Answer:
So visible wavelength which is possible here is
416 nm and 693.3 nm
Step-by-step explanation:
As we know that for normal incidence of light the path difference of the reflected ray is given as
![2\mu t + (\lambda)/(2) = \Delta x](https://img.qammunity.org/2020/formulas/physics/high-school/bx1v16vighpqzrshdhqwcii9wu7n4f4iwu.png)
so here we can say that for maximum intensity condition we will have
![\Delta x = N\lambda](https://img.qammunity.org/2020/formulas/physics/college/wshhq4290bi4u2sc03570zfvjd9ka44rph.png)
so we have
![2\mu t + (\lambda)/(2) = N\lambda](https://img.qammunity.org/2020/formulas/physics/high-school/fxmr3szuukk4t0g82wain153f3wn4fiyrb.png)
now for visible wavelength we have
for N = 1
![2\mu t = (\lambda)/(2)](https://img.qammunity.org/2020/formulas/physics/college/lnqi52ulp78zuntuj5gmuuk00bimup6gt4.png)
![\lambda = 4\mu t](https://img.qammunity.org/2020/formulas/physics/high-school/a8ccuew60fbsd7p3bqwgmbo46hzeh6x7pr.png)
![\lambda = 4((4)/(3))(390 nm)](https://img.qammunity.org/2020/formulas/physics/high-school/i5fcy8gkmoqrf0zngtujgzrnoz0621e82p.png)
![\lambda = 2080 nm](https://img.qammunity.org/2020/formulas/physics/high-school/hq6k58kpj3jtj1nek6lrttw81asn2xw3qp.png)
for N = 2
![\lambda = (4\mu t)/(3)](https://img.qammunity.org/2020/formulas/physics/high-school/3yzgovuy129jz8ljs3f5hvl0wzziooicxg.png)
![\lambda = (4((4)/(3))(390 nm))/(3)](https://img.qammunity.org/2020/formulas/physics/high-school/44o9vmtx74opaw4u4gir1a1pjlplrjacyj.png)
![\lambda = 693.3 nm](https://img.qammunity.org/2020/formulas/physics/high-school/hwuotc0qpxle6anbbq0vr9s31by5788uax.png)
for N = 3
![\lambda = (4\mu t)/(5)](https://img.qammunity.org/2020/formulas/physics/high-school/24uuakne44wu0rdmhzosmvcv6v3wypffz0.png)
![\lambda = (4((4)/(3))(390 nm))/(5)](https://img.qammunity.org/2020/formulas/physics/high-school/oruvv8tuu8emdauybdu45t1dyikffj1y58.png)
![\lambda = 416 nm](https://img.qammunity.org/2020/formulas/physics/high-school/a5euihweu09nmg5o12i8r68es13dsct2wv.png)