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Sagot :
The wavelengths of visible light are really not interfere constructively affected with by reflected waves.
What does constructive interference entail?
When the maximum of two waves are combined together in a way that makes the amplitude of the resulting constructive wave equal to the sum of the separate waves, this is known as constructive interference, or doppler effect is the interference.
Briefing :
The constructive interference for two waves, in same phase can be find out as,
λ = [tex]\frac{2'nd}{m}[/tex]
Here, m=1,2,3,....
The range of wavelengths of visible light given from 550 nm to 750 nm. It has to be found out that, for which visible wavelengths of light, do the reflected waves interfere constructively,
Visible wavelength of light is a range of electromagnetic spectrum, which can be visible by human eyes. This ranges from 550 nm to 700 nm (approx). In the problem given, the range of wavelengths of visible light is from 550 nm to 750 nm.
Put the value of the 2nd wavelength (750 nm) in the above formula for m equal to 1 as,
λ = [tex]\frac{750}{1}[/tex]
λ = 750nm
For m equal to 2,
λ =[tex]\frac{750}{2}[/tex]
λ = 375nm
Both the values does not fall under the range of visible light. Hence, The visible wavelengths of light do not reflected waves interfere constructively.
To know more about interfere constructively visit:
https://brainly.com/question/2166481
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