Question

If we make $\mathit{d}\mathbf{=}\mathit{a}$ in Fig. 36-50, the two slits coalesce into a single slit of width $\mathbf{2}\mathit{a}$. Show that Eq. 36-19 reduces to give the diffraction pattern for such a slit.

Short answer

It is proved that putting $d=a$ in the double slit diffraction intensity reduces it to a single slit diffraction intensity of slit width$2a$.

Step-by-step solution

Given data

Two slits of width $a$ and slit separation$d$ .

Diffraction from double sit and single slit

The intensity at angle $\theta$from light of wavelength$\lambda$passing through two slits of width$a$ and separation$d$is given by

role="math" localid="1663144245716" $\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{m}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\pi dsin\theta }}{\mathbf{\lambda }}\mathbf{\right)}\frac{\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\pi }\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{\lambda }}\mathbf{\right)}}{{\mathbf{\left(}\frac{\mathbf{\pi }\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{\lambda }}\mathbf{\right)}}^{\mathbf{2}}}$ …(i)

The intensity at angle $\theta$ from light of wavelength$\lambda$passing through a single slit of width$a$ is given by

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{m}}\frac{{\mathbf{sin}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\pi asin\theta }}{\mathbf{\lambda }}\mathbf{\right)}}{{\mathbf{\left(}\frac{\mathbf{\pi dsin\theta }}{\mathbf{\lambda }}\mathbf{\right)}}^{\mathbf{2}}}$ …(ii)

Here, $I_m$is the intensity of the central maxima.

Determining the double slit diffraction intensity  

For $d=a$equation (i) becomes

$$\begin{gathered} I=I_m{\cos }^2\left(\frac{\mathrm{\pi asin\theta }}{\lambda }\right)\frac{{\sin }^2\left({\displaystyle \frac{\mathrm{\pi asin\theta }}{\mathrm{\lambda }}}\right)}{{\left(\frac{\mathrm{\pi asin\theta }}{\mathrm{\lambda }}\right)}^2} \\ =I_{m}4{\cos }^2\left(\frac{\mathrm{\pi asin\theta }}{\lambda }\right)\frac{{\sin }^2\left({\displaystyle \frac{\mathrm{\pi asin\theta }}{\mathrm{\lambda }}}\right)}{4{\left(\frac{\mathrm{\pi asin\theta }}{\mathrm{\lambda }}\right)}^2} \\ =I_m\frac{{\sin }^2\left({\displaystyle \frac{\pi 2\mathrm{asin\theta }}{\mathrm{\lambda }}}\right)}{{\left(\frac{\pi 2\mathrm{asin\theta }}{\mathrm{\lambda }}\right)}^2} \end{gathered}$$

This is equal to equation (ii) with$a\to 2a$.

Thus the double slit diffraction pattern reduces to a single slit diffraction pattern of slit width $2a$ .