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SCORPION-xisa [38]
3 years ago
13

8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?

Physics
1 answer:
Sidana [21]3 years ago
8 0

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

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Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p
Bad White [126]

Answer:

Diode Lasers  

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250  

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is  

4. The optical gain bandwidth (as measured between half intensity points) will  

normally depend on the pumping current (diode current) but for this problem  

assume that it is 2 nm.  

(a) What is the mode integer m of the peak radiation?  

(b) What is the separation between the modes of the cavity? Please express your  

answer as Δλ.  

(c) How many modes are within the gain band of the laser?  

(d) What is the reflection coefficient and reflectance at the ends of the optical  

cavity (faces of the InGaAsP crystal)?  

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction  

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the  

beam is a Gaussian beam with the waist located at the output. And the beam  

waist size is approximately the x-y dimensions of the cavity.)  

Solution:  

(a) The wavelength λ of a cavity mode and length L are related by  

n

mL

2

λ = , where m is the mode number, and n is the refractive index.  

So the mode integer of the peak radiation is  

1290

1055.1

10250422

6

6

= ×

××× == −

−

λ

nL

m .  

(b) The mode spacing is given by nL

c f 2

=Δ . As

λ

c f = , λ

λ

Δ−=Δ 2

c f .  

Therefore, we have nm

nL f

c

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .  

(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the  

bandwidth could fit in two possible modes.  

For mode integer of 1290, nm

m

nL 39.1550

1290

10250422 6

= ××× ==

−

λ

Take m = 1291, nm

m

nL 18.1549

1291

10250422 6

= ××× ==

−

λ

Or take m = 1289, nm

m

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

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