Respuesta :
Answer:
None of the wavelength is in the visible range
Explanation:
Constructive interference of the reflected waves for different wavelengths can be estimated using:
λ[tex]_{m}[/tex] = 2nd/m
where m is 1,2,3, ...
Therefore:
m=1, λ[tex]_{1}[/tex] = 750 nm
m=2, λ[tex]_{1}[/tex] = 750/2 = 375 nm
The limits of eye's sensitivity is between 430 nm and 690 nm. Beyond this range, the eye's sensitivity drops to approximately 1% of its maximum value.
The visible wavelengths of light do not reflected waves interfere constructively.
What is interference constructively?
To interference constructively, or constructive interference is the interference, when the maxima of two waves added together in such a way that the amplitude of this constructive wave is equal to the sum of the individual waves.
The constructive interference for two waves, in same phase can be find out as,
[tex]\lambda=\dfrac{2'nd}{m}[/tex]
Here, m=1,2,3,....
The range of wavelengths of visible light given from 380 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 400 nm to 700 nm (approx). In the problem given, the range of wavelengths of visible light is from 380 nm to 750 nm.
Put the value of the 2nd wavelength (750 nm) in the above formula for m equal to 1 as,
[tex]\lambda=\dfrac{750}{1}\\\lambda=750\rm nm[/tex]
For m equal to 2,
[tex]\lambda=\dfrac{750}{2}\\\lambda=375\rm nm[/tex]
Both the values does not fall under the range of visible light. Hence, The visible wavelengths of light do not reflected waves interfere constructively.
Learn more about the interference constructively here;
https://brainly.com/question/2166481
