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Ganezh [65]
3 years ago
6

Parallel wave fronts are incident on an opening in a barrier. Which diagram shows the configuration of wave fronts and barrier o

pening that will result in the greatest diffraction of the waves passing through the opening? [Assume all diagrams are drawn to the same scale.]

Physics
2 answers:
Novosadov [1.4K]3 years ago
5 0
By looking at the diagram you posted above and reading the given information. Diagram number 2 shows the configuration of wave fronts and barrier opening. This graph number 2 shows the greatest diffraction of the waves through the opening. <span />
Ulleksa [173]3 years ago
4 0

Explanation :

It is given that, parallel wave fronts are incident on an opening in a barrier. Diffraction of the waves depends on the size of opening and the wavelength of light used as per the following relation :

y\propto\dfrac{\lambda}{a}

where,

\lambda is the wavelength of light

and a is the opening size (from fig)

So, it is clear that diffraction of waves depends directly on the wavelength of light and inversely on the size of opening.

Hence, the correct option is (2).

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4 years ago
An audio amplifier provides an alternating rms emf of 15.0V a loudspeaker connected to the amplifier has a resistance of 10.4 w
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Current in the speaker will be 1.442 A

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So emf of the audio amplifier V=15volt

Resistance of the amplifier R=10.4ohm

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From ohm's ;aw we know that V = IR , here I is current and R is resistance

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4 0
3 years ago
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo
Mariulka [41]

Answer:

 λ = 605.80 nm

Explanation:

These double-slit experiments the equation for constructive interference is

          d sin θ = m λ

where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.

In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm

Let's use trigonometry to find the angle

         tan θ = y / L

as the angles are very small

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

we substitute in the first equation

         d y / L = m λ          

          λ = d y / m L

let's calculate

           λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)

           λ = 6.05805 10⁻⁷ m

let's reduce to nm

          λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)

          λ = 605.80 nm

8 0
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The focal length is the distance from the middle of the lens to the what?
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It's the distance between the middle of the lens to the focus.

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4 years ago
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