Page:Elementary Text-book of Physics (Anthony, 1897).djvu/412

398 For light the values of $$\lambda$$ are between 0.00039 and 0.00076 mm., and if $$x$$ be taken as 1000 mm., $$\lambda^2$$ will be very small in comparison with $$x\lambda,$$ and may be omitted. The above formulas then become, if $$\sqrt{x \lambda}$$ be represented by $$l,$$ and the several portions into which the wave front is divided are  Taking now the pairs of which the effects at $$P$$ are opposite in sign, we find $$Aa$$ a little more than twice $$ab,$$ while $$bc$$ and $$cd$$ are nearly equal. It is evident, also, that for portions beyond $$d$$ adjacent pairs will be still more nearly equal, and the effect at $$P,$$ therefore, of each pair of segments beyond $$b$$ almost vanishes. The effect at $$P$$ is then almost wholly due to that portion of $$Aa$$ that is not neutralized by $$ab.$$ But, taking the greatest value of $$\lambda,$$ $$Aa = \sqrt{x\lambda} = \sqrt{0.76} = 0.87 \text{mm.},$$ a very small distance. Hence, under the conditions assumed, the effect at any point $$P$$ is due to that portion of the wave-front near the foot of the perpendicular let fall from $$P$$ on the wave-front. It may be demonstrated by experiment that the portions of the wave beyond $$Aa$$ neutralize each other. Suppose a screen $$mn$$ in the position shown in Fig. 94. The point $$P$$ will be in shadow. If the darkness at $$P$$ is due