Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is:

<p>To find the power of a combination of two thin convex lenses, we use the formula for the equivalent focal length $ f_{\text{eq}} $ of the lens system when the lenses are placed coaxially with a certain distance $ d $ between them:</p> <p>$ \frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} $</p> <p>Given:</p> <p><p>$ f_1 = 30 \, \text{cm} $ </p></p> <p><p>$ f_2 = 10 \, \text{cm} $</p></p> <p><p>$ d = 10 \, \text{cm} $ (distance between the two lenses)</p></p> <p>First, convert the focal lengths from centimeters to meters:</p> <p><p>$ f_1 = 0.3 \, \text{m} $</p></p> <p><p>$ f_2 = 0.1 \, \text{m} $</p></p> <p>Plug in the values into the equation:</p> <p>$ \frac{1}{f_{\text{eq}}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{(0.3)(0.1)} $</p> <p>Calculate each term:</p> <p>$ \frac{1}{0.3} = \frac{10}{3} \approx 3.33 $</p> <p>$ \frac{1}{0.1} = 10 $</p> <p>$ \frac{0.1}{(0.3)(0.1)} = \frac{0.1}{0.03} = \frac{10}{3} \approx 3.33 $</p> <p>Substitute these calculated values back into the formula:</p> <p>$ \frac{1}{f_{\text{eq}}} = 3.33 + 10 - 3.33 $</p> <p>$ \frac{1}{f_{\text{eq}}} = 10 $</p> <p>Thus, the power of the lens combination is:</p> <p>$ \text{Power} = \frac{1}{f_{\text{eq}}} = 10 \, \text{D} $</p> <p>Therefore, the power of the lens system is 10 diopters (D).</p>