Question 705034

SCQMEDIUM

The region $y > 0$ has a constant electrostatic potential $V_1$ and $y < 0$ has a constant electrostatic potential $V_2 \neq V_1$ . A charged particle with momentum $\vec{p_1}$ is incident at an angle $\theta_1$ on the interface of the two regions. If the particle has momentum $\vec{p_2}$ in the region $y < 0$ , then the angle $\theta_2$ is given by

(A)

$\cos^{-1}(\frac{p_2}{p_1} \cos \theta_1)$

(B)

$\cos^{-1}(\frac{p_1}{p_2} \cos \theta_1)$

(C)

$\sin^{-1}(\frac{p_2}{p_1} \sin \theta_1)$

(D)

$\sin^{-1}(\frac{p_1}{p_2} \sin \theta_1)$

Detailed Solution

Since the potential is constant in each region, the force (which is the gradient of potential) is zero everywhere except at the interface (a delta-function potential). The tangential component of momentum parallel to the interface ( $x$ -axis) is conserved.

Thus, $p_1 \sin \theta_1 = p_2 \sin \theta_2$ .

Solving for $\theta_2$ , we get

$\theta_2 = \sin^{-1}(\frac{p_1 \sin \theta_1}{p_2})$ .

Free Exam

Boost Your Exam Preparation!

Move beyond just reading solutions. Access our comprehensive Test Series, original Mock Tests, and interactive learning modules. Many premium tests are completely free!

Original Mocks & Regular Test Series
Real NTA-like Interface with Analytics
Many Free Tests Available