A ray of light is incident on a prism `ABC` of `mu = sqrt(3)` as shown in Fig. Find the angle of incidence for which the deviation of light by the prism `ABC` is minimum.
By what angle should the second prism be rotated so that final ray suffers net minimum deviation ?

Answer

(a)`60^(@) (b) 60^(@)`

Explanation

(a) At minimum deviation , `r_(1) = r_(2) = 30^(@)`
`therefore ` From Snell's law `mu = ("sin"i_(r))/("sin"r_(1)) "or" sqrt3 = ("sin"i_(1))/("sin"30^(@))`
`therefore "sin"i_(1) = sqrt(3)/(2) "or" i_(1) = 60^(@)`
(b) In the position shown net deviation suffered by the ray of light should be minimum.
Therefore , the `II^(nd)` prism should be rotated by `60^(@)` (anticlock wise)