Chapter 19, Problem 62A

(a)

To determine

To Calculate: The resolution of the lens.

Answer to Problem 62A

  $5.4\mu \text{m}$

Explanation of Solution

Given:

Aperture of the camera, $50\text{mm}$

F-ratio of the lens, $\frac{f}{8}$

Diameter of the lens, $d=6.25\text{mm}$

Distance of the CCD detector, $L=\text{50}\text{.0 mm}$

Wavelength of the light used, $\lambda =550\text{nm}$

Formula Used:

Resolution of lens can be obtained by:

  $R=\frac{1.22\times \lambda L}{d}$

Here, d is the diameter of the lens, $\lambda$ is the wavelength and Lis the distance to the CCD detector.

Calculations:

Substitute the values and solve:

  $\begin{array}{c}R=\frac{1.22\times \lambda L}{d}\\ =\frac{1.22\times 550\times {10}^{-9}\text{m}\times 50\times {10}^{-3}\text{m}}{6.25\times {10}^{-3}\text{m}}\\ =5.4\mu \text{m}\end{array}$

Conclusion:

Thus, the resolution of the lens is $5.4\mu \text{m}$ .

(b)

To determine

To Compare: The size of the pixel with width of the central spot.

Explanation of Solution

Given:

Side of a pixel, $a=7.6\text{μm}$

Resolution of the lens, $R=5.4\mu \text{m}$ .

Formula Used:

Width of the central spot can be obtained by:

  $w=2R$

Here, R is the resolution.

Calculations:

Substitute the values and solve:

  $\begin{array}{c}w=2R\\ =2\left(\text{5}\text{.4}\text{μm}\right)\\ =10.8\mu \text{m}\end{array}$

Now, take the ratio of the width of central spot and size of the pixel:

  $\begin{array}{c}\frac{w}{a}=\frac{10.8\text{μm}}{7.6\text{μm}}\\ =1.42\end{array}$

Conclusion:

Thus, in ratio of width of the central spot and size of the pixel is 1.42.