Problem set 13

1. A rod of length $l_0$ makes an angle $\theta_0$ with the y-axis in its rest frame, while the rest frame moves to the right along the x-axis with relativistic speed $v$ with respect to lab frame. If $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$, the angle $\theta$ in the lab frame is
1. $\theta=\tan^{-1}{(\gamma\tan\theta_0)}$
2. $\theta=\tan^{-1}{(\gamma\cot\theta_0)}$
3. $\theta=\tan^{-1}{(\frac{1}{\gamma}\tan\theta_0)}$
4. $\theta=\tan^{-1}{(\frac{1}{\gamma}\cot\theta_0)}$

The x and y components of length $l_0$ are given by $l_x=l_0\sin\theta_0$, $l_y=l_0\cos\theta_0$ As the frame is moving along x-axis to the right, there is contraction of length in $l_x$ only $$l_x'=(l_0\sin\theta_0)\sqrt{1-v^2/c^2}$$ $$l_y'=l_0\cos\theta_0$$ \begin{align*} \tan\theta&=\frac{l_y'}{l_x'}\\ &=\frac{l_0\cos\theta_0}{(l_0\sin\theta_0)\sqrt{1-v^2/c^2}}\\ &=\gamma\cot\theta_0 \end{align*} $$\theta=\tan^{-1}(\gamma\cot\theta_0)$$
2. A particle of mass $m$ moves in a potential $V(x)=\frac{1}{2}m\omega^2x^2+\frac{1}{2}m\mu v^2$, where $x$ is the position coordinate, $v$ is the speed, and $\omega$ and $\mu$ are constants. The canonical (conjugate) momentum of the particle is
1. $p=m(1+\mu)v$
2. $p=mv$
3. $p=m\mu v$
4. $p=m(1-\mu)v$
\begin{align*} L&=T-V(x)\\ &=\frac{1}{2}m\dot x^2-\frac{1}{2}m\omega^2x^2-\frac{1}{2}m\mu \dot v^2\\ &=\frac{1}{2}m\dot x^2-\frac{1}{2}m\omega^2x^2-\frac{1}{2}m\mu \dot x^2 \end{align*} Conjugate momentum is given by \begin{align*} p&=\frac{\partial L}{\partial\dot x}\\ &=m\dot x-m\mu\dot x=m(1-\mu)v \end{align*}
3. A solid sphere of radius $R$ carries a uniform volume charge density $\rho$. The magnitude of electric field inside the sphere at a distance $r$ from center is
1. $\frac{r\rho}{3\epsilon_0}$
2. $\frac{R\rho}{3\epsilon_0}$
3. $\frac{R^2\rho}{r\epsilon_0}$
4. $\frac{R^3\rho}{r^2\epsilon_0}$
According to Gauss law , if the surface encloses a charge $q$ within it, then net flux through the surface is $q/\epsilon_0$. The flux is given by $$\phi_E=\int_S\vec E.d\vec A=\frac{q}{\epsilon_0}$$ If $\rho$ is the volume charge density, the charge enclosed within a distance $r$ is $q=\frac{4}{3}\pi r^3\rho$ and $$\int_S\vec E.d\vec A=E(4\pi r^2)$$ $$E(4\pi r^2)=\frac{4}{3\epsilon_0}\pi r^3\rho$$ $$E=\frac{\rho r}{3\epsilon_0}$$
4. The electric field $\vec E(\vec r,t)$ for a circularly polarized electromagnetic wave propagating along the positive $z$ direction is
   1. $E_0(\hat x+\hat y)exp[i(kz-\omega t)]$
   2. $E_0(\hat x+i\hat y)exp[i(kz-\omega t)]$
   3. $E_0(\hat x+i\hat y)exp[i(kz+\omega t)]$
   4. $E_0(\hat x+\hat y)exp[i(kz+\omega t)]$
   Plane wave travelling along the positive z-axis is given by $$\vec E(z,t)=\vec E_0 cos(kz-\omega t)$$ Plane wave travelling along the negative z-axis is given by $$\vec E(z,t)=\vec E_0 cos(kz+\omega t)$$ A beam linearly polarized along the x-axis and traveling in the positive z-direction can be represented by:$$\vec E(z,t)= E_0 cos(kz-\omega t)\hat x$$ Circularly polarized light can be represented by the expressions: \begin{align*} \vec E_{RCP}(z,t)&= E_0 \left(cos(kz-\omega t)\hat x\right.\\ &{}\left.+sin(kz-\omega t)\hat y\right) \end{align*} \begin{align*} \vec E_{LCP}(z,t)&= E_0 \left(cos(kz-\omega t)\hat x\right.\\ &\left.-sin(kz-\omega t)\hat y\right) \end{align*} Consider option (A) \begin{align*} \vec E&=E_0(\hat x+\hat y)exp[i(kz-\omega t)]\\ &=E_0(\hat x+\hat y)\left(cos(kz-\omega t)\right.\\ &{}\left.+isin(kz-\omega t)\right)\\ &=E_0\left[cos(kz-\omega t) \hat x\right.\\ &\left.{}+cos(kz-\omega t)\hat y\right]\\ &{}+iE_0\left[sin(kz-\omega t)\hat x\right.\\ &{}\left.+sin(kz-\omega t)\hat y\right] \end{align*} Hence real part of $\vec E$ is \begin{align*} Re \vec E&=E_0\left[cos(kz-\omega t) \hat x\right.\\ &{}\left.+cos(kz-\omega t)\hat y\right] \end{align*} which represents plane polarized light.
   Now consider the option (B) \begin{align*} \vec E&=E_0(\hat x+i\hat y)exp[i(kz-\omega t)]\\ &=E_0(\hat x+i\hat y)\left(cos(kz-\omega t)\right.\\ &\left. +isin(kz-\omega t)\right)\\ &=E_0\left[cos(kz-\omega t) \hat x\right.\\ &\left.-sin(kz-\omega t)\hat y\right]\\ &+iE_0\left[sin(kz-\omega t)\hat x\right.\\ &\left.+cos(kz-\omega t)\hat y\right] \end{align*} \begin{align*} Re \vec E&=E_0\left[cos(kz-\omega t) \hat x\right.\\ &\left.-sin(kz-\omega t)\hat y\right] \end{align*} Which represents a circularly polarized light, as \begin{align*} E_x^2+E_y^2&=E_0^2cos^2(kz-\omega t)\\ &+E_0^2sin^2(kz-\omega t)\\ &=E_0^2 \end{align*} is equation of circle. Hence (B) is correct option.
5. An unbiased coin is tossed $n$ times. The probability that exactly $m$ heads will come up is
1. $\frac{n}{2^m}$
2. $\frac{1}{2^n}\frac{n!}{m!(n-m)!}$
3. $\frac{1}{2^m}\frac{n!}{m!(n-m)!}$
4. $\frac{m}{2^n}$
This problem is equivalent to 1D random walk problem. In a single toss probability getting head is $p=\frac{1}{2}$ and probability of getting tail is $q=\frac{1}{2}$. In a sequences of $n$ tosses probability of getting $m$ heads and $n-m$ tails is given by $$p^mq^{(n-m)}=\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}$$ However, total combinations of $m$ times head up and $(n-m)$ times tail up are given by $$\frac{n!}{m!(n-m)!}$$ Hence, probability of getting $m$ heads is \begin{align*} P&=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}\\ &=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^n \end{align*} Hence, answer is (B).