Solution

Physical phenomenon:
We are observing multiple-beam interference (as in a Fabry-Perot interferometer). The incident plane wave partially passes through the first plate, reaches the second, where it again partially passes and partially reflects back. The reflected wave returns to the first plate, reflects from it, and goes to the second one again. At the exit of the system, multiple waves emerge having traveled different optical paths.
The path difference between the directly transmitted wave and the wave that underwent two internal reflections is $\Delta = 2d$ (round trip between the plates).
When the distance $d$ changes, the path difference changes, leading to a periodic alternation of constructive (maxima) and destructive (minima) interference.

Determining the wavelength from the graph:
Condition for a transmission minimum: $2d = \left(m + \frac{1}{2}\right)\lambda$.
Condition for the adjacent transmission maximum: $2d = (m + 1)\lambda$.
Consequently, when transitioning from an intensity minimum to the adjacent maximum, the optical path difference $2d$ changes by $\lambda/2$. It follows that the distance $d$ itself changes by an amount $\Delta d = \lambda/4$.

Let's carefully examine the graph provided in the problem.
On the x-axis, the distance between these points is:
$$\Delta d = (2 - 1) \cdot 10^{-5}\text{ cm} = 10^{-5}\text{ cm}$$
Since this is the distance from a minimum to the adjacent maximum, we equate it to $\lambda/4$:
$$\frac{\lambda}{4} = 10^{-5}\text{ cm} \implies \lambda = 4 \cdot 10^{-5}\text{ cm}$$
(which corresponds to $400\text{ nm}$ — the boundary of visible violet light).