91Ó°ÊÓ

The light’s wavelength is 620 nm

The aperture’s diameter is $7.4\mathrm{μ³¾}$

The separation between screen and aperture is $4.5\mathrm{m}$

According to the Rayleigh criterion for the limit to resolution, two images are only resolvable when the center of one's diffraction pattern is directly over the first minimum of the other's diffraction pattern.

$\mathbf{³\S ¾\pm ²Ôθ}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{λ}}{\mathbf{d}}$

Hereis the angle of diffraction, $\mathit{λ}$is the wavelength of light used and dis the diameter of the aperture.

According to Rayleigh,

$$\begin{gathered} \mathrm{²õ¾\pm ²Ôθ}=1.22\frac{\mathrm{λ}}{\mathrm{d}} \\ =1.22\frac{620\mathrm{nm}}{7.4\mathrm{μ³¾}} \\ =1.22\frac{620×{10}^9\mathrm{m}}{7.4×{10}^6\mathrm{m}} \\ =0.102 \end{gathered}$$

As,$\mathrm{θ}$ is verysmall

$$\begin{gathered} \mathrm{²õ¾\pm ²Ôθ}≈\mathrm{θ} \\ =0.102 \\ ≈\mathrm{³Ù²¹²Ôθ} \end{gathered}$$

Now the radius of the disk is

$$\begin{gathered} \mathrm{r}=\mathrm{R³Ù²¹²Ôθ} \\ =(4.5\mathrm{m})×0.102 \\ =0.46\mathrm{m} \end{gathered}$$

Therefore, diameter

$$\begin{gathered} \mathrm{d}=2\mathrm{r} \\ =2×0.46\mathrm{m} \\ =0.92\mathrm{m} \end{gathered}$$ .

Hence, the diameter of the disk is 0.92 m .