Question

An amateur photographer attempts to build a custom zoom lens using a converging lens followed by a diverging lens. The two lenses are separated by a distance \(x=50 . \mathrm{mm}\) as shown below. If the focal length of the first lens is \(2.0 \cdot 10^{2} \mathrm{~mm}\) and the focal length of the second lens is \(-3.0 \cdot 10^{2} \mathrm{~mm},\) what will the effective focal length of this compound lens be? What will \(f_{\text {eff }}\) be if the lens separation is changed to \(1.0 \cdot 10^{2} \mathrm{~mm}\) ?

Short answer

Answer: The approximate effective focal lengths of the compound lens system are 166.7 mm for the first case, when the lenses are separated by 50 mm, and 125.0 mm for the second case, when the lenses are separated by 100 mm.

Step-by-step solution

Understanding the Lens Maker's Formula

To calculate the effective focal length of a compound lens system, the following formula is used: \( \frac{1}{f_{eff}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \) where \(f_{eff}\) is the effective focal length of the compound lens, \(f_1\) is the focal length of the first lens, \(f_2\) is the focal length of the second lens, and \(d\) is the distance between the two lenses. We will use this formula in the subsequent steps to find the effective focal length for both cases.

Calculating the effective focal length for the first case

In this case, the given focal lengths and lens separation are: \(f_1 = 2.0 \cdot 10^{2} \text{mm},\ f_2 = -3.0 \cdot 10^{2} \text{mm}\), and \(d = 50 \mathrm{~mm}\). Using these values in the lens maker's formula: \( \frac{1}{f_{eff}} = \frac{1}{2.0 \cdot 10^2} - \frac{1}{3.0 \cdot 10^2} - \frac{50}{(2.0 \cdot 10^2)(-3.0 \cdot 10^2)} \) Now, calculate the effective focal length: \( f_{eff} = \frac{1}{\frac{1}{2.0 \cdot 10^2} - \frac{1}{3.0 \cdot 10^2} - \frac{50}{(2.0 \cdot 10^2)(-3.0 \cdot 10^2)}} \) \( f_{eff} \approx 166.7 \ mm \) So, the effective focal length for the first case is approximately \(166.7 \ mm\).

Calculating the effective focal length for the second case

In this case, the focal lengths are the same, but the lens separation has changed: \(d = 1.0 \cdot 10^{2} \mathrm{~mm}\) Using the updated lens separation in the lens maker's formula: \( \frac{1}{f_{eff}} = \frac{1}{2.0 \cdot 10^2} - \frac{1}{3.0 \cdot 10^2} - \frac{100}{(2.0 \cdot 10^2)(-3.0 \cdot 10^2)} \) Now, calculate the effective focal length: \( f_{eff} = \frac{1}{\frac{1}{2.0 \cdot 10^2} - \frac{1}{3.0 \cdot 10^2} - \frac{100}{(2.0 \cdot 10^2)(-3.0 \cdot 10^2)}} \) \( f_{eff} \approx 125.0 \ mm \) So, the effective focal length for the second case is approximately \(125.0 \ mm\). In conclusion, the effective focal length of the compound lens is approximately \(166.7 \ mm\) for the first case and \(125.0 \ mm\) for the second case when the lens separations are \(50 \ mm\) and \(100 \ mm\), respectively.