In a certain two-slit interference pattern, 10 bright fringes lie within the second side peak of the diffraction envelope and diffraction minima coincide with two-slit interference maxima. What is the ratio of the slit separation to the slit width?

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 $m_1=1$

role="math" localid="1663102529320" $$\begin{gathered} \mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}{\mathit{m}}_{\mathbf{1}}\mathit{\lambda } \\ \mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{\lambda } \end{gathered}$$ …… (1)

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

role="math" localid="1663102556769" $\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}{\mathit{m}}_{\mathbf{2}}\mathit{\lambda }$ ……. (2)

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

Combine equations (1) and (2).

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

It is given that there are 10 bright fringes in one side, which means $m_2=11$.

The ratio of the slit separation to the slit width will be as follows.

$$\begin{gathered} 11=\frac{d}{a} \\ \frac{d}{a}=11 \end{gathered}$$

Therefore, the ratio of the slit separation to the slit width is 11.