Problem 27.50

27.50.
First, I called the thickness of the plastic t. Then the added distance that the light must travel from the lower slit to the zero order maximum compared to that of the higher slit is given by $\Delta r$. As light advances through a distance t in air, it goes through $\frac{t}{\lambda_{air}}$ cycles. As light goes through the same amount of thickness of the sheet, t, it goes through $\frac{t}{\lambda_{air}/n}$ cycles. (Where n is the index of refraction of the plastic.) Thus, the sheet introduces a phase difference of
$\phi = 2\pi \left(\frac{nt}{\lambda_{air}} - \frac{t}{\lambda_{air}}\right)$
We want the case that the difference in path length travelled by the lower beam is balanced by the changing number of optical cycles gone through by the upper beam. The corresponding difference in the optical path length is given by
$\Delta r = \phi(\frac{\lambda_{air}}{2\pi})$
$\Delta r = 2\pi \left(\frac{nt}{\lambda_{air}} - \frac{t}{\lambda_{air}}\right) (\frac{\lambda_{air}}{2\pi})$
With some cancellation, we see that
$\Delta r = (n-1) t$
Note that the wavelength of the light does not appear in the equation. Also, since the angle between the center of the two slits and the first zero order maximum is approximately the same as the angle from either of the slits to the same maximum, we can approximate the angle $\theta$ by
$\tan \theta = \frac{\Delta r}{d}$
$\tan \theta = \frac{y'}{d}$
Substituting in for y' gives us
$y' = \Delta r (\frac{L}{d})$
$y' = \frac{t (n-1) L}{d}$
$y' = \frac{(5 \times 10^{-5} m)(1.50 - 1)(1.00 m)}{3 \times 10^{-4} m}$
y' = .0833 m
y' = 8.33 cm