Answer
In situations (c) and (d), Equation 35-36 corresponds to the reflections yielding maxima.
Work Step by Step
Let the indices of refraction be $n_1, n_2,$ and $n_3$ from top to bottom.
If $n_2 \gt n_1$, there is a phase shift of $\frac{1}{2}~\lambda$ due to reflection of the ray reflecting at the interface between layer 1 and layer 2.
If $n_3 \gt n_2$, there is a phase shift of $\frac{1}{2}~\lambda$ due to reflection of the ray reflecting at the interface between layer 2 and layer 3.
(a)
At the interface between layer 1 and layer 2, there is a phase shift of $\frac{1}{2}~\lambda$
At the interface between layer 2 and layer 3, there is a phase shift of $\frac{1}{2}~\lambda$
For a maximum to occur, the path difference should be $~~2L = (m)~\frac{\lambda}{n_2}$
(b)
At the interface between layer 1 and layer 2, there is no phase shift.
At the interface between layer 2 and layer 3, there is no phase shift.
For a maximum to occur, the path difference should be $~~2L = (m)~\frac{\lambda}{n_2}$
(c)
At the interface between layer 1 and layer 2, there is no phase shift.
At the interface between layer 2 and layer 3, there is a phase shift of $\frac{1}{2}~\lambda$
For a maximum to occur, the path difference should be $~~2L = (m+\frac{1}{2})~\frac{\lambda}{n_2}$
(d)
At the interface between layer 1 and layer 2, there is a phase shift of $\frac{1}{2}~\lambda$
At the interface between layer 2 and layer 3, there is no phase shift.
For a maximum to occur, the path difference should be $~~2L = (m+\frac{1}{2})~\frac{\lambda}{n_2}$
In situations (c) and (d), Equation 35-36 corresponds to the reflections yielding maxima.
