59 
much less light energy in order to obtain the same reaction, than 
with light coming vertically from above. Therefore the intensity and 
the energy numbers must be divided by a certain coefficient. 
Now Voer found that already after a few minutes the rate of growth 
began to fall off. This led after about 25 minutes to a minimum, 
while after 30—40 minutes the zero point was again passed. Then 
an acceleration of growth occurred, leading to a maximum, so that 
at about 60 minutes after the beginning of illumination the first 
rate of growth had been reestablished. After that the growth oscil- 
lated for a long time, at first with considerable bnt later with grad- 
ually decreasing maxima and minima. 
I have now calculated from Voer’s tables’), how many u the 
plants grew less during the first retardation period than if they had 
continued their growth in the dark. The following figures therefore 
give, in u, the total retardation of growth until the zero-point is 
again reached and the acceleration of growth begins. It should always 
be remembered that the intensity numbers must be divided by a coef- 
ficient before being comparable to those of horizontally incident light. 
TABLE I. 
Duration of illumination. 
Intensity. 
1 min. 3 min. 15 min. continuous. 
16 M.C. | 39 83 
64 M.C. 53 
100 M.C. 102 104 90? 
500 M.C. 122 123 
1000 M.C. 294 76 147 
1500 M.C. 89 
Since in unilateral illamination the front is always more strongly 
illuminated than the back, the former is retarded more than the 
latter, if the retardation of growth increases with increasing intensity. 
Consequently a positive phototropic reaction occurs, provided the 
difference in growth retardation is sufficient to give a visible cur- 
vature. If the retardation of growth diminishes again with increasing 
intensity, a negative reaction will oceur, for in this case the back 
side is retarded more than the front. It is seen from the table that 
the first and second columns will give positive curvatures. We cannot 
1) Voer le. Tables 8, 9 and 11. 
