Consider the potential field V(x, y) which is 0 and -Vo (V0 > 0) respectively in the regions
of y greater and less than zero . Let θ and θ' be the angles of incidence and refraction of
the particle with the y-axis at the point of incidence as it crosses the x-axis . The ratio
sin(θ) / sin(θ ') is given (in terms of Δ = Vo / E) by
sinθ/sinθ'=v/v'[where v=velocity of particle in 1 & v'in 2]
The Attempt at a Solution
(v'2 /v2 )-1=Vο/(1/2mv2)
Which is similar to option b but in question it is sinθ/sinθ'