Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are illuminated by normally incident light of wavelength λ. Required:
a. First we ignore diffraction effects due to the slit width. At what angles θ from the central maximum will the next four maxima in the two-slit interference pattern occur?
b. If the intensity at θ = 0 is I0, what is the intensity at each of the angles in part (a)?
c. Which double-slit interference maxima are missing in the pattern?

(a)  The next four angles from the central maxima are sin⁻¹  λ  / d, sin⁻¹  λ  / 2d , sin⁻¹  λ  / 3d, and sin⁻¹  λ  / 4d respectively.

(b)   With the values of part a are the expression of intensity of each maximum can be calculated accordingly.

(c)   The third order maxima (m = 3) is missing from the pattern obtained through the double-slit experiment.

a)

In the interference phenomenon the maxima are given by the expression

d sin θ = m λ

Here,

d is the slit separation and m is the order of interference.

The maximum for m = 1 is at the angle is,

θ = sin⁻¹  λ  / d

For the second maximum m = 2

θ = sin⁻¹ ( λ  / 2d)

For the third maximum m = 3

θ = sin⁻¹ ( λ  / 3d)

For the fourth maximum m = 4

θ = sin⁻¹ ( λ  / 4d)

Thus, we can conclude that the next four angles from the central maxima are sin⁻¹  λ  / d, sin⁻¹  λ  / 2d , sin⁻¹  λ  / 3d, and sin⁻¹  λ  / 4d respectively.

b)

Considering the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

I = I₀ cos² (Ф) (sin x / x)²

Where Ф is the phase angle, and its value is,

 Ф = π d sin θ /λ

x = π  a sin θ /λ

Here a is the width of the slits

Thus, with the values of part a are introduced in the expression the  intensity of each maximum can be calculated accordingly.

c)

The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present.

For the maximum interference:        

d sin θ = m λ

For the first diffraction minimum:    

a sin θ = λ

Dividing both the expressions as,

d / a = m

Since, d = 3a. Then,

3a / a = m

m = 3

Thus, we can conclude that the third order maxima (m = 3) is missing from the pattern obtained through the double-slit experiment.

Learn more about the interference here:

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a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

        θ = sin⁻¹ ( λ  / 3d)

the fourth maximum m = 4

       θ = sin⁻¹ ( λ  / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

       I = I₀ cos² (Ф) (sin x / x)²

       Ф = π d sin θ /λ

       x = pi a sin θ /λ

where a is the width of the slits

with the values of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference       d sin θ = m λ

first diffraction minimum    a sin θ = λ

we divide the two expressions

                       d / a = m

In our case

                   3a / a = m

                    m = 3

order three is no longer visible

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