Answer:
So visible wavelength which is possible here is
416 nm and 693.3 nm
Explanation:
As we know that for normal incidence of light the path difference of the reflected ray is given as
[tex]2\mu t + \frac{\lambda}{2} = \Delta x[/tex]
so here we can say that for maximum intensity condition we will have
[tex]\Delta x = N\lambda[/tex]
so we have
[tex]2\mu t + \frac{\lambda}{2} = N\lambda[/tex]
now for visible wavelength we have
for N = 1
[tex]2\mu t = \frac{\lambda}{2}[/tex]
[tex]\lambda = 4\mu t[/tex]
[tex]\lambda = 4(\frac{4}{3})(390 nm)[/tex]
[tex]\lambda = 2080 nm[/tex]
for N = 2
[tex]\lambda = \frac{4\mu t}{3}[/tex]
[tex]\lambda = \frac{4(\frac{4}{3})(390 nm)}{3}[/tex]
[tex]\lambda = 693.3 nm[/tex]
for N = 3
[tex]\lambda = \frac{4\mu t}{5}[/tex]
[tex]\lambda = \frac{4(\frac{4}{3})(390 nm)}{5}[/tex]
[tex]\lambda = 416 nm[/tex]
