2 (154) Our second calculation will be of the time fa for the light to go from b to the mirror C. As before, during time fa the mirror C moves to the right a distance ut 3 to the position C; in the same time, the light travels a distar hypotenuse of a triangle, which is BC. For this right triangle we have (ct3)2=L2+(ut3)2 23-u213=(c2-n2)r3, For the return trip from C the distance is the same, as can be seen from the symmetry of the figure; therefore the return time is also the same, and the total time is 2f 3. With a little rearrangement of the form we can write 2L We are now able to compare the times taken by the two beams of light. In expressions(15.4)and(15.5)the numerators are identical, and represent the time that would be taken if the apparatus were at rest. In the denominators, the term u"/c will be small, unless u is comparable in size to c. The denominators represent the modifications in the times caused by the motion of the apparatus. And behold these modifications are not the same-the time to go to C and back is a little les than the time to E and back, even though the mirrors are equidistant from B, and all we have to do is to measure that difference with precisio Here a minor technical point arises--suppose the two lengths L are not exactly equal? In fact, we surely cannot make them exactly equal. In that case re simply turn the apparatus 90 degrees, so that bC is in the line of motion and BE is perpendicular to the motion. Any small difference in length then becomes