At some point on say, the receiving screen, light emanating from the left side of the slit will be out of phase (a difference of 1/2 wavelengths) from light coming from the center of the slit.

Thus for every point that is left of the center of the slit, there will be a point on the right side of the slit that is out of phase,

There will be no light on the screen at that particular point and thus there will be a dark fringe there.

That is the basic explanation for the appearance of dark and bright fringes on the receiving screen.

4 0

1 year ago

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. I

oksano4ka [1.4K]

Answer:

The deviation in path is $4.39 \times 10^{-3}$

Explanation:

Given:

Velocity $v = 1 \times 10^{6}$ $\frac{m}{s}$

Electric field $E = 500$ $\frac{V}{m}$

Distance $x = 1 \times 10^{-2}$ m

Mass of electron $m = 9.1 \times 10^{-31}$ kg

Charge of electron $q = 1.6 \times 10^{-19}$ C

Time taken to travel distance,

$t = \frac{x}{v}$

$t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }$

$t = 10^{-8}$ sec

Acceleration is given by,

$F = qE$

$ma = qE$

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }$

$a = 8.77 \times 10^{13}$ $\frac{m}{s^{2} }$

For finding the distance, we use kinematics equations.

$y = vt + \frac{1}{2} at^{2}$

Where $v = 0$ because here initial velocity zero

$y = \frac{1}{2} at^{2}$

$y = \frac{1}{2} \times 8.77 \times 10^{13 } \times (10^{-2} )^{2}$

$y = 4.39 \times 10^{-3}$ m

Therefore, the deviation in path is $4.39 \times 10^{-3}$

6 0

2 years ago