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                                    A light of wavelength `6000 A` in air, enters a medium with refractive index 1.5 Inside the medium its frequency is….Hz and its wavelength is ….`A` | 
                            
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Answer» Correct Answer - A::D Frequency remains the same `f=c/(lamda)=(3xx10^8)/(6000xx10^-10)=5xx10^14 Hz` and `lamda_2=(lamda_1)/(mu)=6000A/(1.5)=4000 A`  | 
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