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pentagon [3]
4 years ago
12

A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is r

educed, what happens to the width of the central bright fringe? Assume the angles involved remain small. A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. The central bright fringe becomes narrower. The effect cannot be determined unless the distance between the slit and the screen is known. The central bright fringe becomes wider. The central bright fringe remains the same size.
Physics
1 answer:
anygoal [31]4 years ago
5 0

Answer:

Explanation:

General guidance

Concepts and reason

The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.

Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.

Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.

Fundamentals

The condition for diffraction maximum in the case of single slit diffraction is as follows:

sin Θ=λ/α

Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .

The incorrect options are as follows:

• The central bright fringe remains in same size.

The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.

• The effect cannot be determined unless the distance between the slit and screen is known.

Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.

• The central bright fringe becomes narrower.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.

If the width of the slit increases, then the central bright fringe width also increases.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.

The condition for diffraction maximum is as follows:

sin  Θ=λ/α

The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

Therefore,

• The central bright fringe becomes wider is correct.

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.  

Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.

If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.

Answer

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.

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What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the moveme
soldi70 [24.7K]

Answer:

The energy is 4.57x10^{-19} J and the wavelength is 4.34x10^{-7}m for the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 2 to the orbit with n = 5.

<em>In what part of the electromagnetic spectrum do we find this radiation? </em>

In the Ultraviolet part of the electromagnetic spectrum.

Explanation:

The energy of the absorbed photon can be known by the difference in energy between the two states in which the transition is happening (In this case from n = 2 to n = 5):

E = E_{upper}-E_{lower}   (1)

The permitted energy for the atom of hydrogen, according with the Bohr's model, is defined as:

E_{n} = -\frac{13.606 eV}{n^{2}}   (2)

Or it can be expressed in Joules, since 1eV = 1.60x10^{-19}J

E_{n} = -\frac{2.18x10^{-18} J}{n^{2}}   (3)

Where the value -2.18x10^{-18} represents the energy of the ground state¹ and n is the principal quantum number.

For the case of n = 2:

E_{2} = -\frac{2.18x10^{-18} J}{(2)^{2}}

E_{2} = -5.45x10^{-19} J

For the case of n = 5:

E_{5} = -\frac{2.18x10^{-18} J}{(5)^{2}}

E_{5} = -8.72x10^{-20} J

Replacing those values in equation (1) it is gotten:

E = -8.72x10^{-20} J-(-5.45x10^{-19} J )

E = 4.57x10^{-19} J

The wavelength can be determined by means of the Rydberg formula:

\frac{1}{\lambda} = R(\frac{1}{n_{l}^{2}}-\frac{1}{n_{u}^{2}})  (4)

Where R is the Rydberg constant, with a value of 1.097x10^{7}m^{-1}

For this particular case n_{l} = 2 and n_{u} = 5:

\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(\frac{1}{(2)^{2}}-\frac{1}{(5)^{2}})

\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(0.21)

\frac{1}{\lambda} = 2303700m^{-1}

\lambda = \frac{1}{2303700m^{-1}}

\lambda = 4.34x10^{-7}m

So the energy is 4.57x10^{-19} J and the wavelength is 4.34x10^{-7}m for the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 2 to the orbit with n = 5.

<em>In what part of the electromagnetic spectrum do we find this radiation? </em>

In the Ultraviolet part of the electromagnetic spectrum.

Key terms:

¹Ground state: State of minimum energy.  

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Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB
aliina [53]

The moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

<h3>What is moment of inertia?</h3>

The moment of inertia is the amount of rotation obtained  by an object when it is in state of motion or rest.

Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB = 2.83 kg, mC = 3.39 kg.

The coordinates  of Ball A :(2,1) Ball B :(8,2) and Ball C: (5,8) and axis is at (4,3)

Here, for the moment of inertia of each ball

For ball A

IA = 1.09 x ((4 - 2)^2 + (3-1)^2)

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IB = 2.83 * ((4 - 8)^2 + (3 - 2)^2)

IB = 48.11 Kg.m^2

for the ball C

IC = 3.39 * ((4 - 5)^2 + (3 - 8)^2)

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Total moment of inertia = 8.72 + 48.11 +  88.14

Total moment of inertia = 144.97 Kg.m^2

Thus, the moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

Learn more about moment of inertia.

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What is the weight in newtons of a 10 kg mass on the earth's<br> surface?
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Answer:

98 kg

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Explanation:

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Answer:

5 seconds

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d² = x² + y²

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d² = 1521 - 3120t + 1600t² + 51984 + 13680t + 900t²

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The cheetah can run a total distance of:

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The time t at this distance is:

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False

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