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Goshia [24]
3 years ago
9

Laser light of wavelength lambda passes through a thin slit of width a and produces its first dark fringes at angles of +/- 30 d

egree with the original direction of the beam. The slit is then reduced in size to a circle of diameter a. When the same laser light is passed through the circle, the first dark fringe occurs at
A) +/- 66.9 degree.
B) +/- 45.0 degree.
C) +/- 37.6 degree.
D) +/- 36.6 degree.
E) +/- 15.0 degree.
Physics
1 answer:
Scilla [17]3 years ago
7 0

Answer:

C- ±37.6°

Explanation:

This is the case of a single slit diffraction. In this case, the dark fringe occurs at

sin θ = ± mλ/a

where m = 1

θ = ± 30°

Hence we have sin ± 30° = ± λ/a

±0.5 = ± λ/a

Hence, we have a = 2λ

For a circular aperture, the condition for the first dark fringe is

D = diameter of circle = a = 2λ

so we have sin θ1 = 1.22λ/a

sin θ1 = 1.22λ/2λ

hence sin θ1 = 0.61

θ1 = sin⁻¹0.61 = ±37.6°

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1.33

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Mathematically, from the given information;

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where;

n_{water} =1.33\\ \\ n_{air} = 1.00

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A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration ?
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By substitution,

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Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, the
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Complete Image

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by  \Delta L_{thick}and \Delta L_{thin}, respectively.The ratio ΔLthick rod/ΔLthin rod is:

a.) =1

b.) <1

c.) >1

Answer:

The ratio is less than 1 i.e \frac{1}{2}   option B is correct

Explanation:

The Young Modulus of a material is generally calculated with this formula

               E = \frac{\sigma}{\epsilon}

 Where \sigma is the stress = \frac{Force}{Area}

             \epsilon is the strain = \frac{\Delta L}{L}

  Making Strain the subject

              \epsilon = \frac{\sigma}{E}

now in this question we are that the same tension was applied to both wires so

      \frac{\sigma}{E} would be constant

Hence

                 \frac{\Delta L}{L} = constant

for the two wire we have that

                  \frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2}

      Looking at young modulus formula

                E = \frac{\frac{F}{A} }{\frac{\Delta L}{L} }

                    E * \frac{\Delta L }{L}  = \frac{F}{A}

                  A * \frac{\Delta L}{L}  = \frac{F}{E}

Now we are told that a comprehensive force is applied to the wire so for this question

                \frac{F}{E} is constant

And given that the length are the same

so  

     A_1 \frac{\Delta L_{thin}}{L_{thin}} = A_2 \frac{\Delta L_{thick}}{L_{thick}}

Now we are told that one is that one rod is twice as thick as the other

So it implies that one would have an area that would be two times of the other

  Assuming that

           A_2 = 2 A_1

So

       A_1 \frac{\Delta L_{thin}}{L_{thin}} = 2 A_1 \frac{\Delta L_{thick}}{L_{thick}}

     \frac{\Delta L_{thin}}{L_{thin}} = 2 \frac{\Delta L_{thick}}{L_{thick}}

From the question the length are equal

      \Delta L_{thin} =2  \Delta L_{thick}

So  

       \frac{\Delta L_{thick}}{\Delta L_{thin} } = \frac{1}{2}

Hence the ratio is less than 1

       

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