Suppose that the central diffraction envelope of a double-slit diffraction pattern contains 11 bright fringes and the first diffraction minima eliminate (are coincident with) bright fringes. How many bright fringes lie between the first and second minima of the diffraction envelope?

There are mainly two phenomena occurs in double slit experiment. The first one is interference due to the difference in two paths, and the second one is diffraction in the single slit.

For the first minima in the diffraction pattern,

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}{\mathit{m}}_{\mathbf{1}}\mathit{\lambda }$

The angular locations of the bright fringes of the double-slit interference pattern is given by,

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}{\mathit{m}}_{\mathbf{2}}\mathit{\lambda }$ …… (1)

Here, d is the slit separation, and a is single slit width.

For the 1^st minima , $m_1=1$then rewrite the equation (1) as follows.

$a\sin \theta =\lambda$ …… (2)

Combine equation (1) and (2).

$m_2=\frac{d}{a}$

Given that the central diffraction envelope of a double-slit diffraction pattern contains 11 bright fringes. For this, the value of$m_2$will be as follows.

$$\begin{gathered} m_2=\frac{11-1}{2} \\ =5 \end{gathered}$$

This gives the number of bright fringes on one side of central diffraction maxima in central diffraction envelope. Therefore, the 1^st diffraction minima occurs just before $m_2=6$.

For the 2^nd minima $m_1=2$, then rewrite the equation (1) as follows.

$a\sin \theta =2\lambda$ ….. (3)

From equations (1) and (3),

$$\begin{gathered} m_2=\frac{2d}{a} \\ =2\left(5\right) \\ =10 \end{gathered}$$

The 2^nd diffraction minima occur just before $m_2=11$. So, in between 1^st and 2^nd minima of the diffraction envelop, there are the fringes from$m_2=6$ to$m_2=10$ for a total 5 bright fringes in the double slit interference pattern.

Therefore, the total number of bright fringes lies between the first second minima is 5.

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