Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 cm from the slits. The bright bands are 1.00 cm apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

Angle of the minimum fringes due to the single-slit diffraction:

$a\sin \theta =m_{min}$ (1)

Angle of the bright fringes due to the double-slit interference:

$d\sin \theta =m_{max}$ (2)

Given: $\lambda =500.0nm=500.0*{10}^{-9}m$

$$\begin{gathered} R=90.0cm=90.0*{10}^{-2}m \\ \Delta y=1.0cm=1.0*{10}^{-2}m \end{gathered}$$

Angle of the minimum fringes due to the single-slit diffraction:

$a\sin \theta =m_{min}$

Angle of the bright fringes due to the double-slit interference:

$d\sin \theta =m_{max}$

Divide the equation (2) by (1),

$\frac{d}{a}=\frac{m_{max}}{m_{min}}$

The third bright is for $m_{max}$and since it is the first missing bright fringe, so it meets the first dark fringe of the single slit diffraction which is for $m_{min}$

Therefore,

d = 3a

(3)

Now, the distance from the central maxima to the third minimum is given by,

$y_3=3\Delta y$

Now, find the angle since it is the same angle for both equations (1) and (2) and plug the given,

$$\begin{gathered} tan{\theta }_3=\frac{y_3}{R}=\frac{3\Delta y}{R} \\ \Rightarrow \theta =ta{n}^{-1}\left(\frac{3\Delta y}{R}\right)=ta{n}^{-1}\left(\frac{3*1.0*{10}^{-2}}{90*{10}^{-2}}\right)=1.{91}^{\circ } \end{gathered}$$

Now, solve (1) forand plug the given,

$a=\frac{\lambda }{\sin \theta }=\frac{500*{10}^{-9}}{\sin 1.91°}15.0\mu m$

Solve (2) for d and plug the given,

$d=\frac{3\lambda }{\sin \theta }=\frac{3*500*{10}^{-9}}{\sin 1.91°}45.0\mu m$

Thus, the width and separation of the two slits are $15.0\mu m\mathrm{and}45.0\mathrm{\mu m}$.