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

1. Apr 5, 2017

### pallab

1. The problem statement, all variables and given/known data
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
region1
region2

(a)√(1+2Vο/E)
(b)√(1+Vο/E)
(c)1+Vο/E
(d)1+2Vο/E

2. Relevant equations
nsinθ=n'sinθ'
sinθ/sinθ'=v/v'[where v=velocity of particle in 1 & v'in 2]

3. The attempt at a solution
1/2mv2=E=1/2mv'2-Vο
1/2mv'2-1/2mv2=Vο
(v'2 -v2)/v2=2Vο/mv2
(v'2 /v2 )-1=Vο/(1/2mv2)
∴sinθ'/sinθ=√(1+Vο/E)

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

Last edited: Apr 5, 2017
2. Apr 5, 2017

### TSny

Are you sure this equation is correct? If Vo is positive, which would be larger: v or v'? θ or θ'?

3. Apr 6, 2017

### pallab

I have just used snell's law.
θ∝v
but I am not sure about increase or decrease of kinetic energy in the presence of potential +ve or -ve.

4. Apr 6, 2017

### pallab

I have just used snell's law.
θ∝v
but I am not sure about increase or decrease of kinetic energy in the presence of potential +ve or -ve.

5. Apr 6, 2017

### TSny

This problem deals with a particle moving from a region of zero potential energy to another region of constant potential energy -Vo.

For this situation, Snell's law as written for light $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$ does not apply. Have you covered how Snell's law is modified for the particle situation?

Last edited: Apr 6, 2017
6. Apr 6, 2017

No.
No.

7. Apr 6, 2017

### TSny

Then you will need to derive "Snell's law" for the particle.

As a start, consider the x and y components of the velocity of the particle.
Does vx change when the particle passes through the origin?
Does vy change?