Answer
$\lambda=6.3\times10^{-7}m$
$f = 4.8\times10^{14}Hz$
Work Step by Step
The spacing on the screen between bright fringes is addressed in Example 24-1.
For light shining through 2 slits separated by a distance d, the spacing between successive fringes on a screen a distance $\mathcal{l}$ away is $\Delta x = \frac{\lambda \mathcal{l}}{d}$.
Solve for the wavelength.
$$\lambda=\frac{d\Delta x}{\mathcal{l}}=\frac{(0.048\times10^{-3}m)(0.085m)}{6.50m}$$
$$\lambda=6.3\times10^{-7}m$$
Find the frequency.
$$f=\frac{c}{\lambda}=\frac{3.00\times10^{8}m/s}{6.277\times10^{-7}m }=4.8\times10^{14}Hz$$
