91Ó°ÊÓ

Angles of the bright fringes in double-slit experiments:

$\mathbf{sin}\mathbf{}\mathbf{θ}\mathbf{=}\frac{\mathbf{³¾Î»}}{\mathbf{d}}$

Intensity is given by

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

Where

$\mathbf{Ï•}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€»å²õ¾\pm ²Ô}\mathbf{}\mathbf{θ}}{\mathbf{λ}}$

And

$\mathbf{β}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€²¹}\mathbf{}\mathbf{²õ¾\pm ²Ôθ}}{\mathbf{λ}}$

Given: d =3 a

The angle of the bright fringes in double-slit experiments is given by,

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

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

Also, m =0 is for the bright maxima.

So,

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{θ}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{d}} \\ \mathbf{⇒}{\mathit{θ}}_{\mathbf{1}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\mathbf{}\left(\frac{\mathrm{λ}}{d}\right) \end{gathered}$$ (1)

Similarly, for other three angles,

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{θ}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}\mathbf{λ}}{\mathbf{d}}\mathbf{⇒}{\mathit{θ}}_{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\mathbf{}\left(\frac{2\mathrm{λ}}{d}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(2\right) \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{θ}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{3}\mathbf{λ}}{\mathbf{d}}\mathbf{⇒}{\mathit{θ}}_{\mathbf{3}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\mathbf{}\left(\frac{3\mathrm{λ}}{d}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(3\right) \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{θ}}_{\mathbf{4}}\mathbf{=}\frac{\mathbf{4}\mathbf{λ}}{\mathbf{d}}\mathbf{⇒}{\mathit{θ}}_{\mathbf{4}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\mathbf{}\left(\frac{4\mathrm{λ}}{d}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(4\right) \end{gathered}$$

Now, the intensity is given as,

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

Whereas

$\mathit{ϕ}\mathbf{=}\frac{\mathbf{2}\mathbf{π}\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}}{\mathbf{λ}}$

And

$\mathit{β}\mathbf{=}\frac{\mathbf{2}\mathbf{π}\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{θ}}{\mathbf{λ}}$

Therefore,

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\frac{\mathrm{π}dsin\mathrm{θ}}{\mathrm{λ}}\right){\left(\frac{sin\frac{\mathrm{π}asin\mathrm{θ}}{\mathrm{λ}}}{\frac{\mathrm{π}asin\mathrm{θ}}{\mathrm{λ}}}\right)}^{\mathbf{2}}$

So, for all the four angles, plug the$\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{θ}$ value,

$$\begin{gathered} {\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\frac{\mathrm{π}d\mathrm{λ}}{\mathrm{λ}d}\right){\left(\frac{sin\frac{\mathrm{π}a\mathrm{λ}}{\mathrm{λ}d}}{\frac{\mathrm{π}a\mathrm{λ}}{\mathrm{λ}d}}\right)}^{\mathbf{2}} \\ \mathbf{⇒}{\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\mathrm{π}\right){\left(\frac{sin\frac{\mathrm{π}a}{d}}{\frac{\mathrm{π}a}{d}}\right)}^{\mathbf{2}} \end{gathered}$$

From the given,

$$\begin{gathered} \mathbf{⇒}{\mathbf{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\mathrm{π}\right){\left(\frac{sin\frac{\mathrm{π}}{3}}{\frac{\mathrm{π}}{3}}\right)}^{\mathbf{2}} \\ \mathbf{⇒}{\mathit{I}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.684}{\mathit{I}}_{\mathbf{0}} \end{gathered}$$

Similarly for all the angles,

$$\begin{gathered} {\mathit{I}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{171}{\mathit{I}}_{\mathbf{0}} \\ {\mathit{I}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}} \\ {\mathit{I}}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{043}{\mathit{I}}_{\mathbf{0}} \end{gathered}$$

Since d =3a, the missing bright fringe is the third one in sequence. In other words, after two bright interference, the third is a dark one. As above, angle one and angle two are bright but angle three is dark and then angle four is bright and so on.

In the figure provided, there is a dark fringe each4 bright fringes. This means that the fourth interference is missing while in this case the third one is missing.

Thus, the angles are $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\left(\frac{\mathrm{λ}}{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\mathbf{}\left(\frac{2\mathrm{λ}}{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\mathbf{}\left(\frac{3\mathrm{λ}}{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{−}\mathbf{1}}\mathbf{}\left(\frac{4\mathrm{λ}}{d}\right)$The intensity at each of the angles is $\mathbf{0}\mathbf{.}\mathbf{684}{\mathit{I}}_{\mathbf{0}}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{171}{\mathit{I}}_{\mathbf{0}}\mathbf{,}\mathbf{0}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{043}{\mathit{I}}_{\mathbf{0}}$. The third interference maxima is missing in the pattern. Fourth interference is always dark, while in this case, the third interference is the missing one.