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2 years ago

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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

lot the peak closest to 632.8 nm for 4 values of R = 0.99, 0.90, 0.75 and 0.6. For each case find the spectral width δλm, the finesse F and Q using the equations given in the book or in class. How accurate are our equations in predicting δλm? (You may want to use a graphing software for this problem.)

1 answer:

Bad White [126]2 years ago

8 0

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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Answer:

Δ P = 13.24 Pa

Explanation:

Given that

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The change in the height of column ,Δh = 1.5 mm

The pressure given as

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Change in the pressure

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Now by putting the values

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Δ P = 13.24 Pa

Therefor the change in the pressure will be 13.24 Pa.

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2 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

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Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

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Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad

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Answer:

468449163762.0812 W

Explanation:

m = Mass = $\rhoV$

V = Volume = $\dfrac{4}{3}\pi r^3$

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

$\rho$ = Density = 19 g/cm³

c = Speed of light = $3\times 10^8\ m/s$

A = Area = $\pi R^2$

I = Intensity = $\dfrac{P}{4\pi r^2}$

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

$F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}$

According to the question

$F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W$

The power required of the light source is 468449163762.0812 W

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2 years ago

Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot, isothermal metal substrate at Th = 300°C. The proper

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Answer:

74.52s

Explanation:

The solution is shown in the picture below

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2 years ago

Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.

Katena32 [7]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created is $P_{avg} = F * v_{avg}$

Explanation:

From the question we are told that

The that the average power is mathematically represented as

$P_{avg} = \frac{W }{\Delta t }$

Where W is is the Workdone which is mathematically represented as

$W = F * s$

Where F is the applies force and s is the displacement due to the force

So

$P_{avg} = \frac{F *s }{\Delta t }$

Now this displacement can be represented mathematically as

$s = v_{avg} * \Delta t$

Where $v_{avg }$ is the average velocity and $\Delta t$ is the time taken

So

$P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }$

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