Answer
1.3 m.
Work Step by Step
The maximum and minimum wavelengths are 750 nm and 410 nm. Calculate the first-order angles for them using equation 24–4.
$$\theta = sin^{-1}(\frac{m \lambda}{d})$$
The reciprocal of the number of lines per centimeter is the slit separation d, in cm.
$$\theta_1 = sin^{-1}((410\times10^{-7}cm)(7800lines/cm))=18.65^{\circ}$$
$$\theta_2 = sin^{-1}((750\times10^{-7}cm)(7800lines/cm))=35.80^{\circ}$$
The distance from the centerline to the first maximum is the distance to the screen, L, multiplied by the tangent of the first-order angles we just calculated.
$$ x_2 = L(tan \theta_2)=3.40m(tan 35.80^{\circ})=2.452 m$$
$$ x_1 = L(tan \theta_2)=3.40m(tan 18.65^{\circ})=1.1475 m$$
The width of the spectrum is the difference.
$$\Delta x = 1.3m$$
