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kupik [55]
1 year ago
15

a single rectangular slit of width of 50 is illuminated by red light with a wavelength of 600 nm. a screen is placed a certain d

istance behind the slit so that the central bright band of the diffractio pattern is 18 mm wide
Physics
1 answer:
katovenus [111]1 year ago
8 0

A single rectangular slit of width of 50 is illuminated by red light with a wavelength of 600 nm and a screen is placed a certain distance behind the slit so that the central bright band of the diffraction pattern is 18 mm wide then the width of slit is 1.667 mm .

<h3>How is the width calculated?</h3>

Given ,

λ = 600 × 10⁻⁹

D = 50

d = 18 mm

The position of central  maxima in the diffraction pattern is

w = Dλ / d

∴ w = (50 × 600  × 10⁻⁹ ) / (18  × 10⁻³ )

∴ w = 30000 × 10⁻⁹ / 18 × 10⁻³

∴ w = 30 × 10⁻⁶ / 18 × 10⁻³

∴ w = 1.667 × 10⁻³

∴ w = 1.667 mm

Hence, the width of the central bright fringe  of the diffraction pattern will be 1.667 mm.

<h3>What is the central bright band of diffraction?</h3>
  1. The width of the central diffraction peak is found to be inversely proportional to the slit width.
  2. Increasing the width magnitude a decreases the angle θ at which the intensity first goes to zero, narrowing the central band.
  3. Reducing the width of the slit also increases the angle θ and widens the central band.
  4. The width and spacing of the bright bands on the screen behind the slit depend on the distance between the slit and the screen.
  5. Therefore, the light intensity distribution becomes diffracted light.

Can learn more about the slit experiment from brainly.com/question/14703580

#SPJ4

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A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
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Answer:

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A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor
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Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

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The  formula for the induced emf is

E = \frac{\Delta  \phi}{\Delta  t}

\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

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substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

we plug in the values in the equation above

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E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

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