An electromagnetic wave of frequency 5.0 x 1014Hz, is incident on the surface of water of refractive index \(\frac{4}{3}\). Taking the speed of the wave in air as 3.0 x 108ms-1, calculate the wavelength of the wave in water.
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Correct Answer: Option E
Explanation:
\(_an_w = \frac{\lambda_a}{\lambda_w}\)
\(\frac{4}{3} = \frac{\lambda_a}{\lambda}\)
\(\lambda_w = \frac{v}{f}\)
= \(\frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 0.6 \times 10^{-6}\)
\(\lambda_w = \frac{3}{4} \lambda_a = \frac{3}{4} \times 0.6 \times 10^{-6}\)
= 4.5 x 10-7m
\(_an_w = \frac{\lambda_a}{\lambda_w}\)
\(\frac{4}{3} = \frac{\lambda_a}{\lambda}\)
\(\lambda_w = \frac{v}{f}\)
= \(\frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 0.6 \times 10^{-6}\)
\(\lambda_w = \frac{3}{4} \lambda_a = \frac{3}{4} \times 0.6 \times 10^{-6}\)
= 4.5 x 10-7m