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wolverine [178]
3 years ago
8

Two sources of light of wavelength 720 nm are 10 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for

their diffraction patterns to be resolved by Rayleigh's criterion?
Physics
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

The value is  y  =  0.00732 \ m    

Explanation:

From the question we are told that

    The wavelength of each source is  \lambda  = 720 \ nm =  720 *10^{-9} \  m

     The distance from the pinhole D = 10 \ m

     The diameter is d =  1.2 mm = \frac{1.2}{1000} = 0.0012 \ m

Generally from Rayleigh's criterion  we have that the distance between the sources of light for their diffraction patterns is mathematically represented as '

        y  =  \frac{ 1.22 *  \lambda *  D}{d}

=>      y  =  \frac{ 1.22 * 720 *10^{-9}*   10 }{ 0.0012}

=>      y  =  0.00732 \ m      

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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
Avogadro‘s number was calculated by determining The number of atoms in
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Answer:

12 grams of the isotope carbon-12.

Explanation:

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7 0
2 years ago
chuck wagon travels with a constant velocity of 0.5 mile/minutes. determine the total distance traveled by Chuck Wagon during 12
n200080 [17]

Given:

Velocity: 0.5 mile/minute

Time: 12 minute

Now we know that speed and velocity have the same magnitude. Hence speed=velocity=0.5 mile/min

Substituting the given values in the above formula we get

Distance = 0.5 x 12= 6 miles

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3 years ago
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snow_tiger [21]
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3 years ago
If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

                         θ_1  = 90° -  ∅

                         θ_1  = 90° -  20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

8 0
3 years ago
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