Physics Help - Diffraction

A thin sheet of plastic of index 1.77 is inserted between two panes of glass to reduce infrared (of wavelength 700 nm) losses.
What thickness is necessary to produce constructive interference in the reflected EM radiation?
Answer in units of nm.

Any help is greatly appreciated!!

For reflected radiation the ray which hits the leading edge has an inverted phase because the plastic has a higher refractive index from glass.
At the other side the reflected ray is right way up ( passing from heavy to light)
So there is a half wavelength phase change by the material.

The wavelength of light is inversely proportional to the refractive index so if you mean that 700nm is the wavelength in air then it is 700/1.77 in the plastic.

Let the thickness be d
For constructive interference we need a further half wavelength path difference ( to add to the half already calculated by the reflection)

2d = 1/2 lambda
d = 1/4 lambda = 1/4 * 700/1.77 = 99 nm

So that is the required thickness of the film