91Ó°ÊÓ

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

$a\sin \mathrm{θ}=m_{min}$ (1)

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

$d\sin \mathrm{θ}=m_{max}$ (2)

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

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

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

$a\sin \mathrm{θ}=m_{min}$

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

$d\sin \mathrm{θ}=m_{max}$

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,

$y_3=3\mathrm{Δ}y$

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

$$\begin{gathered} tan{\mathrm{θ}}_3=\frac{y_3}{R}=\frac{3\mathrm{Δ}y}{R} \\ ⇒\mathrm{θ}=ta{n}^{-1}\left(\frac{3\mathrm{Δ}y}{R}\right)=ta{n}^{-1}\left(\frac{3*1.0*{10}^{-2}}{90*{10}^{-2}}\right)=1.{91}^{∘} \end{gathered}$$

Now, solve (1) forand plug the given,

$a=\frac{\mathrm{λ}}{\sin \mathrm{θ}}=\frac{500*{10}^{-9}}{\sin 1.91°}15.0\mathrm{μ}m$

Solve (2) for d and plug the given,

$d=\frac{3\mathrm{λ}}{\sin \mathrm{θ}}=\frac{3*500*{10}^{-9}}{\sin 1.91°}45.0\mathrm{μ}m$

Thus, the width and separation of the two slits are $15.0\mathrm{μ}m\mathrm{and}45.0\mathrm{μ³¾}$.