WAEC - Physics (2016 - No. 29)
\(\frac{V}{L}\)
\(\frac{5VL}{2}\)
\(\frac{5L}{4V}\)
\(\frac{5V}{4L}\)
توضيح
L = \(\frac{5\lambda}{4}\)
\(\lambda = \frac{4L}{5}\)
But \(\lambda = f\lambda\)
f = \(\frac{v}{\lambda} = \frac{v}{\frac{4L}{5}}\)
= \(\frac{5v}{4L}\)
\(\lambda = \frac{4L}{5}\)
But \(\lambda = f\lambda\)
f = \(\frac{v}{\lambda} = \frac{v}{\frac{4L}{5}}\)
= \(\frac{5v}{4L}\)