Chapter 19, Problem 50A

To determine

The length of the kaleidoscope.

Answer to Problem 50A

The length of the kaleidoscope is $0.4\text{ m}$ .

Explanation of Solution

Given:

The diameter of the eye hole at the end is $D=7.0\text{ mm}$ .

The separation distance between objects is $x_{\text{obj}}=40\text{ μm}$ .

The wavelength is $\lambda =650\text{ nm}$ .

Formula used:

The expression for the Rayleigh criterion is,

  $x_{\text{obj}}=\frac{1.22\lambda L_{\text{obj}}}{D}$

Here, $\lambda$ is the wavelength, $L_{\text{obj}}$ is the distance from the circular aperture to the objects, and $D$ is the diameter of the circular aperture.

Calculation:

The length of the kaleidoscope is,

  $\begin{array}{l}{x}_{\text{obj}}=\frac{1.22\lambda L_{\text{obj}}}{D}\\ L_{\text{obj}}=\frac{D{x}_{\text{obj}}}{1.22\lambda }\\ L_{\text{obj}}=\frac{\left(40\text{ μm}\times \frac{{10}^{-6}\text{ m}}{1\text{ μm}}\right)\left(7.0\text{ mm}\times \frac{{10}^{-3}\text{ m}}{1\text{ mm}}\right)}{1.22\times \left(650\text{ nm}\times \frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)}\\ L_{\text{obj}}=0.4\text{ m}\end{array}$

Conclusion:

Thus, the length of the kaleidoscope is $0.4\text{ m}$ .