# Dielectric constant of a rod

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1. Dec 28, 2017

### Pushoam

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

2. Relevant equations

3. The attempt at a solution
$n_1 = \sqrt{ \epsilon _1}$ .....(1) ,

$n_2 = 1$ .....(2)

$\frac { \sin {\theta_i}}{ \sin {\theta_l} } = \frac { n_1}{n_2} = n_1$ .....(3)

$\cos{\theta_1} = \frac { n_2}{n_1} = \frac1{ n_1}$ .....(4)

According to the question, the dielectric constant should be such that even when the incident angle is slightly less than $\frac { \pi} 2$ , the ray should come out of the rod without getting absorbed.

So, $\theta _i \leq \frac { \pi} 2$ .....(5)

Taking $\theta _i = \frac { \pi} 2$ , .....(6) I don't know why I am taking this.

$\frac1{ \sin {\theta_l} }= n_1$ .....(7)

$\cos{\theta_1} = \frac1{ n_1}$ .....(8)

From (7) and (8), I get

$n_1 = \sqrt{ 2}$ .....(9)

From (1), $\epsilon _1 = n_1^2$ .....(10)

So, that dielectric constant = 2, option (c).

Is this correct?

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2. Dec 28, 2017

Snell's law is written $n_1 \sin(\theta_1)=n_2 \sin(\theta_2)$. For this case $n_1=\sqrt{\epsilon_{r1}}$, and $n_2 =1$. $\\$ $\theta_1$ is the angle of incidence inside the material, measured from the normal. There is a $\theta_1$ for which the left side of the equation is equal to 1. What happens if $\theta_1$ is such that the left side of the equation is greater than 1 ? Can you get a solution for the right side in that case, to determine the emerging angle of the refracted ray? $\\$ Editing: This problem is slightly tricky: If $\theta_i=90^o$,(the steepest angle of incidence at the entry point), $\theta_r=\sin^{-1}(1/n_1)$. The resulting $\theta_1=90^o-\theta_r$. I'll let you try to finish it up. Meanwhile, the file you uploaded is apparently the wrong one.

Last edited: Dec 28, 2017
3. Dec 28, 2017

### haruspex

As far as I can make out, this matches what @Pushoam has done. Have you identified an error in the working?

4. Dec 28, 2017

The OP's statement after equation (6) was somewhat confusing. Meanwhile, I was expecting to see the statement $n_1 \sin(\theta_1)>1$ for total internal reflection. In addition, $\sin(\theta_1)=\cos(\theta_r)=\frac{\sqrt{n_1^2-1}}{n_1}$. Thereby $\sqrt{n_1^2-1}>1$, so that $n_1^2=\epsilon_{r1}>2$. $\\$ I didn't see these details in the OP's solution. $\\$ (It was difficult to answer the question by @haruspex without providing the solution.) $\\$ From what I can see, the OP gets the right answer, but the algebraic steps to the answer are not readily apparent. Perhaps the OP performed a similar algebra, but too many steps were omitted to readily tell how the OP arrived at the answer. :) $\\$ In any case, I think the OP @Pushoam might find these details useful.