91Ó°ÊÓ

Equation for the first minimum of the dark fringes:

$\mathbf{a}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{θ}\mathbf{=}\mathbf{³¾Î»}$

Intensity is given by

$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{0}}\left({\cos }^2\frac{\mathrm{ϕ}}{2}\right){\left(\frac{\sin {\displaystyle }\frac{\mathrm{β}}{2}}{\frac{\mathrm{β}}{2}}\right)}^{\mathbf{2}}$

Where

$\mathbf{β}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€²¹}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{θ}}{\mathbf{λ}}$

Since the bright fringe contains5 fringes, so the minimum fringe that makes an interference of total dark is for $\mathit{m}\mathbf{=}\mathbf{Â\pm }\mathbf{3}$.

The equation is given by

$\mathit{a}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$

Hence,

$\mathit{a}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{θ}\mathbf{=}\mathit{λ}$ (1)

While it is the third bright fringe in the double-slit pattern,

$\mathit{d}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{θ}\mathbf{=}\mathbf{3}\mathit{λ}$ (2)

Now, the intensity is given by

$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{0}}\mathbf{(}{\mathbf{cos}}^{\mathbf{2}}\mathbf{}\frac{\mathbf{ϕ}}{\mathbf{2}}\mathbf{)}{\mathbf{\left(}\frac{\mathbf{sin}{\displaystyle }\frac{\mathbf{β}}{\mathbf{2}}}{\frac{\mathbf{β}}{\mathbf{2}}}\mathbf{\right)}}^{\mathbf{2}}$

Where

$\mathit{β}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€²¹}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{θ}}{\mathbf{λ}}$

Thus,

$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{0}}\mathbf{(}{\mathbf{cos}}^{\mathbf{2}}\mathbf{}\frac{\mathbf{Ï•}}{\mathbf{2}}\mathbf{)}{\mathbf{\left(}\frac{\mathbf{sin}{\displaystyle }\frac{\mathbf{Ï€²¹}\mathbf{}\mathbf{²õ¾\pm ²Ôθ}}{\mathbf{λ}}}{\frac{\mathbf{Ï€²¹}\mathbf{}\mathbf{²õ¾\pm ²Ôθ}}{\mathbf{λ}}}\mathbf{\right)}}^{\mathbf{2}}$

Now, for $\mathit{m}\mathbf{=}\mathbf{Â\pm }\mathbf{3}$, the intensity is zero.

$$\begin{gathered} \mathbf{0}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\frac{\mathrm{ϕ}}{2}\right){\left(\frac{sin\frac{\mathrm{π}asin\mathrm{θ}}{\mathrm{λ}}}{\frac{\mathrm{π}asin\mathrm{θ}}{\mathrm{λ}}}\right)}^{\mathbf{2}} \\ \mathbf{⇒}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\frac{\mathbf{π}\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}}{\mathbf{λ}}}{\frac{\mathbf{π}\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}}{\mathbf{λ}}}\mathbf{=}\mathbf{0} \\ \mathbf{⇒}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\frac{\mathbf{π}\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}}{\mathbf{λ}}\mathbf{=}\mathbf{0} \end{gathered}$$

Plug the value,

$\begin{array}{c}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\left(\frac{3\mathrm{π}asin\mathrm{θ}}{dsin\mathrm{θ}}\right)\mathbf{=}\mathbf{0}\\ \mathbf{⇒}\frac{\mathbf{3}\mathbf{a}}{\mathbf{d}}\mathbf{=}\mathbf{1}\\ \mathbf{⇒}\mathbf{d}\mathbf{/}\mathbf{a}\mathbf{=}\mathbf{3}\mathbf{.0}\end{array}$

Now, the fringe for the central maximum fringe are for

$\mathit{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{Â\pm }\mathbf{1}\mathbf{,}\mathbf{Â\pm }\mathbf{2}$

So,

$\mathit{a}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{θ}\mathbf{=}\mathbf{2}\mathit{λ}$ (3)

And

$\mathit{d}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$ (4)

Divide (4) by (3) and plug d /a,

$$\begin{gathered} \frac{\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}}{\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}}\mathbf{=}\frac{\mathbf{m}\mathbf{λ}}{\mathbf{2}\mathbf{λ}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{⇒}\frac{\mathbf{m}}{\mathbf{2}}\mathbf{=}\mathbf{3} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{⇒}\mathit{m}\mathbf{=}\mathbf{Â\pm }\mathbf{6} \end{gathered}$$

This means that the second bright fringe envelope is contained between$\mathit{m}\mathbf{=}\mathbf{Â\pm }\mathbf{3}$ and $\mathit{m}\mathbf{=}\mathbf{Â\pm }\mathbf{6}$.

Implies that this envelope only contains two bright fringes which are for $\mathit{m}\mathbf{=}\mathbf{Â\pm }\mathbf{4}\mathbf{,}\mathbf{Â\pm }\mathbf{5}$.

Thus, the ratio d /a must be 3.0. There are3.0 fringes contained with the first diffraction maximum on one side of the central maximum.