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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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2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗
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Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

   The tension is  T =  48.255 \ N

b

   The time taken is  t =  0.226 \ s

c

   The position for maximum velocity is  

        S = 0  

d

     The maximum velocity is  V_{max} =0.384 \ m/s

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

    The mass of the bob  is m_b  =  5 \ kg

     The angle is  \theta = 10^o

     The length of the string is  L =  0.5 \ m

The tension on the string is mathematically represented as

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substituting values

             T =  5 * 9.8 cos(10)

             T =  48.255 \ N

The motion of the bob is mathematically represented as

             S =  A sin (w t  + \frac{\pi }{2} )

     =>   S =  A sin (wt)

Where  w is the angular speed

and  \frac{\pi}{2} is the phase change

At initial position S =  0

   So   wt  = cos^{-1} (0)

          wt  = 1

Generally w can be mathematically represented as

          w =  \frac{2 \pi }{T}

Where T is the period of oscillation which i mathematically represented as

          T  =  2 \pi \sqrt{\frac{L}{g} }

So  

      t   = \frac{1}{w}

       t   = \frac{T}{2 \pi}

       t =  \sqrt{\frac{L}{g} }

substituting values

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Looking at the equation

         wt =  1

We see that maximum velocity of the bob will be at  S = 0  

i. e     w =  \frac{1}{t}

The maximum velocity is mathematically represented as

          V_{max}   =  w A

Where A is the amplitude which is mathematically represented as

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So

      V_{max} = \frac{2 \pi }{T }  L sin \theta

      V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} }   L sin \theta  Recall   T  =  2 \pi \sqrt{\frac{L}{g} }

     V_{max} = \sqrt{gL} sin \theta

substituting values        

       V_{max} = \sqrt{9.8 * 0.5 } sin (10)

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