Physics Vidyapith ✍️

What is the energy density in the electromagnetic wave in free space? The total energy stored in electromagnetic waves per unit volume due to the electric field and the magnetic field is called energy density in the electromagnetic wave in free space. $U=\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$ Derivation of Energy density in electromagnetic waves in free space: The energy per unit volume due to the electric field is $U_{E}= \frac{1}{2} \overrightarrow{E}.\overrightarrow{D} \qquad(1)$ The energy per unit volume due to the magnetic field is $U_{B}= \frac{1}{2} \overrightarrow{B}.\overrightarrow{H} \qquad(2)$ The total energy density of electromagnetic waves is $U=U_{E}+U_{B} \qquad(3)$ Now substitute the value of $U_{E}$ and $U_{B}$ in equation$(3)$ then we get $U=\frac{1}{2} \left( \overrightarrow{E}.\overrightarrow{D}+\overrightarrow{B}.\overrightarrow{H} \right)$ $U=\frac{1}{2} \left( \overrightarrow{E}.\epsilon_{0}\overrightarrow{E}+\overri

Poynting Vector: The rate of flow of energy per unit area in plane electromagnetic wave is known as Poynting vector. It is represented by $\overrightarrow{S}$. It is a vector quantity. $\overrightarrow{S}=\overrightarrow{E} \times \overrightarrow{H}$ $\overrightarrow{S}=\frac{1} {\mu_{0}} (\overrightarrow{E} \times \overrightarrow{B})$ Poynting Theorem (Work energy theorem): The most important aspect of electrodynamics is: Energy density stored with an electromagnetic wave Energy Flux associated with an electromagnetic wave To derive the energy density and energy flux. We consider the conservation of energy in small volume elements in space. The work done per unit volume by an electromagnetic wave: $W=\overrightarrow{J}.\overrightarrow{E} \qquad(1)$ This work done also consider as energy dissipation per unit volume. This energy dissipation must be connected with the net decrease in energy density and energy flow out of the volume.

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We know that the electromagnetic wave propagates perpendicular to both electric field and magnetic field which can describe as $\overrightarrow{k} \times \overrightarrow{E}= \omega \overrightarrow{B} \qquad(1)$ If $\hat{n}$ is a unit vector in the direction of the propagation then $\overrightarrow{k}=k \hat{n}$ Substitute these values in equation$(1)$ then we get $k(\hat{n} \times \overrightarrow{E})= \omega \overrightarrow{B}$ $\overrightarrow{B}= \frac{k}{\omega}(\hat{n} \times \overrightarrow{E}) \qquad(2)$ But the value of $k$ and $\omega$ is $k=\frac{2\pi}{\lambda}$ $\omega=2 \pi \nu$ Then value of $\frac{k}{\omega}=\frac{1}{c}$ Now substitute the value of $\frac{k}{\omega}$ in equation$(2)$ then we get $\overrightarrow{B}= \frac{1}{c}(\hat{n} \times \overrightarrow{E})$ The magnitude form of the above equation can be written as $B=\frac{E}{c}$ $\frac{E}{B}=c$ $\frac{E}{\mu_{0}H}=c \qquad (\because B=\mu_{0} H)$ $\frac{

When an electromagnetic wave strikes a surface then its momentum changes. the rate of change of momentum is equal to the applied force. this force acting on the unit area of the surface exerts a pressure called radiation pressure$(P_{rad})$. Let us consider a plane electromagnetic wave incident normally on a perfectly absorbing surface of area $A$ for a time $t$. If energy $U$ is absorbed during this time then momentum $P$ delivered to the surface is given according to Maxwell's prediction by $P=\frac{U}{C} \qquad(1)$ If $S$ is the energy flow per unit area per unit time i.e. Poynting vector then the energy density $U=SAt \qquad(2)$ From equation $(1)$ and equation $(2)$ $P=\frac{SAt}{c}$ $P=UAt \qquad (\because U=\frac{S}{c})$ $\frac{P}{t}=UA \qquad (3)$ If average force $(F)$ acting on the surface, is equal to the average rate of change of momentum $(P)$, is delivered to the surface then $F=\frac{P}{t} \qquad(4)$ Now from equation$(3)$ and equation$(4

