A classical mechanics problem by karandeep singh ludhar

For a double convex lens $\huge\color{#3D99F6}{\frac{1}{f}=(\mu -1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}})}$ Where:
$\color{#3D99F6}{f}$ is the focal length of the double convex lens, $\color{#3D99F6}{\mu}$ is the relative refractive index, $\color{#3D99F6}{{R}_{1}}$ is the radius of curvature of the side which is in direction of the incident ray and $\color{#3D99F6}{{R}_{2}}$ is the radius of curvature of the side which is in direction opposite to that of the incident ray.

If a symmetric convex lens of focal length f is cut along its optical axis, the focal length of each part is 2f.