Show that if two thin lenses are close together \((s,\) the distance between the lenses, is negligibly small), the two lenses can be replaced by a single equivalent lens with focal length \(f_{\mathrm{eq}} .\) Find the value of \(f_{\mathrm{eq}}\) in terms of \(f_{1}\) and \(f_{2}.\)

The lens formula can be written as: $$ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $$ Where \(f\) is the focal length of the lens, \(d_o\) is the object distance, and \(d_i\) is the image distance. We'll use this formula for both lenses while solving this problem.

Apply the lens formula for the first lens, which has a focal length of \(f_1\): $$ \frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_{i1}} $$ Where \(d_{i1}\) is the image distance formed by the first lens.

Apply the lens formula for the second lens, which has a focal length of \(f_2\). Since the image formed by the first lens acts as the object for the second lens, and the distance between the lenses is negligible, we can write the lens formula as: $$ \frac{1}{f_2} = \frac{1}{d_{i1}} + \frac{1}{d_{i2}} $$ Where \(d_{i2}\) is the final image distance formed by the combination of both the lenses.

From step 2, rewrite the equation to express \(d_{i1}\) in terms of \(f_1\) and \(d_o\): $$ d_{i1} = \frac{f_1d_o}{d_o - f_1} $$ Substitute this expression for \(d_{i1}\) in the equation from step 3: $$ \frac{1}{f_2} = \frac{1}{\frac{f_1d_o}{d_o - f_1}} + \frac{1}{d_{i2}} $$ Now, we will apply the lens formula for the equivalent lens with focal length \(f_{\text{eq}}\): $$ \frac{1}{f_{\text{eq}}} = \frac{1}{d_o} + \frac{1}{d_{i2}} $$

Comparing the last equation with the equation for the second lens, we can write: $$ \frac{1}{f_{\text{eq}}} = \frac{1}{f_2} - \frac{1}{\frac{f_1d_o}{d_o - f_1}} $$ Solving for \(f_{\text{eq}}\), we get: $$ \frac{1}{f_{\text{eq}}} = \frac{1}{f_2} - \frac{d_o - f_1}{f_1d_o} $$ Simplify the equation to obtain the desired expression for \(f_{\text{eq}}\) in terms of \(f_1\) and \(f_2\): $$ \frac{1}{f_{\text{eq}}} = \frac{-(f_1 - d_o) + f_2d_o}{f_1d_o} = \frac{f_2d_o - f_1d_o + f_1^2}{f_1d_o} $$ Finally, the required value of \(f_{\text{eq}}\) is given by: $$ f_{\text{eq}} = \frac{f_1d_o}{f_2d_o - f_1d_o + f_1^2} $$ This is the expression for the equivalent focal length of the two thin lenses close together in terms of their individual focal lengths, \(f_1\) and \(f_2\).