Einstein’s Mass-Energy Relation:
Einstein's mass energy relation gives the relation between mass and energy. It is also knows as mass-energy equivalence principle.
According to Newtonian mechanics, Newton’s second law
$f=\frac{dP}{dt}$
Where $P$ is the momentum of the particle. So put $P=mv$ in above equation:
$f=\frac{d}{dt}\left ( mv \right )\quad\quad (1)$
According to theory of relativity, mass of the particle varies with velocity so above equation $(1)$ can be written as:
$f=m \frac{dv}{dt}+v\frac{dm}{dt}\quad\quad (2)$
When the particle is displaced through a distance $dx$ by the applied force $F$. Then the increase in kinetic energy $dk$ of the particle is given by
$dk= Fdx\quad\quad (3)$
Now substituting the value of force $F$ in equation $(3)$
$dk =m\frac{dv}{dt}\cdot dx+v\frac{dm}{dt}\cdot dx \quad (4) $
$dk=mv\cdot dv +v^{2}\cdot dm \:\: (5) \: \left \{ \because \frac{dx}{dt}=v \right \}$
The variation of mass with velocity equation
$m=\frac{m_{\circ}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\quad\quad (6) $
Square both sides in above equation:
$m^{2}=\frac{m_\circ^{2}}{1-\frac{v^{2}}{c^{2}}}$
$m^{2}c^{2}-m^{2}v^{2}=m_\circ ^{2}c^{2}$
Differentiate the above equation which can be written as
$2mdm\cdot c^{2}- 2m \cdot dm \cdot v^{2}-2v\cdot dv\cdot m^{2}=0$
$c^{2}dm-v^{2}dm-vm\cdot dv$
$c^{2}dm=v^{2}dm+mv\cdot dv \quad\quad (7)$
Now substitute the value of $dk$ from equation $(5)$ in equation $(7)$. So above equation can be written as:
$dk = c^{2}dm$
Now consider that the particle is at rest initially and by the application of force it acquires a velocity $v$. The mass of body increase from ${m_{\circ}}$ to $m$. The total kinetic energy acquired by the particle is given by
$dk = \int_{m_\circ}^{m}c^{2}\cdot dm$
$k = c^{2}\left ( m-m_{\circ} \right )$
$k = mc^{2} - m_\circ c^{2}$
$k+m_\circ c^{2} = mc^{2}$
Where $k$ is the kinetic energy of the particle and $m_\circ c^{2}$ is the rest mass-energy of the particle. So The sum of these energies is equal to the total energy of the particle $E$. So
$E= m c^{2}$
Where $E$ is the total energy of the particle.
The above equation is called the mass energy equivalence equation.
Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig
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