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## Homework Statement

Consider the potential field V(x, y) which is 0 and -Vo (V

_{0}> 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

region1

(a)√(1+2Vο/E)

(b)√(1+Vο/E)

(c)1+Vο/E

(d)1+2Vο/E

## Homework Equations

nsinθ=n'sinθ'

sinθ/sinθ'=v/v'[where v=velocity of particle in 1 & v'in 2]

## The Attempt at a Solution

1/2mv

^{2}=E=1/2mv

^{'2}-Vο

1/2mv

^{'2}-1/2mv

^{2}=Vο

(v

^{'2}-v

^{2})/v

^{2}=2Vο/mv

^{2}

(v

^{'2}/v

^{2})-1=Vο/(1/2mv

^{2})

∴sinθ'/sinθ=√(1+Vο/E)

Which is similar to option b but in question it is sinθ/sinθ'

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