Answer:
The wavelength is [tex]\lambda_ 1 = 574.2 nm[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 198 nm = 198 *10^{-9 }\ m[/tex]
The refractive index of the non-reflective coating is [tex]n_m = 1.45[/tex]
The refractive index of glass is [tex]n_g = 1.50[/tex]
Generally the condition for destructive interference is mathematically represented as
[tex]2 * n_m * t * cos (\theta) = n * \lambda[/tex]
Where [tex]\thata[/tex] [tex]\theta[/tex] is the angle of refraction which is 0° when the light is strongly transmitted
and n is the order maximum interference
so
[tex]\lambda = \frac{2 * n * t * cos (\theta )}{n}[/tex]
at the point n = 1
[tex]\lambda _1 = \frac{2 * 1.45 * 198*10^{-9} * cos (0 )}{1}[/tex]
[tex]\lambda_1 = 574.2 *10^{-9}[/tex]
[tex]\lambda_1 = 574.2 nm[/tex]
at n =2
[tex]\lambda _2 = \frac{\lambda _1 }{2}[/tex]
[tex]\lambda _2 = \frac{574.2*10^{-9} }{2}[/tex]
[tex]\lambda _2 = 2.87 1 *10^{-9} \ m[/tex]
[tex]\lambda _2 = 287. 1 nm[/tex]
Now we know that the wavelength range of visible light is between
[tex]390 \ nm \to 700 \ nm[/tex]
So the wavelength of visible light that is been transmitted is
[tex]\lambda_ 1 = 574.2 nm[/tex]
