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Alina [70]
2 years ago
15

The paths of the light waves that interfere to cause first-order lines (2 points) Group of answer choices differ in length by th

e wavelength of the light are parallel lines are the same length differ in length by one-half of the wavelength of the light
Physics
1 answer:
Zarrin [17]2 years ago
6 0

The paths of the light waves that interfere cause first-order lines to differ in length by the wavelength of the light.

The phenomenon of wave interference occurs when two waves meet while traveling in the same medium.

As the two light waves interfere in the first order they interfere by differing the consecutive lengths by the wavelength of the light. The wavelength of the light can be defined as the distance between identical points (adjacent crests) in the adjacent cycles of a wave signal propagated in space or along a wire.

Hence, it can be concluded that the paths of the light waves that interfere cause first-order lines to differ in length by the wavelength of the light.

Learn more about waves here:

brainly.com/question/15663649

#SPJ10

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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

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