We have assumed that the wave associated with a particle in motion is represented by a complex variable quantity called the wave function $\psi(x,t)$. Therefore, it can not have a direct physical meaning. Since it is a complex quantity, it may be expressed as $\psi(x,y,z,t)=a+ib \qquad(1)$ Where $a$ and $b$ are real functions of the variable $(x,y,z,t)$. The complex conjugate of wave function $\psi(x,y,z,t)$ $\psi^{*}(x,y,z,t)=a-ib \qquad(2)$ Multiply equation $(1)$ and equation $(2)$ $\psi(x,y,z,t).\psi^{*}(x,y,z,t)=a^{2}+b^{2} \qquad(3)$ $ \left| \psi(x,y,z,t) \right|^{2}=a^{2}+b^{2} \qquad(4)$ If $\psi \neq 0$ Then the product of $\psi$ and $\psi^{*}$ is real and positive. Its positive square root is denoted by $\left|\psi(x,y,z,t) \right|$, and it is called the modulus of $\psi$. The quantity $ \left| \psi(x,y,z,t) \right|^{2}$ is called the probability density $(P)$. So for the motion of a particle, the probability of finding the particle in the region $d\tau$ w

Derivation of energy flow in the electromagnetic wave in free space: The Poynting vector is given by $\overrightarrow{S}=\overrightarrow{E} \times \overrightarrow{H} \qquad(1)$ $\overrightarrow{S}=\frac{1}{\mu_{0}} ( \overrightarrow{E} \times \overrightarrow{B} ) \qquad(2) \qquad (\because \overrightarrow{B}= \mu_{0} \overrightarrow{H})$ We know that the characteristic impedance equation i.e. $\overrightarrow{B}=\frac{1}{\mu_{0}c}(\hat{n} \times \overrightarrow{E}) \qquad(3)$ Now substitute the value of $\overrightarrow{B}$ in equation$(2)$ $\overrightarrow{S}=\frac{1}{\mu_{0}c} [\overrightarrow{E} \times (\hat{n} \times \overrightarrow{E})]$ $\overrightarrow{S}=\frac{1}{\mu_{0}c} [(\overrightarrow{E}.\overrightarrow{E}) \hat{n}- (\overrightarrow{E}.\hat{n}) \overrightarrow{E})] \qquad(4)$ As $\overrightarrow{E}$ is perpendicular to $\hat{n}$ so $\overrightarrow{E} . \hat{n}=0$ then we get for above equation$(4)$ $\overrightarrow{S}=\frac{1}{\mu_{0}c} E^{2} \hat{n}

Derivation of momentum of electromagnetic wave: Maxwell's had also predicted that electromagnetic waves transport linear momentum in the direction of propagation. Let a particle which has mass $m$ moving with velocity then the momentum of a particle, $\overrightarrow{P}=m\overrightarrow{v} \qquad(1)$ According to mass-energy relation $U=mc^{2}$ Here $U$ - Total energy of the particle $m=\frac{U}{c^{2}} \qquad(2)$ From equation $(1)$ and equation $(2)$ $\overrightarrow{P}=\frac{U}{c^{2}} \overrightarrow{v} \qquad(3)$ If the electromagnetic wave is propagating along the x-axis then $\overrightarrow{v}=c \hat{i}$ Put this value in the above equation $(3)$ $\overrightarrow{P}=\frac{U}{c} \hat{i} \qquad(4)$ We know that the equation of energy flow in electromagnetic wave $\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{n}$ Here wave is propagating along x-axis i.e  $\hat{n}=\hat{i}$ $\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{i} \qquad(5)$

