Relative refractive index of two media is 0.80. In one of them, light has wavelength $6000A^o$and travels at$2.4\times {10}^8\mathrm{m}/\mathrm{s}$. This light is refracted into the second medium. Its frequency in the second medium is

Given that ${\mathrm{\mu }}_1={\mathrm{\mu }}_2=0.8$

$\mathrm{So}, {\mathrm{v}}_1={\mathrm{v}}_2 , {\mathrm{n}}_1={\mathrm{n}}_2 , {\mathrm{\lambda }}_1={\mathrm{\lambda }}_2$

$\left(\because \mathrm{\mu }=\frac{\mathrm{c}}{\mathrm{v}}=\frac{{\mathrm{\lambda }}_0}{\mathrm{\lambda }}\right)$

${\mathrm{\lambda }}_1=6000{\mathrm{A}}^{\mathrm{o}}=6000\times {10}^{-10}\mathrm{m}=6\times {10}^{-7}\mathrm{m}$

${\mathrm{v}}_1=2.4\times {10}^8\mathrm{m}/\mathrm{s}$

$\mathrm{v}=\mathrm{n\lambda }\Rightarrow {\mathrm{v}}_1={\mathrm{n}}_1{\mathrm{\lambda }}_1$

$n_1=\frac{v_1}{{\lambda }_1}=\frac{2.4\times {10}^8}{6\times {10}^{-7}}=\frac{2.4\times {10}^{15}}{6}$

${\mathrm{n}}_1=0.4\times {10}^{15}=4\times {10}^{14} \mathrm{Hz}$

So Frequency in second medium ${\mathrm{n}}_2={\mathrm{n}}_1$

${\mathrm{n}}_2=4\times {10}^{14} \mathrm{Hz}$