Given : $\mathit{\lambda }=600\mathit{n}\mathit{m}$
$\mathit{\mu }=1.33$
(i) In reflection, the ray will reflect back in the same medium as that of incident ray.
Hence, wavelength $\left(\mathit{\lambda }\right)=600\mathit{n}\mathit{m}$
frequency $=\mathit{v}=\frac{\mathit{c}}{\mathit{\lambda }}=\frac{3\times {10}^8}{600\times {10}^{-9}}$
$=0.5\times {10}^{15}\mathit{H}\mathit{z}$
speed $=3\times {10}^8\mathit{m}{\mathit{s}}^{-1}$
(ii) In refraction, the speed and wavelength change while frequency remains same.
Hence, speed $=\mathit{v}=\frac{\mathit{c}}{\mathit{\mu }}=\frac{3\times {10}^8}{1.33}$
$=2.26\times {10}^8\mathit{m}{\mathit{s}}^{-1}$
wavelength $=\mathit{\lambda }=\frac{\mathit{\nu }}{\mathit{v}}=\frac{2.26\times {10}^8}{0.5\times {10}^{15}}$
$=4.52\times {10}^{-7}\mathit{m}$
$=452\mathit{n}\mathit{m}$
frequency $=\mathit{v}=0.5\times {10}^{15}\mathit{H}\mathrm{z}$