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

  • Thread starter pallab
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  • #1
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Homework Statement


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
upload_2017-4-5_19-53-33.png
region2



(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


1/2mv2=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'
 
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Answers and Replies

  • #2
TSny
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sinθ/sinθ'=v/v' [where v=velocity of particle in 1 & v'in 2]
Are you sure this equation is correct? If Vo is positive, which would be larger: v or v'? θ or θ'?
 
  • #3
27
2
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
27
2
Are you sure this equation is correct? If Vo is positive, which would be larger: v or v'? θ or θ'?
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
TSny
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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?
 
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  • #6
27
2
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?
No.
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?
No.
 
  • #7
TSny
Homework Helper
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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?
 

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