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(A) The speed of light in medium $A$ is given by $v_A = f \lambda_A = (5 	imes 10^{14} \ Hz) 	imes (300 	imes 10^{-9} \ m) = 1.5 	imes 10^8 \ m/s$

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(A) The speed of light in medium $A$ is given by $v_A = f \lambda_A = (5 	imes 10^{14} \ Hz) 	imes (300 	imes 10^{-9} \ m) = 1.5 	imes 10^8 \ m/s$. The absolute refractive index of medium $A$ is $\mu_A = \frac{c}{v_A} = \frac{3 	imes 10^8 \ m/s}{1.5 	imes 10^8 \ m/s} = 2$. We are given the ratio of speeds $\frac{v_A}{v_B} = \frac{4}{5}$. Since the refractive index $\mu$ is inversely proportional to the speed of light $v$ $(\mu = \frac{c}{v})$,we have $\frac{\mu_B}{\mu_A} = \frac{v_A}{v_B}$. Substituting the values: $\frac{\mu_B}{2} = \frac{4}{5}$. Therefore,$\mu_B = 2 	imes \frac{4}{5} = \frac{8}{5} = 1.6$.

Light of wavelength $300 \ nm$ in medium $A$ enters into medium $B$ through a plane surface. If the frequency of light is $5 \times 10^{14} \ Hz$ and the ratio of speed in medium $A$ to that in medium $B$ is $\frac{4}{5}$,the absolute refractive index of medium $B$ is:

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