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Sergeeva-Olga [200]
2 years ago
5

A diffraction-limited laser of length L and aperture diameter 0.50 cm generates light of wavelength 700nm . If the beam is direc

ted at the surface and the radius of the illuminated area on the surface is approximately 45cm, how far away is the surface
Physics
1 answer:
Hoochie [10]2 years ago
4 0

The distance of the surface illuminated by the light of the given wavelength is 5,269.32 m.

<h3>Distance of the surface</h3>

The distance of the surface illuminated by the light of the given wavelength is calculated as follows;

D = (rd)/(0.61λ)

where;

  • r is the radius of the illuminated area = 45 cm = 0.45 m
  • d is the diameter of the diameter = 0.5 cm = 0.005 m
  • λ is wavelength = 700 nm = 700 x 10⁻⁹ m

D = (0.45 x 0.005) / (0.61 x 700 x 10⁻⁹)

D = 5,269.32 m

Thus, the distance of the surface illuminated by the light of the given wavelength is 5,269.32 m.

Learn more about wavelength here: brainly.com/question/10728818

#SPJ1

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a measure has true value of 2.00and the measurement value was found to e 1.95.calculate the precision of the measuring instrumen
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The precision of the measuring instruments will be 2.5%.

<h3>What is precision?</h3>

It should be noted that precision simply means how close a test result will be when it's repeated.

In this case, the precision of the measuring instruments will be:

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= 2.5%

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8 0
2 years ago
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3 years ago
If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is
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Answer:

q_2=2.47\times 10^{-4}\ C

Explanation:

The charge on one object, q_1=9.9\times 10^{-5}\ C

The distance between the charges, r = 0.22 m

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Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C

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3 years ago
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