JEE Main · 2021 · 24 Feb 2021 · Shift 2 · EM Waves

In vacuum the wavelength is ${\lambda }_0=\frac{c}{f}=\frac{3\times 10^8}{3\times 10^9}=0.1$ m $=10$ cm. On entering the dielectric the frequency is unchanged but the wave speed drops to $v=\frac{c}{\sqrt{{\epsilon }_r}}=\frac{c}{\sqrt{2.25}}=\frac{c}{1.5}$, so the wavelength shortens by the same factor, $\lambda =\frac{{\lambda }_0}{\sqrt{{\epsilon }_r}}=\frac{10}{1.5}=6.67$ cm. Expressed in the requested unit of $10^{-2}$ cm this is $667\times 10^{-2}$ cm.