An optical detector window is coated with two dielectric layers deposited sequentially on the air-exposed surface. From top to bottom the structure is
air (n0=1.00) → Layer 1 (n1=1.45, thickness d1) → Layer 2 (n2=2.10, thickness d2) → glass window (n3=1.52).
The engineer chooses the layer thicknesses so that each layer has an optical thickness equal to one-quarter of the target wavelength λ0=520nm in air; that is, n1d1=n2d2=λ0/4. Light is incident normally.
(1) Part 1: (a) For each of the three reflections that can reach an observer—air–Layer 1, Layer 1–Layer 2, and Layer 2–glass—indicate whether the reflected wave undergoes a half-wavelength phase change. Provide brief justification for each interface.
(2) Part 2: (b) Using the chosen thicknesses, determine the phase difference (in radians) between the waves reflected from the air–Layer 1 interface and the Layer 1–Layer 2 interface before they recombine in air. Show your work clearly.
(3) Part 3: (c) Explain qualitatively how the presence of the third reflected wave (from the Layer 2–glass interface) affects the overall reflected intensity, and argue why the engineer’s quarter-wave design still achieves a strong reduction in reflection.
(4) Part 4: (d) Suppose both layer thicknesses were accidentally doubled while keeping all indices unchanged. Predict whether the reflected intensity at λ0 would increase, decrease, or stay approximately the same compared with the intended design, and justify your prediction.