Page:Elektrische und Optische Erscheinungen (Lorentz) 133.jpg

the first medium, and second, we replace t by -t. Then we find, that in 1 the light emerges

$\cos \ 2\pi {\frac {t}{T}}$

when in the paths 2 and 4 the incident motions

$a_{1.2}\ \cos \left(2\pi {\frac {t}{T}}+b_{1.2}\right)$

(120)

and

$a_{1.4}\ \cos \left(2\pi {\frac {t}{T}}+b_{1.4}\right)$

(121)

exist.

However, since the light vector, which is generated by the motion (120) in the first path, has the value

$a_{1.2}a'_{2.1}\cos \left(2\pi {\frac {t}{T}}+b_{1.2}-b'_{2.1}\right)$

and also the light vector emerging form (121), is to be replaced by

$a_{1.4}a'_{4.1}\cos \left(2\pi {\frac {t}{T}}+b_{1.4}-b_{4.1}\right)$,

then it is given

$a_{1.2}a'_{2.1}\cos \left(2\pi {\frac {t}{T}}+b_{1.2}-b'_{2.1}\right)+a_{1.4}a'_{4.1}\cos \left(2\pi {\frac {t}{T}}+b_{1.4}-b_{4.1}\right)=$ $=\cos \ 2\pi {\frac {t}{T}}$

From that if follows

$a_{1.2}a'_{2.1}\cos \left(b_{1.2}-b'_{2.1}\right)+a_{1.4}a{}_{4.1}\cos \left(b_{1.4}-b_{4.1}\right)=1$,

(122)

and

$a_{1.2}a'_{2.1}\sin \left(b_{1.2}-b'_{2.1}\right)+a_{1.4}a{}_{4.1}\sin \left(b_{1.4}-b_{4.1}\right)=0$,

(123)

§ 100. The following remark leads to a simple relation. If we start by a condition, at which the incident light follows path 1, and if we take the mirror image with respect to the second plane of symmetry (§ 96), then we arrive at a condition, at which the light is incident in 2. Consequently it has to be

$a_{2.1}=a_{1.2},\ b_{2.1}=b_{1.2}$

(124)

and in the same way

$a'_{2.1}=a'_{1.2},\ b'_{2.1}=b'_{1.2}$.

(125)