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avanturin [10]
3 years ago
9

Using a 679 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

hat the 14 th dark fringe is 8.01 cm away from the center of the central maximum. How far is the screen located from the slit?
Physics
1 answer:
sveta [45]3 years ago
7 0

To solve the problem, apply the concepts related to constructive and destructive interference. For this purpose, we will start by defining the distance from the dark fringe from the central maximum in terms of the slit separation, the number of fringe the wavelength and the distance of the screen to the slit. Mathematically this is,

y_n = \frac{(2n-1)\lambda D}{2d}

D = \frac{2dy_n}{(2n-1)\lambda}

Here,

d = Slit separation

n = Number of fringe

y_n = Distance of the dark fringe from the central maximum

D = Distance of the screen to the slit

\lambda = Wavelength

Our values are given as,

\lambda = 679nm = 679*10^{-9}m

d = 1.1mm = 1.1*10^{-3}m

y_{14} = 8.01cm = 8.01*10^{-2} m

n = 14

Replacing,

D = \frac{2(1.1*10^{-3})(8.01*10^{-2})}{(2(14)-1)(679*10^{-9})}

D = 9.61m

Therefore the screen is located from the slit to 9.61m

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Answer:

D. 22 Hz and 42 Hz

Explanation:

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  • <em><u>The frequency of the beat produced is equivalent to </u></em><em><u>the difference between the individual frequencies of the two waves involved.</u></em>
  • <em><u>Therefore; in this case since the frequency of the beat is 20 Hz, that is from 20 beats per second.</u></em>
  • We need to find a pair from the choices whose frequency difference is 20 Hz.
  • This happens to be choice D. 22 Hz and 42 Hz,  that is 42 Hz - 22 Hz = 20 Hz
8 0
3 years ago
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A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)

\phi = 1.89 \times 10^{-7}t

(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

magnitude of voltage is

e = 1.89 x 10^-7 V

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8 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

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solve the following system by any method 2x - 6y = 24 ... -5x + 6y = -6 ... A. (0,-6) B. (4,-1) C. (-6,-6) D. (6,-1)
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Using elimination, the answer is C. 
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