Orange light of wavelength 6600 Å travelling in air gets refracted in water. If the speed of light in air is 3xx10^(8)ms^(-1) and refractive index of water is 4/3, find : (i) the frequency of light in air, (ii) the speed of light in water, and (iii) the wavelength of light in water.

Orange light of wavelength 6600 Å travelling in air gets refracted in

1.	Orange light of wavelength 6600 Å travelling in air gets refracted in water. If the speed of light in air is 3xx10^(8)ms^(-1) and refractive index of water is 4/3, find : (i) the frequency of light in air, (ii) the speed of light in water, and (iii) the wavelength of light in water.
Answer» Solution :Given: `lamda_(air)=6600Å=6600xx10^(-10)m` `c=3XX10^(8)ms^(-1),mu=4//3` (i) From relation `c=flamda` (in air) Frequency of LIGHT in air `f_(air)=(c)/(lamda_(air))` or `f_(air)=(3xx10^(8))/(6600xx10^(-10))` `=454xx10^(14)Hz` (ii) From relation `mu=(c)/(V)` Speed of light in water `V=(c)/(mu)=(3xx10^(8))/(4//3)` `=225xx10^(8)ms^(-1)`. (iii) Since the frequency of light remains unchanged in refraction, so `f_(water)=f_(air)=454xx10^(14)Hz` Now speed of light in water `V=f_(water)xxlamda_(water)` `THEREFORE` Wavelength of light in water `lamda_(water)=(V)/(f_(water))=(225xx10^(8))/(454xx10^(14))` `=495xx10^(-7)m` or 4950 Å Alternative: `lamda_(water)=(lamda_(air))/(mu_(water))=(6600Å)/(4//3)=4950Å`