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Light of wavelength 300 nm in medium A enters into medium B througha plane surface. If frequency of light is `5xx10^(14)` Hz and `v_(a)//v_(b)=4//5`, then find absolute refractive indices of media A and B. |
Answer» Speed `v_(A)=vlamda_(A)=5xx10^(14)xx300xx10^(-9)=1.5xx10^(8)`m/s Absolute refractive index, `mu_(A)=c/(v_(A))=(3xx10^(8))/(1.5xx10^(8))=2` We have, `muprop1/v` `therefore (mu_(A)/mu_(B))=v_(B)/v_(A) rArr 2/mu_(B)=5/4` Refracticvve index, `mu_(B)=0.8xx2=1.6` |
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