The refractive index of the material of a prism is `sqrt((3)/(2)` and the refractive angle is `90^(@)` . Calculate the angle of minimum deviation and the corresponding angle of incidence. Show that the limiting angle of incident for getting emergent ray is `45^(@)`.

Explanation

Here, refractive index of the material of the prism, `mu = sqrt((3)/(2), "angle of prism", A = 90^(@)`.
`"we know", mu = (sin""(A + delta_(m))/(2))/(sin""(A)/(2)) or, sqrt((3)/(2) ) = (sin""(A + delta_(m))/(2))/(sin45^(@))`
`or, " " sin""(A + delta_(m))/(2) = sqrt((3)/(2)) xx (1)/(sqrt(2)) = (sqrt(3))/(2) = sin60^(@)`
`therefore " " (A + delta_(m))/(2) = 60^(@) or, A + delta_(m) = 120^(@)`
`"For minimum deviation," i_(1) = i_(2) and r_(1) = r_(2)`
`therefore " " delta_(m) = i_(1) + i_(2) - A or, 30^(@) = 2i_(1) - 90^(@)`
`or, " " 2i_(1) = 120^(@) or, i_(1) = 60^(@)`
`therefore` For minimum deviation, angle of incidence = `60^(@)`
To obtain the emergent ray let i be the limiting angle of incidence. Then,
`sin i = sqrt(mu^(2) - 1)sinA - cos A`
`= sqrt((3)/(2) - 1) * sin90^(@) - cos90^(@) = (1)/(sqrt(2)) = sin45^(@)`
`therefore " " i = 45^(@)`