Description of Normalization of the wave function of a particle in one dimension box or infinite potential well:
We know that the wave function for the motion of the particle along the x-axis is
$\psi_{n}(x)= A \: sin \left( \frac{n \pi x}{L} \right) \quad \left\{ Region \quad 0 \lt x \lt a \right\}$
$\psi_{n}(x)= 0 \quad \left\{ Region \quad 0 \gt x \gt a \right\}$
The total probability that the particle is somewhere in the box must be unity. Therefore,
$\int_{0}^{L} \left| \psi_{n}(x)\right|^{2}dx =1$
Now substitute the value of the wave function in the above equation. Then
$\int_{0}^{L} \left| A \: sin \left( \frac{n \pi x}{L} \right) \right|^{2}dx =1$
$\int_{0}^{L} A^{2} \: sin^{2} \left( \frac{n \pi x}{L} \right) dx =1$
$ \frac{A^{2}}{2}\int_{0}^{L} \left[ 1- cos \left( \frac{2n \pi x}{L} \right) \right] dx =1$
$ \frac{A^{2}}{2} \left[ x - \left( \frac{L}{2n\pi} \right) sin \left( \frac{2n \pi x}{L} \right) \right]_{0}^{L} =1$
$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( \frac{2n \pi L}{L} \right) \right] =1$
$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( 2n \pi \right) \right] =1$
$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( 2n \pi \right) \right] =1$
$ \frac{A^{2}}{2} \left[ L - 0 \right] =1 \qquad(\because sin2n\pi =0)$
$ \frac{A^{2} L}{2} =1$
$ A= \sqrt{\frac{2}{L}}$
Hence, the normalized wave function
$\psi_{n}(x)=\sqrt{\frac{2}{L}} sin \left( \frac{n \pi x}{L} \right)$
The absolute square $\left| \psi_{n}(x) \right|^{2}$ of the wave function $\psi_{n}(x)$ gives the probability density. Hence
$\left| \psi_{n}(x) \right|^{2} = \frac{2}{L} sin^{2} \left( \frac{n \pi x}{L} \right)$
The wave function for the particle in a box can be viewed in analogy with standing waves on a string. The wave function for a standing wave that has nodes at endpoints is of the form $\psi_{n}(x)= A \: sin \left( \frac{n \pi x}{L} \right)$. The condition for a standing wave can also be expressed in terms of wavelength.
$\lambda_{n}=\frac{2 \pi}{k_{n}}$
$\lambda_{n}=\frac{2 \pi}{\frac{n \pi}{L}} \qquad \left( \because k_{n}=\frac{n \pi}{L} \right)$
$\lambda_{n}=\frac{2 L}{n}$
$L= \frac{n \: \lambda_{n}}{2}$
So,
$L= \frac{\: \lambda_{1}}{2} \qquad \left( for \: n=1 \right)$
$L= \lambda_{2} \qquad \left( for \: n=2 \right)$
$L= \frac{3 \: \lambda_{3}}{2} \qquad \left( for \: n=3 \right)$
$L= 2 \lambda_{4} \qquad \left( for \: n=4 \right)$
Geo structure of wave function $\psi_{n}(x)$ and wave function's density $\left| \psi_{n}(x) \right|^{2}$.
Variation of the wave function and probability of finding the particle in a one-dimensional box:
We know that normalised wave function $\psi_{n}(x)$
$\psi_{n}(x)=\sqrt{\frac{2}{L}} sin \left( \frac{n \pi x}{L} \right)$
The probability density of wave function $\left| \psi_{n}(x) \right|$
$\left| \psi_{n}(x) \right|^{2} = \frac{2}{L} sin^{2} \left( \frac{n \pi x}{L} \right)$
Maximum Condition:
The values of $\psi_{n}(x)$ and $\left| \psi_{n}(x) \right|^{2}$ will be maximum. When
$sin \left( \frac{n \pi x}{L} \right)=1$
$sin \left( \frac{n \pi x}{L} \right )=sin \frac{\left( 2m+1 \right) \pi}{2}$
$ \frac{n \pi x}{L} =\left( 2m+1 \right) \frac{ \pi}{2}$
$ x =\left( 2m+1 \right) \frac{ L}{2n}$
Minima Condition:
The values of $\psi_{n}(x)$ and $\left| \psi_{n}(x) \right|^{2}$ will be minima. When
$sin \left( \frac{n \pi x}{L} \right)=0$
$sin \left( \frac{n \pi x}{L} \right)= \sin \: m\pi$
$ \frac{n \pi x}{L} = \: m\pi$
$x=m\left( \frac{L}{n} \right)$
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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