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:
\[\boxed{20 \text{ cm}}.\]
In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)
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