Answers

2015-11-06T17:13:25+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
If the width of the slits are A₁ and A₂, the amplitude will also be A1 and A2.

The maximum amplitude, A' will happen when constructive interference happens
A' = (A₁ + A₂)
The maximum amplitude, A will happen when destructive interference happens
A = (A₁ - A₂)

We know that intensity is proportional to the square of amplitude.
 \frac{I_1}{I_2} = \frac{A'^2}{A^2} = \frac{(A_1+A_2)^2}{(A_1-A_2)^2} \\ \\  \frac{9}{25} =(\frac{A_1+A_2}{A_1-A_2} )^2\\ \\  \sqrt{ \frac{9}{25} }=\frac{A_1+A_2}{A_1-A_2}  \\ \\\frac{A_1+A_2}{A_1-A_2} = \frac{3}{5} \\ \\ \frac{A_1}{A_2}= \frac{5+3}{5-3}  = \frac{8}{2} \\ \\ \boxed{A_1:A_2= 4:1}
1 4 1
Slit widths A1, A2. Then Amplitudes are in the same ratio. Not equal to A1 and A2.
2015-11-07T02:52:20+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
      In a Single Slit Fraunhofer Diffraction pattern, let us have the wavelength lambda of the light comparable to the slit width a.  Let the distance L between the screen and the slits.   L >> a.  We take the amplitude of the wave emerging from a slit, as being proportional (nearly) to slit width of the slit.   Proof/reason  is given below:

Let slit widths be a1 and a2 and the amplitudes from the two slits be A1 and A2. 
Then:   a1 / a2 = A1 / A2.
The intensity I in a light wave is proportional to amplitude square A².

Now in the double slit Young's Diffraction Experiment, the Intensity of the wave resulting from interference of two waves is given as:

    I =  I1 + I2 + 2 √I1 √I2 Cos Ф
    A² = A1² + A2² + 2 A1 A2 Cos Ф

At Maximum intensity,  Ф = 0.  So    A = A1 + A2
At minimum intensity,  Ф = π.  So    A =  A1 - A2,    Assume A1 > A2.

We have :  (A1 - A2)² / (A1 + A2)² = 9/25
         =>  (A1 - A2 ) / (A1 + A2 ) =  3 /5
         =>  2 A1 = 8 A2
         =>  A1 : A2  =  4 : 1

Since the amplitudes are nearly proportional to slit widths,
               w1 : w2 = 4 : 1

============
proof:

    The Amplitude A  of the wave that is emitted by the slit at an angle  theta  is the result of sum of all wave fronts from all points from y = 0 to a, along the slit width.

Sum using integration on  [0, a] of  A Sin[kL+ ky Sin θ - ωt] dy.
              (L + y Sinθ)  gives the path length of wave.

  so  y(t) = [ (a A Sin α) / α]  Sin (K L + α - ω t)            --- eq of wave emitted at  y.
             = a A Sinc (α)  * Sin (K L + α - ω t)
                       where   α = π a Sinθ / λ  = 1/2 * k a Sinθ

Thus amplitude of a light wave emitted by a single slit of width a, is proportional to  (sin Ca), where C is a constant.   We will take it as proportional to a.

===
1 5 1