Answer
$401nm$ and $515nm$
Work Step by Step
We know that
$\lambda=\frac{2ntcos\theta_r}{m+\frac{1}{2}}$
for $m=0$ and plugging in other values, we obtain:
$\lambda=\frac{2(1.33)(800nm)cos(32.12^{\circ})}{(0+\frac{1}{2})}=3605nm$
for $m=1$ and plugging in other values, we obtain:
$\lambda=\frac{2(1.33)(800nm)cos(32.12^{\circ})}{(1+\frac{1}{2})}=1202nm$
for $m=2$ and plugging in other values, we obtain:
$\lambda=\frac{2(1.33)(800nm)cos(32.12^{\circ})}{(2+\frac{1}{2})}=721nm$
for $m=3$ and plugging in other values, we obtain:
$\lambda=\frac{2(1.33)(800nm)cos(32.12^{\circ})}{(3+\frac{1}{2})}=515nm$
for $m=4$ and plugging in other values, we obtain:
$\lambda=\frac{2(1.33)(800nm)cos(32.12^{\circ})}{(4+\frac{1}{2})}=401nm$
for $m=5$ and plugging in other values, we obtain:
$\lambda=\frac{2(1.33)(800nm)cos(32.12^{\circ})}{(5+\frac{1}{2})}=328nm$
The wavelengths which lie in the range of visible wavelength are $401nm$ and $515nm$; therefore, the wavelengths which are constructively reflected are $401nm$ and $515nm$.
