After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at±19.0° with theoriginal direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is 1/10 the maximum intensity on the screen?

For double-slit experiment, that the position of the dark fringes is given by

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\left(m+\frac{1}{2}\right)\mathit{\lambda }$

Intensity of fringe is given by

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\varphi }}{\mathbf{2}}\mathbf{\right)}$

We see that, rearranging the formula,

$\frac{\mathrm{d}}{\mathrm{\lambda }}=(\mathrm{m}+\frac{1}{2})\times \frac{1}{\mathrm{sin\theta }}$

Note, m=0

$\frac{d}{\lambda }=\frac{1}{2\sin \theta }$

Input$\mathrm{\theta }=19.0°$

$$\begin{gathered} \frac{d}{\lambda }=\frac{1}{2\sin \left(19.0\right)} \\ \frac{d}{\lambda }=1.54 \end{gathered}$$

Hence the ratio is 1.54

We know that intensity is given by

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_0{\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right) \\ \mathrm{I}={\mathrm{I}}_0{\cos }^2\left(\frac{\mathrm{\pi dsin\theta }}{\mathrm{\lambda }}\right) \end{gathered}$$

We need to find the angle in which the intensity is one-tenth the maximum intensity i.e.

$\mathrm{I}=\frac{1}{10}{\mathrm{I}}_0$

Plug this into

$$\begin{gathered} \frac{1}{10}{\mathrm{I}}_0={\mathrm{I}}_0{\cos }^2\left(\frac{\mathrm{\pi dsin\theta }}{10}\right) \\ \frac{1}{10}{\mathrm{I}}_0={\cos }^2\left(\frac{\mathrm{\pi dsin\theta }}{10}\right) \\ \sqrt{\frac{1}{10}}=\cos \left(\frac{\mathrm{\pi dsin\theta }}{10}\right) \\ {\cos }^{-1}\left(\sqrt{\frac{1}{10}}\right)=\frac{\mathrm{\pi dsin\theta }}{\mathrm{\lambda }} \end{gathered}$$

Plug the value of and

${\cos }^{-1}\left(\sqrt{\frac{1}{10}}\right)=1.54\pi \sin \theta$

Solve for theta we get

$\theta ={\sin }^{-1}\left[\frac{{\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{10}}}\right)}{1.54\pi }\right]$

We get$\theta =\pm 15.0°$

Hence, The smallest angle is$\pm 15.0°$