Explain why a bloomed lens observed by reflection in daylight appears purple?

I got this question word for word from the book Physics for advanced level by Jim Breithaupt. I dont quite understand wat it means to have a bloomed lens.





Also with diffraction you get constructive and destructive interference, and you represent this with a intensity graph. When are the Intensity for all orders the same height on the graph and when do they decrease as the order (fringe) number increases?





Sorry if i dont make sense i dont know how i can write better than thisExplain why a bloomed lens observed by reflection in daylight appears purple?
Blooming is a treatment given to lenses to increase the amount of light that they transmit. Its not true that a lens will look purple if bloomed - the colour will depend on the type of blooming - but purple is a common outcome of blooming.





The process works by coating the lens with a thin layer of material of a lower refractive index than the glass but a higher one than air. Any light reflected at the glass-bloom interface has a 180 degree phase change, and at the bloom-glass interface again a 180 degree phase change. If the bloom layer is the right thickness this makes the two cancel - so there is no net reflection and more light power is transmitted. It will work perfectly for light of one frequency only when the layer thicknes is a multiple of the wavelength/2.





Of course, not ALL of the light reflected at the glass-bloom boundary is then reflected at the bloom-air boundary - some escapes. The colour of this light will be tinged by the thickness of the layer that is chosen. Typically this is chosen to match green light (half way through the spectrum and the peak sensitivity of the eye). This means that the reflected light will have less green in it and more red and blue. And guess what colour that makes.