Derivation of Fringe width of the wedge-shaped thin film:
The distance between two consecutive bright (or dark) fringes is called the fringe width.
If the $n^{th}$ bright fringe is formed at a distance $x_{n}$ from the edge of the wedge shaped film where the thickness is $t_{n}$. So the path difference for $x_{n}$ bright fringe:
Derivation of interference of light due to a wedge-shaped thin film:
Interference of light due to wedge-shaped thin film
The wedge-shaped film is bound by two plane surfaces inclined at angle $\alpha$.
The thickness t of the film varies uniformly from zero at the edge to its maximum value at the other end.
A Ray of light ab incident on the film will be partially reflected along be and partially transmitted along $BC$. The ray $BC$ will be partially reflected along $CD$ which will be again partially transmitted along $BF$. The two rays $BE$ and $DF$ in reflected light diverge.
The ray $DF$ suffers a phase change of $\pi$ due to reflection from a denser medium at $C$. Whereas there is not any change in phase due to reflection for ray $BE$.
$\Delta= 2\mu t \: cos \: (\alpha + r) \pm \frac{\lambda}{2}$
For Constructive Interference:
$\Delta= n \lambda$
$\Delta= 2\mu t \: cos \: (\alpha + r) \pm \frac{\lambda}{2}= n \lambda$
$ 2\mu t \: cos \: (\alpha + r)= (2n \pm 1) \frac{\lambda}{2}$
For Destructive Interference:
$\Delta= (2n \pm 1) \frac{\lambda}{2}$
$\Delta= 2\mu t \: cos \: (\alpha + r) \pm \frac{\lambda}{2}=(2n \pm 1) \frac{\lambda}{2}$
$\Delta= 2\mu t \: cos \: (\alpha + r) =2n \lambda$
Derivation of interference of light due to thin-film:
Let's consider a Ray of light $AB$ incident on a thin film of thickness $t$ and the refractive index of a thin film is $\mu$
The ray $AB$ is partially reflected and partially transmitted at $B$. The transmitted BC is against partially transmitted and partially reflected at $C$. The reflected ray $CD$ is partially reflected and partially refracted at $D$.
Propagation of light ray in thin film
The interference pattern in reflected light will be due to ray $BF$ and $DH$ which are coherent as they are both derived from the same Ray $AB$.
The interference pattern in transmitted light will be due to ray $CI$ and $EJ$.
The path difference between $BF$ and $DH$ ray will be
$\Delta=\mu(BC+CD)-BG \qquad(1)$
The triangle $\Delta BCK$ and $\Delta CDK$ are congruent because
The ray $BF$ suffers phase change of $\pi$ due to reflection from the denser medium at $B$. Therefore the path difference of $\frac{\lambda}{2}$ is introduced between two rays due to reflection-
$\Delta= 2\mu t \: cos\:r \pm \frac{\lambda}{2}$
Constructive Interference due reflected ray:
for constructive interference
$\Delta=n \lambda$
so from the above equations, we get
$2\mu t \: cos\: r \pm \frac{\lambda}{2}=n \lambda$
$2\mu t \: cos\: r = (2n \pm 1)\frac{\lambda}{2}$
Destructive Interference due to reflected ray:
for destructive interference
$\Delta=(2n \pm 1)\frac{\lambda}{2}$
so from the above equations, we get
$2\mu t \: cos\: r \pm \frac{\lambda}{2}=(2n \pm 1)\frac{\lambda}{2}$
$2\mu t \: cos\: r =n \lambda$
Interference in transmitted ray:
The ray $CI$ and $EJ$ in transmitted Ray have the same path difference as a reflected ray. There is not any change in phase for $CI$ due to reflection as it gets transmitted at $C$. The ray $EJ$ also does not undergo phase change due to reflection as it is reflected from the rarer medium at $C$ and $D$.
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
Similarly, the energy density of electromagnetic waves in free space in terms of the magnetic field $B$ can be written as:
$U= \frac{B^{2}}{\mu_{0}} $
The average value of energy density in the electromagnetic waves in free space:
Now we will find the average value of energy density in the electromagnetic wave in free space from the above equation $U= \epsilon_{0} E^{2} $. So we get
$\left< U \right> = \epsilon_{0} \left< E^{2} \right>$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\frac{E_{rms}^{2}}{Z_{0}} .\hat{n}}{\epsilon_{0} E_{rms}^{2}}$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\epsilon_{0} Z_{0}}$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\sqrt{\epsilon_{0} \mu_{0}}} \qquad(\because z_{0}= \sqrt{\frac{\mu_{0}}{\epsilon_{0}}})$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\hat{n} c \qquad(\because c= \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$
$ \left< \overrightarrow{S} \right>=\hat{n} c \left< U \right> $
The energy flow per unit area per unit time in an electromagnetic wave is the product of energy density, speed of light, and the direction of propagation.The ratio of the energy densities of the electric field and magnetic field:
So from above equation $U_{E}=\epsilon_{0} E^{2}$ and equation $U_{B}=\frac{B^{2}}{\mu_{0}}$, we can find the ratio between them i.e.
