Answer
$493nm$ and $658nm$
Work Step by Step
We know that
$\lambda_{vaccum}=\frac{2n_f t}{m}$.....eq(1)
We plug in the values for $m$; that is, $m=1,2,34,5,6.....$
for $m=1$ eq(1) becomes
$\lambda_{vacuum}=\frac{2(1.33)(742\times 10^{-9}m)}{1}=1973.7nm$
for $m=2$
$\lambda_{vacuum}=\frac{(2)(1.33)(742\times 10^{-9}m)}{2}=986.8nm$
for $m=3$
$\lambda_{vacuum}=\frac{(2)(1.33)(742\times 10^{-9}m)}{3}=658nm$
for $m=4$
$\lambda_{vacuum}=\frac{(2)(1.33)(742\times 10^{-9}m)}{4}=493nm$
for $m=5$
$\lambda_{vacuum}=\frac{(2)(1.33)(742\times 10^{-9}m)}{5}=394.7nm$
for $m=6$
$\lambda_{vacuum}=\frac{(2)(1.33)(742\times 10^{-9}m)}{6}=329nm$
Now the visible wavelengths that are constructively reflected are $493nm$ and $658nm$.
