Hand written solution required

Paragraph for Question Nos. 7 to 9 When a wave reaches at boundary of any medium it get partially refracted and partially reflected. Direction of refracted as well as reflected ray can be given by Snell's law. sin i sin r v. = velocity of incident wave longitudinal wave v. = velocity of reflected/refracted wave cast iron For the reflected wave of same type, v. = v as the wave returns steel to the same medium. Thus sin i = sin r => i=r. It is observed that when a longitudinal wave reaches at a surface, it generates longitudinal and transverse wave. These two waves get reflected and transmitted in direction suggested by Snell's law. Given that velocity of longitudinal wave in steel = 4000 m/s velocity of longitudinal wave in cast iron = 3000 m/s velocity of transverse wave in steel = 3200 m/s velocity of transverse wave in cast iron = 1600 m/s Longitudinal waves come at boundary of cast iron and steel from cast iron at angle of incidence i. 7. If the longitudinal wave is incident at an angle of 30%, then the angle between the refracted longitudinal and reflected transverse wave will be :- (A) sin - - sin~ 4 (B) sin -1 WIN + sin ~ 15 15 (C) * -sin-' - sin -1 4 (D) - sin 15 6