The angle of minimum deviation for prism of angle `pi//3 is pi//6`. Calculate the velocity of light in the material of the prism if the velocity of light in vacuum is `3 xx 10^8 ms^-1`.

Explanation

`A= (pi)/( 3 ) = 60^(@) , delta _(m) = (pi)/(6) = 30^(@) xx 10^(8) ms^(-1) , V = ?`
` :' mu = (sin .(A+ delta )/( 2))/(sin .(A)/(2 )) = ( c )/(v )`
or `v= c (sin .(A)/(2 ))/(sin .(A+delta )/(2 )) = (3 xx 10^(8) xx sin .(60^(@))/(2)))/(sin .(60^(@)+ 30^(@))/(2 ))`
` = (3xx 10^(8) sin 30^(@))/( sin 45^(@))`
`= (3xx 10^(8) xx 0.5000)/( 0.7071)`
`= 2.12 xx 18^(8) ms^(-1)`