Correct Answer - A::B::C::D
Given that, for sodium light,
`1 = 589nm = 589 xx 10^(-9) m`
(a) `f_a = (3xx10^8)/(589xx 10^(-9))`
`= 5.09 xx 10^(14) sec^(-1)`
(b) `mu_a = (lambda_w)/(lambda_q)`
`rArr (1)/(1.33) = (lambda_w)/(589 xx 10^(-9))`
`rArr = lambda_w = 443 nm`
(c )` f_w = f_a`
`= 5.09 xx 10^(14) sec^(-1)`
[Frequency does not change]
(d) `(mu_a)/(mu_w) = (upsilon_w)/(upsilon_a)`
`= rarr upsilon_w = (mu_a upsilon_a)/(mu_w)`
`=(3xx 10^8)/((1.33)) = 2.25 xx 10^8 m//sec.`