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kolbaska11 [484]
3 years ago
14

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

the line density of the grating were doubled? Refer to the diffraction grating equations in your answer.
Physics
1 answer:
jolli1 [7]3 years ago
5 0

Answer:

a) The distance between the ineas doubles, intensity decreases with distance

b) The distance between the ineas doubles

Explanation:

The diffraction pattern of a grid is given as a percentage

                d sin θ = m λ

where d give the distance between two consecutive lines, θ it is at an angle, λ the wavelength and m is an integer that determines the order of diffraction, let's not forget that the entire spectrum is at a value of m and then it is repeated.

Let's apply this to our case

a) distance from grid to observation screen doubles

Here we have two effect:

* the energy of the source is constant, it must be distributed over a surface, therefore the intensity decreases with distance

* The other factor can be found using trigonometry

          tan θ= y / L

where y is the distance from the central maximum to the line under study and L is the distance to the screen

           

In general, diffraction experiments cover very small angles

             tan θ = sin θ/ cos θ = sin θ

we substitute

          sinθ = y / L

we subtitle into the diffraction equation

          d y / L = m λ

          y = L / d m λ

          L = 2 L₀

          y = 2 L₀ m λ / d

we see that by doubling the distance to the screen the lines we are seeing are separated by double

b) When the density of lines doubles, it means that in the same distance I have twice as many lines, therefore the distance between two consecutive lines is reduced by half

          d = d₀o / 2

          y = (L m λ) / d

          y = (L m λ/ d₀) 2

we see that The distance between the ineas doubles

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A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
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