Answer
$8.72\times10^{-5}m$.
Work Step by Step
Section 24–9 discusses the Michelson interferometer. The change in path length, for the beam that strikes the movable mirror, is two times the distance that the mirror moves. That is because the beam goes toward the mirror, hits it, then returns.
One fringe shift occurs when the path length changes by one wavelength $\lambda$, and so this corresponds to the mirror moving by $\frac{1}{2}\lambda$.
Let N represent the number of fringe shifts observed when the mirror moves by a distance $\Delta x$.
$$N=\frac{\Delta x}{0.5\lambda}$$
The problem tells us that when the mirror moves a distance equal to the thickness of the foil, 296 fringe shifts are observed. Solve for the distance.
$$\Delta x=\frac{1}{2}N\lambda=\frac{1}{2}(296)(589\times10^{-9}m)=8.72\times10^{-5}m$$
This is the thickness of the foil.
