Question #90ee0

Jan 30, 2017

It is defined as deflection produced per volt of deflecting voltage and has units $m \cdot {V}^{-} 1$.

Explanation: Internal structure of a CRT is as shown in the figure above. It has five main parts listed below

1. Electron gun.
2. Deflection plate system.
3. Fluorescent screen.
4. Glass envelope.
5. Base.

Electrostatic deflection sensitivity of CRT is associated with the deflection system as in the figure below.

A beam of electrons produced, focused and accelerated by the electron gun and its components enters the two deflecting plates with velocity ${v}_{\text{o} x}$. A voltage of ${E}_{d}$ is applied across the plates so that electric field produced is along $+ y$-axis as shown. If ${V}_{a}$ is the voltage of the anode with respect to cathode, the kinetic energy gained by the electron of charge is $e {V}_{a}$ which is also equal to $\frac{1}{2} m {v}_{\text{o} x}^{2}$. Hence we hav the equation
$\frac{1}{2} m {v}_{\text{o} x}^{2} = e {V}_{a}$ ......(1)
Let $d$ be distance between the plates. Electric field between the plates is given by $\vec{E} = {E}_{d} / \mathrm{dh} a t y$.
Electrical Force on a electron of charge $e$ is ${\vec{F}}_{y} = e {E}_{d} / \mathrm{dh} a t y$
If $m$ is the mass of charge then acceleration ${a}_{y} = e {E}_{d} / \left(\mathrm{dm}\right) \hat{y}$.

Writing kinematic equation for $y$ deflection produced in time $t$ and using the equation
$s = u t + \frac{1}{2} a {t}^{2}$ and inserting above values and noting that initial defection is $0$ we get
$y = \frac{1}{2} \left(e {E}_{d} / \left(\mathrm{dm}\right)\right) {t}^{2}$ ......(2)
As the velocity along the $x$ direction is constant therefore we can write for displacement $x$
$x = {v}_{\text{o} x} t$ .....(3)
Inserting value of $t$ from (3) in (2) we get
$y = \frac{1}{2} \left(e {E}_{d} / \left(\mathrm{dm} {v}_{\text{o} x}^{2}\right)\right) {x}^{2}$ .....(4)

This is an equation of parabolic path followed by electron in the deflection plate region. Just after leaving the deflection plates the electron beam travels in a straight line path and hits the screen at $y = L$.

If ${l}_{d}$ is length of deflection plates slope of electron at the exit can be calculated by differentiating equation (4) and evaluating slope at the $x = {l}_{d}$. Which is
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(e {E}_{d} / \left(\mathrm{dm} {v}_{\text{o} x}^{2}\right)\right) {l}_{d}$
Using equation (1) this can be rewritten as
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{E}_{\mathrm{dl}} _ d}{2 {\mathrm{dV}}_{a}}$ .....(5)
For a small deflection in the deflecting plate region, $\angle \theta$ can be approximated as shown in the figure. As such
$\frac{\mathrm{dy}}{\mathrm{dx}} = \tan \theta = \frac{D}{L}$ .....(6)

where $L$ is as shown in the figure. It is straight portion of the electron beam at the exit point extended backwards which meets the $x$-axis at the mid point of length ${l}_{d}$.

Using (5) and (6) we get
$\frac{D}{L} = \frac{{E}_{\mathrm{dl}} _ d}{2 {\mathrm{dV}}_{a}}$
$\implies \frac{D}{E} _ d = \frac{L {l}_{d}}{2 {\mathrm{dV}}_{a}}$