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.
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:
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. According to Modified Maxwell's Forth equation:
This equation represents the conservation of energy principle. It is also known as Poynting theorem. Here Negative signs of work done to represent that electromagnetic flow with energy flux as continuity energy density. So this equation is also known as the continuity equation.
Where
$ \overrightarrow{\nabla}.\overrightarrow{S}$ $\rightarrow$ The flow of energy
$U$ $\rightarrow$ Energy density of electromagnetic filed
The term $\frac{E}{H}$ has dimensions of the impedance and is known as characteristic impedance or intrinsic impedance of free space. It is represented by $(Z_{0})$.
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 get
$F=UA \qquad(5)$
The radiation pressure $(P_{rad})$ exerted on the surface is
$P_{rad}=\frac{F}{A} \qquad(6)$
Now substitute the value of $F$ from equation$(5)$ in equation$(6)$ then we get
$P_{rad}=\frac{UA}{A}$
$P_{rad}=U$
Hence, the radiation pressure exerted by a normally incident play electromagnetic wave on a perfect absorber is equal to the energy density of the wave.
For a perfect reflector or for a perfect reflecting surface, the radiation after reflection has momentum equal in magnitude but opposite in direction to the incident radiation. Then the momentum imparted to the surface will therefore be twice as on perfect absorber i.e.
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 $\psi(x,y,z,t)$
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 $d\tau$ will be:
Here $P$ are the probability that tells us that the particle will be found in a volume element $d\tau(=dx.dy.dz)$ surrounding the point at position $(x,y,z)$ at time $t$.
For the motion of a particle in one dimension, the probability of finding the particle in the region $dx$ will be:
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,
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
$\begin{Bmatrix}
V(x)=0 &for \: 0\leq x \leq L \\
V(x)=\infty & for \: 0> x > L \\
\end{Bmatrix}$
The general solution of the above differential equation $(2)$
$\psi(x)= A sin(kx)+ B cos(kx) \qquad(4)$
The wave function $\psi(x)$ should be zero everywhere outside the box since the probability of finding the particle outside the box is zero. Similarly, the wave function $\psi(x)$ must also be zero at walls of the box because the probability density $[\psi(x)]^{2}$ must be continuous. Thus, the boundary conditions for this problem is that
(i) $\psi(x)=0$ For $x=0$
(ii) $\psi(x)=0$ For $x=L$
Now applying the boundary condition in equation$(4)$ i.e.
(i) At $x=0$ the wave function $\psi(0)=0$
Now we get
$\psi(0)= A sin(k.0)+ B cos(k.0)$
$A sin(k.0)+ B cos(k.0)=0 \qquad (\because \psi(0)= 0)$
$B=0$
Hence substitute the value of $B$ in equation$(4)$ ,
$\psi(L)= A sin(kx) \qquad(5)$
Now applying the second boundary condition:
(ii) At $x=L$ the wave function $\psi(L)=0$, we get
$\psi(x)= A sin(kL) \qquad(6)$
This equation will satisfy only for certain values of $k$, say $k_{n}$. Since $A$ can not be taken zero hence
$sin(k_{n}L)=0 $
$sin(k_{n}L)=sin(n\pi) $
$k_{n}L=n\pi $
$k_{n}=\frac{n\pi}{L} \qquad(7)$
Thus for each allowed values of $k_{n}$ there is a wave function $\psi(x)$ given as, using equation$(5)$ and equation$(7)$
$\psi_{n}(x)=A sin(\frac{n\pi x}{L})$
This is the expression of the wave function or eigen function for a particle in a box.
Now, from equation $(3)$ and equation$(7)$, we get
This is the expression of energy or eigen value for a particle in a box.
In general, the expression for this energy is written as:
$E_{n}=\frac{n^{2} h^{2}}{8mL^{2}}$
For different values of $n$ energy values can be written as
For $n=1$
$E_{1}=\frac{h^{2}}{8mL^{2}}$
It is known as zero-point energy or ground energy state
For $n=2$
$E_{2}=\frac{2^{2} h^{2}}{8mL^{2}}=2^{2}E_{1}$
For $n=3$
$E_{3}=\frac{3^{2} h^{2}}{8mL^{2}}=3^{2}E_{1}$
For $n=4$
$E_{4}=\frac{4^{2} h^{2}}{8mL^{2}}=4^{2}E_{1}$
So generalized form of the above equation can be written as
$E_{n}=n^{2}E_{1}$
Some of the possible energies for a particle in a box are shown on an energy-level diagram in the figure below.
Possible Energies for a particle in a box
The energy levels have a spacing that increases with increasing $n$ and thus the particle in a box can take only certain discrete energy values, called Eigen-values. This means that the energy levels of a particle in a box are quantized but according to classical mechanics, the particle may take any continuous range of energy values between zero and infinity.
