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Oksana_A [137]
2 years ago
12

Light from a sodium lamp passes through a diffraction grating having 1000 slits per millimeter. The interference pattern is view

ed on a screen of 1.00m behind the grating. Two bright yellow fringes are visible 72.88cm and 73.00cm from the central maximum.
What are the wavelengths of these two fringes?
Physics
1 answer:
sashaice [31]2 years ago
3 0

Answer:

The wavelengths of these two fringes is  589.61 x 10∧9

Explanation:

d = slit width = 0.001/1000 = 1 x 10-6 m

X1 = Position of first fringe = 0.7288 m

X2 = Position of second fringe = 0.73 m

D = distance of the screen

tan\theta1 = X1/D   \Rightarrow\theta1 = tan-1(X1/D ) = tan-1(0.7288/1 ) = 36.085

tan\theta2 = X2/D   \Rightarrow\theta2 = tan-1(X2/D ) = tan-1(0.73/1 ) = 36.129

position of first bright fringe is given as

d Sin\theta1 = n\lambda1

for bright fringe , n = 1

d Sin\theta1 = \lambda1

\lambda1 = (1 x 10-6 ) Sin36.085 = 589 x 10-9

position of second bright fringe is given as

d Sin\theta2 = n\lambda2

for bright fringe , n = 1

d Sin\theta2 = \lambda2

\lambda2 = (1 x 10-6 ) Sin36.129 = 589.61 x 10-9

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Answer:

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Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

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Slotting values into equation (3).

= 10/2.5.

= ✓4.

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A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
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Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

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That would be our first equation

Now, we need to analyze its vertical motion

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Knowing v_{o} in our first equation (1)

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\frac{1}{2}(9.8)t^2=16tan(60)-8

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t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

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v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

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Answer:

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