PHYSICS COLLEGE

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Express your answer in micrometers(not in nanometers).

Answers

Answer 1

Answer:

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=(y)/(D)\n\Rightarrow \theta=tan^(-1){(y)/(D)}\n\Rightarrow \theta=tan^(-1){(4.84* 10^(-3))/(3)}\n\Rightarrow \theta=0.09243\ ^(\circ)$

$sin\theta=(\lambda)/(d)\n\Rightarrow d=(\lambda)/(sin\theta)\n\Rightarrow d=(598* 10^(-9))/(sin0.09243)\n\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=(\lambda')/(2)\n\Rightarrow \lambda'=2dsin\theta\n\Rightarrow \lambda'=2* 0.00037066* sin0.09243\n\Rightarrow \lambda'=1.196* 10^(-6)\n\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

Answer 2

Answer:

Final answer:

The wavelength of light that will produce the first-order dark fringe at the same point on the screen is the same as the original wavelength of the light, which is 598 nm (0.598 μm).

Explanation:

To find the wavelength of light that will produce the first-order dark fringe at the same point on the screen, we can use the equation dsinθ = nλ, where d is the separation between the slits, θ is the angle of the fringe, n is the order of the fringe, and λ is the wavelength of the light.

In this case, the first-order bright fringe is located at a distance of 4.84 mm from the center of the central bright fringe. Since this is a first-order fringe, n = 1.

Plugging in the values, we have (0.120 mm)(sinθ) = (1)(λ). Rearranging the equation gives sinθ = λ/0.120 mm.

Since the first-order dark fringe is located at the same point as the first-order bright fringe, the angle of the first-order dark fringe can be calculated by taking the sine inverse of λ/0.120 mm.

Finally, to find the wavelength of light that will produce the first-order dark fringe at this point, we can rearrange the equation to solve for λ: λ = (0.120 mm)(sinθ).

Now, substitute the known values into the equation to calculate the wavelength of light:

λ = (0.120 mm)(sinθ) = (0.120 mm)(sin sin^-1(λ/0.120 mm)) = λ.

The wavelength of light that will produce the first-order dark fringe at this point on the screen is the same as the original wavelength of light, which is 598 nm. Converting this value to micrometers, we get 0.598 μm.

Learn more about interference pattern here:

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If a coil stays at rest in a very large static magnetic field, no emf is induced in this coil. Group of answer choices True False

Answers

Answer:

True

Explanation:

Faraday's Law says that there is a emf induced in a conductor when the vector flux of the magnetic field across it changes in time.
This can be true due to one of two facts, either the magnitude of the magnetic field changes in time, or the area through which the flux occurs changes due to the movement of the object.
In this case, due to the magnetic field is constant, and the coil stays at rest, there is no possible change in flux, so emf induced is zero.

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

Answers

Answer

given,

Tension of string is F

velocity is increased and the radius is not changed.

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone

= $(mv^2)/(r)$

now centrifugal force is balanced by tension

T = $(mv^2)/(r)$

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1 = 50 F, T2 = 120 F at the inlets and T3 = 80 F at the exit. The pipe diameters are d1 = d3 = 2" Sch 40 and d2 = 1¼" Sch 40. If the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?

Answers

Answer:

m2=3.2722lbm/s

Explanation:

Hello!

To solve this problem follow the steps below

1. Find water densities and entlapies in all states using thermodynamic tables.

note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)

through prior knowledge of two other properties, such as pressure and temperature.

D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3

D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3

D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3

h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm

h2=Enthalpy(Water;T=120;x=0)=88 BTU/lbm

h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm

2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit

m1+m2=m3

3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out

m1h1+m2h2=m3h3

18.05(m1)+88(m2)=48.03(m3)

divide both sides of the equation by 48.03

0.376(m1)+1.832(m2)=m3

4. Subtract the equations obtained in steps 3 and 4

m1 + m2 = m3

0.376m1 + 1.832(m2) =m3

--------------------------------------------

0.624m1-0.832m2=0

solving for m2

(0.624/0.832)m1=m2

0.75m1=m2

5. Mass flow is the product of density by velocity across the cross-sectional area

m1=(D1)(A)(v1)

internal Diameter for 2" Sch 40=2.067in=0.17225ft

$A=(\pi )/(4) D^2=(\pi )/(4) (0.17225)^2=0.0233ft^2$

m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s

6.use the equation from step 4 to find the mass flow in 2

0.75m1=m2

0.75(4.3629)=m2

m2=3.2722lbm/s

A curve of radius 35 m isbanked; therefore, no friction
is required at a speed of 7
m/s of a car. What is the?
banking angle

Answers

Answer:

The banking angle is 23.98 degrees.

Explanation:

We have,

Radius of a curve is 35 m

Speed of a car is 7 m/s

It is required to find the banking angle. At equilibrium, net force is equal to the centripetal force between vehicle and the road such that the banking angle is given by :

$\tan\theta=(v^2)/(rg)$

g is acceleration due to gravity

$\tan\theta=(7^2)/(35* (22)/(7))\n\n\theta=\tan^(-1)\left(0.445\right)\n\n\theta=23.98^(\circ)$

So, the banking angle is 23.98 degrees.

A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.Calculate di/dt
di/dt = _________.

Answers

The value of di/dt from the given values of the solenoid electric field is;

di/dt = 7.415 A/s

We are given;

Number of turns; N = 450 per m

Radius; r = 1.17 cm = 0.0117 m

Electric Field; E = 8.2 × 10⁻⁶ V/m

Position of electric field; r' = 3.45 cm = 0.0345 m

According to Gauss's law of electric field;

∫| E*dl | = |-d∅/dt |

Now, ∅ = BA = μ₀niA

where;

n is number of turns

i is current

A is Area

μ₀ = 4π × 10⁻⁷ H/m

Thus;

E(2πr') = (d/dt)(μ₀niA) (negative sign is gone from the right hand side because we are dealing with magnitude)

Since we are looking for di/dt, then we have;

E(2πr') = (di/dt)(μ₀nA)

Making di/dt the subject of the formula gives;

di/dt = E(2πr')/(μ₀nA)

Plugging in the relevant values gives us;

di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)

di/dt = 7.415 A/s

Read more at; brainly.com/question/14003638

Answer:

$(di)/(dt) = 7.31 \ A/s$

Explanation:

From the question we are told that

The number of turns is $N = 450 \ turns$

The radius is $r = 1.17 \ cm = 0.0117 \ m$

The position from the center consider is x = 3.45 cm = 0.0345 m

The induced emf is $e = 8.20 *10^(-6) \ V/m$

Generally according to Gauss law

$\int\limits { e } \, dl = \mu_o * N * (di)/(dt ) * A$

=> $e * 2\pi x = \mu_o * N * (d i )/(dt ) * A$

Where A is the cross-sectional area of the solenoid which is mathematically represented as

$A = \pi r ^2$

=> $e * 2\pi x = \mu_o * N * (d i )/(dt ) * \pi r^2$

=> $(di)/(dt) = (2e * x )/(\mu_o * N * r^2)$ ggl;

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) \ N/A^2$

=> $(di)/(dt) = (2 * 8.20*10^(-6) * 0.0345 )/( 4\pi * 10^(-7) * 450 * (0.0117)^2)$

=> $(di)/(dt) = 7.31 \ A/s$

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s.A. What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?B. What frequency frecede is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and receding from it?