Determine the change in wavelength of light during its passage from air to glass. Refractive Index of Glass with respect to Air is 1.5 and the frequency of light is ` 5 xx 10 ^( 14 ) ` Hz. Also find the wave number of light in glass (velocity of light in air ` c = 3xx 10 ^( 8 ) m//s ` ).

Explanation

Given ` : mu _ g = 1.5, c = 3 xx 10 ^( 8 ) m//s, n = 5 xx 10 ^( 14 ) ` Hz
we know that, ` c = n lamda _ a `
` therefore lamda _ a = ( c ) / ( n ) = ( 3xx 10 ^( 8)) /( 5 xx 10 ^( 14)) = 6000 ` Å
` mu _ g = ( c ) /( v _ g ) = ( n lamda _ a ) /( n lamda _ g ) `
` therefore lamda _ g = ( lamda _ a ) /( mu _ g ) = ( 6000 ) /( 1.5 ) = 4000 ` Å
` therefore lamda _ a - lamda _ g = 6000 - 4000 = 2000 Å `
Wave number in glass,
` bar ( lamda _ g ) = ( 1 ) / ( lamda _ g ) `
` therefore bar (lamda _ g ) = ( 1 ) /( 4 xx 10^( - 7 ) ) = 2.5 xx 10 ^( 6 ) ` per m.