Answer
481 nm.
Work Step by Step
The path difference is a whole number multiple of the wavelength since we have constructive interference (equation 24–2a).
$$d sin \theta = m \lambda$$
The distance on the screen from the centerline is given by $x=\ell tan \theta$ (figure 24–7c). For small angles, we have the sine and tangent functions are approximately equal.
$$d sin \theta \approx d tan \theta =d\frac{x}{\ell} =m \lambda$$
$$x_1=\frac{\lambda_1 m \ell}{d}$$
$$x_2=\frac{\lambda_2 m \ell}{d}$$
$$\Delta x=x_1-x_2=\frac{\lambda_1 m \ell}{d}-\frac{\lambda_2 m \ell}{d}$$
Finally, m = 2 for second-order fringes. Solve for the second wavelength.
$$\lambda_2=\lambda_1-\frac{d \Delta x}{m \ell}$$
$$\lambda_2=\lambda_1-\frac{d \Delta x}{m \ell}$$
$$\lambda_2=650\times10^{-9}m-\frac{(6.6\times10^{-4}m)(1.23\times10^{-3}m)}{2 (2.40m)}\approx 481 nm$$
