Calculation of true thickness of the bed

 

11) Electric current density incident at an angle 40⁰ from vertical at the horizontal interface between two layers with resistivity $\rho_{1}$=100Ωm and $\rho_{2}$ = 500Ωm(from layer 1 to layer 2 ). The current density will enter into the second layer at an angle ______degrees from vertical.

 (Thanks to Chandrasekhar, ANU)

Solution:

Given that $\rho_{1}$=100Ωm

            $\rho_{2}$=200Ωm

Incident angle $\theta_{1}$=40⁰.

Refracting angle$\theta_{2}$=?

  According to the law of refraction

$\frac{\tan\theta_{1}}{\tan\theta_{2}}=\frac{\rho_{2}}{\rho_{1}}$

 By substituting the given values in the above formulae we get

$\frac{\tan40}{\tan\theta_{2}}=\frac{500}{100}$

 

$\frac{\tan40}{\tan\theta_{2}}=5$

 

$\tan\theta_{2}=\frac{\tan40}{5}$

 

$\tan\theta_{2}=\frac{0.839}{5}$

 

$\tan\theta_{2}=0.1678$

 

$\theta_{2}=\arctan(0.1678)$

 

$\theta_{2}=9.526$⁰