Answer:
The answer is "[tex]1 \times 10^{-6}\ s[/tex]"
Explanation:
[tex]v = velocity = 55.5 \ \frac{cm}{s} = 55.5 \times 10^{-2} \ \frac{m}{s}\\\\\lambda = wavelength = 555 \ nm = 555 \times 10^{-9}\ m\\\\\upsilon = frequency\\\\v = \lambda \upsilon \\\\\upsilon = \frac{v}{ \lambda}\\\\[/tex]
[tex]\upsilon= 55.5 \times 10^{-2} \ \frac{m}{s} \div 555 \times 10^{-9}\ m\\\\\ \ \ \ \ = \frac{10^6}{s}[/tex]
In 1 second, [tex]10^6[/tex] cycles take place.
1 wavelength cycle included. Thus, [tex]10^6[/tex] cycles include [tex]10^6[/tex] wavelengths.
[tex]10^6[/tex] wavelengths 1 second until point A [tex]10^6[/tex]
1 wavelength is [tex]\frac{1}{10^6}\ sec = 10^{-6}[/tex]seconds after A.
