Question

Diffraction and Interference Combined. Consider the interference pattern produced by two parallel slits of width$\mathbf{\alpha }$ and separation d, in which d =3 a. The slits are illuminated by normally incident light of wavelength $\mathit{\lambda }$. (a) First, we ignore diffraction effects due to the slit width. At what angles$\mathbf{\theta }$ from the central maximum will the next four maxima in the two slit interference pattern occurs? Your answer will be in terms ofd and $\mathbf{\lambda }$. (b) Now we include the effects of diffraction. If the intensity at$\mathbf{\theta }\mathbf{=}\mathbf{0}\mathbf{°}$ is ${\mathbf{I}}_{\mathbf{0}}$, what is the intensity at each of the angles in part (a)? (c) Which double-slit interference maxima are missing in the pattern? (d) Compare your results to those illustrated in Fig. 36.12 c. In what ways are your results different?

Short answer

(a) The angles are$\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{\lambda }{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{2\lambda }{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{3\lambda }{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{4\lambda }{d}\right)$

(b) The intensity at each of the angles is$\mathbf{0}\mathbf{.}\mathbf{684}{\mathit{I}}_{\mathbf{0}}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{171}{\mathit{I}}_{\mathbf{0}}\mathbf{,}\mathbf{0}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{043}{\mathit{I}}_{\mathbf{0}}$

(c) The third interference maxima is missing in the pattern.

(d) Fourth interference is always dark, while in this case, the third interference is the missing one.

Step-by-step solution

Formulas used to solve the question

Angles of the bright fringes in double-slit experiments:

$\mathbf{sin}\mathbf{}\mathbf{\theta }\mathbf{=}\frac{\mathbf{m\lambda }}{\mathbf{d}}$

Intensity is given by

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

Where

$\mathbf{\varphi }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi dsin}\mathbf{}\mathbf{\theta }}{\mathbf{\lambda }}$

And

$\mathbf{\beta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi a}\mathbf{}\mathbf{sin\theta }}{\mathbf{\lambda }}$

Determine the angles

Given: d =3 a

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

$\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }}{\mathbf{d}}$

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

Also, m =0 is for the bright maxima.

So,

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{d}} \\ \mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{\lambda }{d}\right) \end{gathered}$$ (1)

Similarly, for other three angles,

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}\mathbf{\lambda }}{\mathbf{d}}\mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{2\lambda }{d}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(2\right) \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{\theta }}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{3}\mathbf{\lambda }}{\mathbf{d}}\mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{3}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{3\lambda }{d}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(3\right) \\ \mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{\theta }}_{\mathbf{4}}\mathbf{=}\frac{\mathbf{4}\mathbf{\lambda }}{\mathbf{d}}\mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{4}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{4\lambda }{d}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(4\right) \end{gathered}$$

Determine the intensity at each angle

Now, the intensity is given as,

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

Whereas

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta }}{\mathbf{\lambda }}$

And

$\mathit{\beta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta }}{\mathbf{\lambda }}$

Therefore,

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\frac{\pi dsin\theta }{\lambda }\right){\left(\frac{sin\frac{\pi asin\theta }{\lambda }}{\frac{\pi asin\theta }{\lambda }}\right)}^{\mathbf{2}}$

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

$$\begin{gathered} {\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\frac{\pi d\lambda }{\lambda d}\right){\left(\frac{sin\frac{\pi a\lambda }{\lambda d}}{\frac{\pi a\lambda }{\lambda d}}\right)}^{\mathbf{2}} \\ \mathbf{\Rightarrow }{\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\left(co{s}^2\pi \right){\left(\frac{sin\frac{\pi a}{d}}{\frac{\pi a}{d}}\right)}^{\mathbf{2}} \end{gathered}$$

From the given,

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

Similarly for all the angles,

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

Determine the missing interference maxima

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.

Compare the results

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 $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{\lambda }{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{2\lambda }{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{3\lambda }{d}\right)\mathbf{,}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{4\lambda }{d}\right)$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.