The angle of minimum deviation measured with a prism is `30^(@)` and the angle of prism is `60^(@)`. The refractive index of prism material is

Explanation

Relation between refractive index and angle of minimum deviation is as follows :
`mu=(sin((A+delta_(m))/(2)))/(sin""(A)/(2))`
Here A is `60^(@)`.
`rArr" "mu=(sin((60^(@)+delta_(m))/(2)))/(sin30^(@))`
`rArr" "mu=2sin((60^(@)+delta_(m))/(2))" ....(i)"`
In this question it is given that one of the rays is found to deviate through `30^(@)`. Hence, we can say for sure that `delta_(m)lt30^(@)`. Applying this condition in equation (i) we can write the following :
`rArr" "mult2sin((60^(@)+30^(@))/(2))`
`rArr" "mult2 sin 45^(@)`
`rArr" "mult2xx(1)/(sqrt2)`
`rArr" "multsqrt2`