Chapter 36, Problem 50CP

Slit 1 of a double-slit is wider than slit 2 so that the light from slit 1 has an amplitude exactly 3 times that of the light from slit 2. Show that Equation 36.9 is replaced by the following equation for this situation:

$I=I_{\max }\left[1+3{\cos }^2\left(\frac{\pi d\sin \theta }{\lambda }\right)\right]$

Begin by assuming that the total magnitude of the electric field at point P on the screen in Figure 36.4 is the superposition of two waves, with electric field magnitudes

$E_1=3E_0\sin \omega t{E}_2=E_0\sin (\omega t+\varphi )$

The phase angle ϕ in E₂ is due to the extra path length traveled by the lower beam in Figure 36.4. You will need to evaluate the integral of the square of the sine function over one period. Refer to Figure 32.5 for an easy way to perform this evaluation. You might find the following trigonometric identities helpful:

$\begin{array}{c}\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\\ \sin (A+B)=\sin A\cos B+\cos A\sin B\\ \cos A=2{\cos }^2\left(\frac{A}{2}\right)-1\end{array}$