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 \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).