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MAVERICK [17]
3 years ago
14

3. A laser with a wavelength of 650 nm passes through a double slit. A pattern is observed on a wall that is 1.5 meters away fro

m the slits. The spacing between 10 bright fringes is measured to be 3 cm. Calculate the separation between the slits. Show your work.
Physics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

Slit separation, d=3.25\times 10^{-4}\ m

Explanation:

It is given that,

Wavelength of laser light, \lambda=650\ nm=650\times 10^{-9}\ m

Distance between pattern and the wall, L = 1.5 m

The spacing between 10 bright fringes is, y = 3 cm = 0.03 m

We need to find the separation between the slits. According to the equation of diffraction law :

d\ sin\theta=n\lambda

Where

d is the separation between the slits

For smaller angles, sin\theta=tan\theta=\dfrac{y}{L}

d(\dfrac{y}{L})=n\lambda

d=\dfrac{n\lambda L}{y}

d=\dfrac{10\times 650\times 10^{-9}\times 1.5}{0.03}

d = 0.000325 m

or

d=3.25\times 10^{-4}\ m

So, the separation between the slits is 3.25\times 10^{-4}\ m. Hence, this is the required solution.

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Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

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After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

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If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

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v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

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When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th
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