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skad
1 month ago
5

How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th

e line density of the grating were doubled? Refer to the diffraction grating equations in your answer.
Physics
1 answer:
Softa [3K]1 month ago
4 0

Answer:

Refer to the explanation

Explanation:

Solution:-

- For this problem, we will assume the grating has a defined line density ( N ) representing the lines per mm.

- The angle formed by each fringe on the screen is given by ( θ ).

- The order of bright/dark spots is characterized by an integer ( n )

- The incident light's wavelength is ( λ )

- Using the diffraction grating relationship from the Young's Experiment is as follows:

                               n*lambda = \frac{sin(theta)}{N}

- The provided formula corresponds to constructive interference.

- We will examine how increasing the distance between the screen and the grating affects the pattern. Let ( L ) be the distance from the grating center to the screen center. The distance ( yn ) indicates the vertical spacing between each fringe on the screen.

- For small angles ( θ ), we use the approximation of sin ( θ ) ≈ tan ( θ ). Thus,

                            sin ( θ ) ≈ tan ( θ ) = [ yn / L ]

- Replacing this approximation in the diffraction grating relation results in:

                            y_n = n*lamda*L*N

- To double the distance from the screen to the grating, we use the relation with ( 2L ):

                            yn ∝ L

Result: The separation between each order of bright and dark fringe doubles. The interference pattern will spread further! Consequently, there will be fewer bright spots visible on the screen since a larger surface area is required to accommodate the entire pattern. The increased distance also diminishes the intensity contrast between bright and dark fringes due to the extended path traveled by light rays. Intensity is inversely related to the square of the travel distance.

- If the line density of the grating ( N ) were doubled, it follows that:

                            yn ∝ N

Result: The spacing between bright and dark fringes would also double. The interference pattern expands further, leading to the necessity of a bigger screen to show the entire pattern.

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Softa [3030]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
3 months ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
Softa [3030]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

\Phi =E\times S\times COS\Theta

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

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 θ=57.94°

4 0
2 months ago
A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air
Sav [3153]

Answer:

Height (h) = 17 m

Velocity (v) = 18.6 m/s

Explanation: This problem can be solved using kinematic motion equations.

Given Data

Initial velocity (u) = 0

Acceleration (a) = g

Time (t) = 1.9 seconds

First, we calculate the height.

s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

Then, we find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is a linear representation described by y=9.8, as it remains constant:

The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):

The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):

These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.

5 0
3 months ago
In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N
Keith_Richards [3271]

Answer:

The temperature of the cooler object was nearly at room temperature. As a result, the system underwent minimal change

Explanation:

In a closed system with two objects at varying temperatures, heat energy typically flows from the hotter object to the cooler one. This transfer is more pronounced when there is a significant temperature disparity between the objects. Conversely, if the temperature difference is minor or negligible, the resulting change will be minimal.

3 0
1 month ago
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