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Gwar [14]
2 years ago
5

laser light hits two very narrow slits that are separated by 0.1mm adn is viewwed on a screen 2m downstream. Sketch on the axis

below what the ray theory of light predicts you will observe for this situation
Physics
2 answers:
Irina-Kira [14]2 years ago
5 0

Answer:

 d y / x = m λ

Explanation:

When the laser beam, which is a coherent light, hits the slits, part of each beam passes through each slit.

When this is observed on a screen that is quite far from the slits, a series of intense linear separated by dark areas. The explanation for this distribution of the light pattern is that when adding the rays that come out of the slits they travel different distances, which introduces a difference in optical path and if this difference is an integer multiple of the wavelength, a bright line

                d sin θ = m λ

Where d is the distance between the slits (0.1 mm)

Also, since the angle of the measurements is small, we can approximate the tangent

          tan θ = y / x = sin θ /sin θ

          sint θ = y / x

Substituting into the equation

          d y / x = m λ

This expression gives the location of the bright lines on the screen

Kisachek [45]2 years ago
3 0

Answer:

d y / x = m λ

Explanation:

<em>When a laser beam, which is  light coherent, hits the slits, part of each beam passes through each slit.</em>

<em>When this is noticed on a screen that is quite far from the slits, a series of intense linear moved apart  by dark areas. </em>

<em>The explanation for this distribution of the light pattern is that when adding the rays that come out of the slits they travel different distances, which makes a difference in optical path and if this difference is an integer multiple of the wavelength </em>

<em>a bright line is defined as</em>

<em />

<em>         d sin  θ = m λ </em>

<em />

<em> d is the distance between the slits (0.1 mm)</em>

<em> since the angle of the measurements is small, the tangent approximated will be</em>

<em>     tan θ = y / x = sin θ /sin θ</em>

<em>            </em>

<em>              sint θ = y / x</em>

<em />

<em>Substituting it into the equation</em>

<em />

<em>          d y / x = m λ is: </em>

<em>The  expression that  gives the location of the bright lines on the screen  </em>

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a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

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a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

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3 years ago
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