Q37E - University Physics with Modern Physics

Question

Monochromatic light with wavelength 620 nm passes through a circular aperture with diameter 7.4  $\mathit{\mu }$m. The resulting diffraction pattern is observed on a screen that is 4.5 m from the aperture. What is the diameter of the Airy disk on the screen?

Step-by-Step Solution

Answer

On the screen, the diameter of the disk is  0.92 m

Step 1: Given Data

The light’s wavelength is 620 nm  

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

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

Step 2: Rayleigh Criterion

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.

  $\mathbf{Sin\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}}$

Here  is the angle of diffraction, $\mathit{\lambda }$is the wavelength of light used and d is the diameter of the aperture.

Step 3: Diameter of the airy disk

According to Rayleigh, 

$$\begin{gathered} \mathrm{sin\theta }=1.22\frac{\mathrm{\lambda }}{\mathrm{d}} \\ =1.22\frac{620 \mathrm{nm}}{7.4 \mathrm{\mu m}} \\ =1.22\frac{620\times {10}^9\mathrm{m}}{7.4\times {10}^6\mathrm{m}} \\ =0.102 \end{gathered}$$ 

As,$\theta$ is very small

$$\begin{gathered} \mathrm{sin\theta }\approx \mathrm{\theta } \\ =0.102 \\ \approx \mathrm{tan\theta } \end{gathered}$$  

Now the radius of the disk is 

$$\begin{gathered} \mathrm{r}=\mathrm{Rtan\theta } \\ =(4.5\mathrm{m})\times 0.102 \\ =0.46 \mathrm{m} \end{gathered}$$  

Therefore, diameter

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

Hence, the diameter of the disk is 0.92 m .