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PolarNik [594]
3 years ago
6

A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the

slit is 0.215 mm wide. The light that passes through the slit creates a diffraction pattern on a screen, which is 1.80 m from the slit. (a) How wide (in mm) is the central maximum (the central, bright fringe), as measured on the screen? ______ mm
(b) How wide (in mm) is either of the two first-order bright fringes, as measured on the screen? ______ mm
Physics
1 answer:
vredina [299]3 years ago
3 0

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light  λ is  580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D  = 1.8 m.

Width of one fringe = \frac{\lambda\times D}{d}

Putting the values we get fringe width

= \frac{580\times10^{-9}\times1.8}{.000315}

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe  is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be  

= 4.86 mm.

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A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels
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Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

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\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

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We know the distance is x=640 m and the time taken t= 40 s, thus:

\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

(c) From [2] we can solve for a:

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Since vo=8 m/s, x=640 m, t=40 s:

\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

v_f=8+0.4\cdot 40

v_f=8+16=24

The final velocity is 24 m/s

3 0
3 years ago
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