1.

The change in the wavelength of light when it travels from air to glass of refractive index 1.5 and the frequency of light `4 xx 10^(14) Hz` is

Answer» Given : `mu_(g) = 1.5`
`v = 4 xx 10^(14) Hz`
`c = 3 xx 10^(8) m//s`
`lambda_(g) = ?`
As we know `c = v lambda_(a)`
`:. lambda_(a) = c/v`
`= (3 xx 10^(8))/(4 xx 10^(4)) = 7500 Å`
`lambda_(g) = (c)/(v_(g)) = (vlambda_(a))/(vlambda_(g))`
`lambda_(g) = (lambda_(a))/(mu_(g))`
`= (7500)/(1.5)`
`= 5000 Å`
`:. lambda_(a) -lambda_(g) = 7500-5000`
`= 2500 Å`
Wave number in glass
`barlambda_(g) = 1/(lambda_(g))`
`barlambda_(g) = (1)/(5 xx 10^(-7))`
` :. lambda_(g) = 2 xx 10^(6)` per m


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