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stealth61 [152]
4 years ago
9

In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi

thin each secondary diffraction peak (between m= + 1 and + 2 in Dsinθ=mλ). Assume the first diffraction minimum occurs at an interference minimum. Express your answer as an integer.
Physics
1 answer:
Ronch [10]4 years ago
6 0

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

no of fringes  contained within each secondary diffraction peak = 6.5

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Answer:

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Explanation:

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The gravitational force is determine by the relation:

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According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

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3 0
3 years ago
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th
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Answer:

58.5 m

Explanation:

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The initial vertical velocity of the ball is

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u = 21.5 m/s is the initial speed

\theta=33.5^{\circ} is the angle

Substituting,

u_y = (21.5) sin 33.5^{\circ} =11.9 m/s

The vertical position of the ball at time t is given by

y = h + u_y t + \frac{1}{2}gt^2

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The horizontal velocity of the ball is

u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

d= u_x t = (17.9)(3.27)=58.5 m

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3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
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In other words a infinitesimal segment dV caries the charge 
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<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
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<span>drop higher order terms </span>
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<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
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<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
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4 years ago
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