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ $\frac{d^{2} \psi(x)}{d x

The electromagnetic wave equations in free space: For electric field vector: $\nabla^{2} \overrightarrow{E}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$ For magnetic field vector: $\nabla^{2} \overrightarrow{B}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{B}}{\partial t^{2}} \qquad(2)$ The wave equation of electric field vector: $\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$ The wave equation of magnetic field vector: $\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$ Now the solution of electromagnetic wave for electric field vector. Differentiate with respect to $t$ of equation $(3)$ $\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$ Again differentiate with respect to $t$ of the above equation:

Time independent Schrodinger wave equation: We know the time dependent Schrodinger wave equation: $i \hbar \frac{\partial \psi(x,t)}{\partial t}= -\frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}}+ V(x) \psi(x,t) \qquad(1)$ The wave function $\psi(x,t)$ is the product of space function $\psi(x)$ and time function $\psi(t)$. So $\psi(x,t)=\psi(x) \psi(t) \qquad (2)$ Now apply the wave function form of equation$(2)$ to time dependent Schrodinger wave equation $(1)$ $i \hbar \psi(x) \frac{d \psi(t)}{d t}= -\frac{\hbar^{2}}{2m} \psi(t) \frac{d^{2} \psi(x)}{d x^{2}}+ V(x) \psi(x) \psi(t) \qquad(3)$ In the above equation $(3)$ ordinary derivatives is used in place of partial derivatives because each of function $\psi(x)$ and $\psi(t)$ depends on only one variable. Now divide the above equation $(3)$ by $\psi(x)\psi(t)$ so $i \hbar \frac{1}{\psi(t)} \frac{d \psi(t)}{d t}= -\frac{\hbar^{2}}{2m} \frac{1}{\psi(x)} \frac{d^{2} \psi(x)}{d x^{2}}+ V(x) \qqu

Time-dependent Schrodinger wave equation: Let a particle of mass $m$ is moving along the positive $x$-direction. So the total energy $E$ of the particle is: $E=\frac{1}{2}mv^{2}+V(x)$ $E=\frac{(mv)^{2}}{2m}+V(x)$ $E=\frac{(P_{x}^{2})^{2}}{2m}+V(x) \qquad(1)$ Since moving particles are associated with the wave function $\psi(x,t)$. So multiply $\psi(x,t)$ on both sides of equation$(1)$ $E\psi(x,t) =\frac{(P_{x}^{2})^{2}}{2m} \psi(x,t)+V(x) \psi(x,t) \qquad(2)$ The wave function $\psi(x,t)$ representing the plane wave associated with the particle is given by: $\psi(x,t)=A e^{\frac{i}{\hbar}(P_{x}.x-Et)} \qquad(3)$ Differentiate with respect to $x$ the above equation $(3)$ $\frac{\partial \psi(x,t)}{\partial x}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})P_{x} \qquad(4)$ Again differentiate the above equation$(4)$ $\frac{\partial^{2} \psi(x,t)}{\partial x^{2}}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})^{2}P_{x}^{2}$ $\frac{\partial^{2} \psi(x

The electromagnetic wave equations in non-conducting media : For electric field vector: $\nabla^{2} \overrightarrow{E}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$ For magnetic field vector: $\nabla^{2} \overrightarrow{B}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{B}}{\partial t^{2}} \qquad(2)$ The wave equation of electric field vector: $\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$ The wave equation of magnetic field vector: $\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$ Now the solution of electromagnetic wave for electric field vector. Differentiate with respect to $t$ of equation $(3)$ $\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$ Again diff

Difference between Stable and Unstable Resonators: The oscillating beam is converged in stable resonator while in unstable resonator is spreads out of the the resonator. In stable resonator laser output is from the centre of optical axis while in unstable resonator laser output comes from the edge of the output mirror. The field is confined to the axis in stable resonator while it is not so in unstable resonator. Stable resonators are used for low power lasers while unstable resonators are used for high power lasers. In stable resonator these remains risk of breakage of the m  irrors while it is reduced to unstable resonators. The mode volume is is small in stable resonators while it is large in unstable resonators. The geometrical losses are large in unstable resonator in comparison to stable resonators. In unstable resonators better beam quality may be achieved in comparison to stable resonators.

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations for free space. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified Form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified Form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$ For free space

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