Correct Answer - (A)-p (B)r (C)-q (d)-s
Critical angle
`2 xx "sin"theta_(C) = 1."sin"90^(@)`
`theta_(C) = sin^(-1)(1)/(2) = (pi)/(6)`
`1."sin"90^(@) = 2 xx "sin"r`
`"sin"r = (1)/(2) implies r = (pi)/(6)`
`therefore 0 lt A -r lt theta_(C)` for all rays refrection
(A) `A = 15 ^(@) implies A-R = -ve` hence (a)
(B) ` A = 45^(@) implies 45- 30^(@) = 15^(@) lt 30^(@)` for same rays `45^(@) - 0^(@) = 45^(@) gt 30^(@)` for same rays
`(C) A = 70^(@) , A -r 70^(@) - 30^(@)= 40^(@) gt theta_(C)(30^(@))`
`70^(@) - 0^(@) = 70^(@) gt theta_(C)` All rays reflected back
(D) ` A = 50^(@) , A -r = 50^(@) - 30^(@) = 20^(@) lt theta_(C)(30^(@))"Refrection" 50^(@)-0^(@) = 50^(@) gt theta_(C)` (Reflection)