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3 years ago

Explain with schematics the operating principle of solid state lasers.

Engineering

1 answer:

alina1380 [7]3 years ago

7 0

Explanation:

A solid state laser contains a cavity like structure fitted with spherical mirrors or plane mirrors at the end filled with a rigidly bonded crystal. It uses solid as the medium. It uses glass or crystalline materials.

It is known that active medium used for this type of laser is a solid material. This lasers are pumped optically by means of a light source which is used as a source of energy for the laser. The solid materials gets excited by absorbing energy in the form of light from the light source. Here the pumping source is light energy.

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Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from the person if the exposed su

Ghella [55]

Answer:

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Explanation:

3 0

3 years ago

Sea B = 5.00 m a 60.0°. Sea C que tiene la misma magnitud que A y un ángulo de dirección mayor que el de A en 25.0°. Sea A ⦁ B =

uranmaximum [27]

Answer:

$\| \vec A \| = 6.163\,m$

Explanation:

Sean A, B y C vectores coplanares tal que:

$\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A})$ , $\vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B})$ y $\vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})$

Donde $\| \vec A \|$ , $\| \vec B \|$ y $\| \vec C \|$ son las normas o magnitudes respectivas de los vectores A, B y C, mientras que $\theta_{A}$ , $\theta_{B}$ y $\theta_{C}$ son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.

Por definición de producto escalar, se encuentra que:

$\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}$

$\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}$

Asimismo, se sabe que $\| \vec B \| = 5\,m$ , $\theta_{B} = 60^{\circ}$ , $\vec A \,\bullet \,\vec B = 30\,m^{2}$ , $\vec B\, \bullet\, \vec C = 35\,m^{2}$ , $\|\vec A \| = \| \vec C \|$ y $\theta_{C} = \theta_{A} + 25^{\circ}$ . Entonces, las ecuaciones quedan simplificadas como siguen:

$30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})$

$35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})]$

Es decir,

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})]$

Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:

$\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}$

$\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}$

Es decir,

$\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}$

$\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}$

Las nuevas expresiones son las siguientes:

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})]$

Ahora se simplifican las expresiones, se elimina la norma de $\vec A$ y se desarrolla y simplifica la ecuación resultante:

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})$

$\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}$

$30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})$

$122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}$

$35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}$

$\tan \theta_{A} = \frac{35.41}{65.6}$

$\tan \theta_{A} = 0.540$

Ahora se determina el ángulo de $\vec A$ :

$\theta_{A} = \tan^{-1} \left(0.540\right)$

La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:

$\theta_{A, 1} \approx 28.369^{\circ}$ y $\theta_{A, 2} \approx 208.369^{\circ}$

Ahora, la magnitud de $\vec A$ es:

$\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}$

$\| \vec A \| = 6.163\,m$

8 0

4 years ago

A 50 mm 45 mm 20 mm cell phone charger has a surface temperature of Ts 33 C when plugged into an electrical wall outlet but not

romanna [79]

Answer:

C = $0.0032 per day

Explanation:

We are given;

Dimension of cell phone; 50 mm × 45 mm × 20 mm

Temperature of charger; T1 = 33°C = 306K

Emissivity; ε = 0.92

convection heat transfer coefficient; h = 4.5 W/m².K

Room air temperature; T∞ = 22°C = 295K

Wall temperature; T2 = 20°C = 293 K

Cost of electricity; C = $0.18/kW.h

Chargers are usually in the form of a cuboid, and thus, surface Area is;

A = (50 × 45) + 2(50 × 20) + 2(45 × 20)

A = 6050 mm²

A = 6.05 × 10^(-3) m²

Formula for total heat transfer rate is;

E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)

Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴

Thus;

E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))

E_t = 0.7406 W = 0.7406 × 10^(-3) KW

Now, we know C = $0.18/kW.h

Thus daily cost which has 24 hours gives;

C = 0.18 × 0.7406 × 10^(-3) × 24

C = $0.0032 per day

6 0

3 years ago

Kkhghghglgghklghghghlk

RideAnS [48]

Answer:

hLDskjdbKSABCLABJC

Explanation:

BECAUSE

3 0

4 years ago

Read 2 more answers

The files provided in the code editor to the right contain syntax and/or logic errors. In each case, determine and fix the probl

Yanka [14]

Question Continuation

public class DebugOne3{

public static void main(String args){

System.out.print1n("Over the river");

System.out.pr1ntln("and through the woods");

system.out.println("to Grandmother's house we go");

}

Answer:

Line 2: Invalid Syntax for the main method. The main method takes the form

public static void main (String [] args) { }

public static void main (String args []) { }

Line 3: The syntax to print is wrong.

To print on a new line, make use of System.out.println(".."); not System.out.print1n();

Line 4:

To print on a new line, make use of System.out.println(".."); not System.out.pr1ntln();

Line 5:

The case of "system" is wrong.

The correct case is sentence case, "System.out.println" without the quotes

The correct program goes, this:

public class DebugOne3{

public static void main(String [] args){

System.out.println("Over the river");

System.out.println("and through the woods");

System.out.println("to Grandmother's house we go");

}

Explanation:

3 0

3 years ago