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Ostrovityanka [42]
3 years ago
5

Monochromatic light with lambda equals 630 space n m is incident on a single slit. The slit width is 0.8 mm. If the distance bet

ween the screen and slit is 1.8 m, what is the width of the central bright fringe on the screen
Physics
1 answer:
krek1111 [17]3 years ago
4 0

Answer:

The angular width is "1.575\times 10^{-3} \ rad" and the linear width is "2.835\times 10^{-3} \ m".

Explanation:

Given:

Wavelength,

\lambda = 630 \ nm

or,

  =630\times 10^{-9} \ m

Slit width,

a = 0.8 \ mm

or,

  =0.8\times 10^{-3} \ m

Distance between slit and screen,

D = 1.8 \ m

As we know,

The central bright fringe's angular height,

⇒ \theta = \frac{2 \lambda}{a}

       = \frac{2\times 630\times 10^{-9}}{0.8\times 10^{-3}}

       =1.575\times 10^{-3} \ rad

and,

The linear width will be:

⇒ x_o=\frac{2D \lambda}{a}

By substituting the values, we get

        =\frac{3\times 1.8\times 630\times 10^{-9}}{0.8\times 10^{-3}}

        =2.835\times 10^{-3} \ m

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Answer:

a) W = 10995.6 J

b) W = - 9996 J

c) Kf = 999.6 J

d) v = 5.77 m/s

Explanation:

Given

m = 60 Kg

h = 17 m

a = g/10

g = 9.8 m/s²

a) We can apply Newton's 2nd Law as follows

∑Fy = m*a     ⇒     T - m*g = m*a     ⇒    T = (g + a)*m

where T is the force exerted by the cable

⇒    T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)

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W = F*d = T*h = (646.8 N)*(17 m)

W = 10995.6 J

b) We use the formula

W = m*g*h     ⇒    W = (60 Kg)(9.8 m/s²)(-17 m)

⇒    W = - 9996 J

c) We have to obtain Wnet as follows

Wnet = W₁ + W₂ = 10995.6 J - 9996 J

⇒    Wnet = 999.6 J

then we apply the equation

Wnet = ΔK = Kf - Ki = Kf - 0 = Kf    

⇒  Kf = 999.6 J

d) Knowing that

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⇒    v = √(2*999.6 J / 60 Kg)

⇒    v = 5.77 m/s

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4 years ago
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