From the lens formula:
\[\frac{1}{f} = \frac{1}{x} + \frac{1}{y}.\]
The plot shows the relationship between y (image distance) and x (object distance). For any specific point on the graph, the values of x and y can be substituted into the lens formula to calculate f.
Alternatively, the product of x and y at corresponding points can be used to directly calculate the focal length. From the lens equation, we know:
\[x \cdot y = f^2.\]
From the graph: - At point C, x = 20 cm and y = 20 cm.
Using \(x \cdot y = f^2\):
\[f^2 = 20 \cdot 20 = 400.\]
Taking the square root:
\[f = \sqrt{400} = 20 \text{ cm}.\]
Final Answer: The focal length of the lens is:
\[\boxed{20 \text{ cm}}.\]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: