Answer
$488nm $ and $627nm $
Work Step by Step
We know that
$\lambda_{vacuum}=\frac{2nt}{m+\frac{1}{2}}$
for $ m=0$
$\lambda_{vacuum}=\frac{2(1.33)(825\times 10^{-9}m)}{0+\frac{1}{2}}=4389nm $
for $ m=1$
$\lambda_{vacuum}=\frac{2(1.33)(825\times 10^{-9}m)}{1+\frac{1}{2}}=1463nm $
for $ m=2$
$\lambda_{vacuum}=\frac{2(1.33)(825\times 10^{-9}m)}{2+\frac{1}{2}}=877.8nm $
for $ m=3$
$\lambda_{vacuum}=\frac{2(1.33)(825\times 10^{-9})m}{3+\frac{1}{2}}=627nm $
for $ m=4$
$\lambda_{vacuum}=\frac{2(1.33)(825\times 10^{-9}m)}{4+\frac{1}{2}}=487.66nm $
for $ m=5$
$\lambda=\frac{2(1.33)(825\times 10^{-9}m)}{5+\frac{1}{2}}=399nm $
Thus, the visible wavelengths that are constructively reflected are $488nm $ and $627nm $
