A convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ( $\mu=$ the refractive index of the material of the lens)

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Aicharya · Accepted Answer

The given, $R_1=R, R_2=-R$ $ f=F $

Lens Maker's formula $ \begin{aligned} \frac{1}{F} & =(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\ \frac{1}{f} & =(\mu-1)\left[\frac{1}{R}+\frac{1}{R}\right] \\ f & =\frac{R}{2(\mu-1)} \\ R & =2 f(\mu-1) \end{aligned} $ Now, it is divided vertically into two identical plano convex lens

$ \begin{aligned} & \frac{1}{f^{\prime}}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\ & \frac{1}{f_1}=(\mu-1)\left[\frac{1}{R}-\frac{1}{\infty}\right] \quad\left[\because R_1=R, R_2=\infty\right] \\ & f_1=\frac{R}{(\mu-1)} \\ & f_1=\frac{2 f(\mu-1)}{(\mu-1)} \\ & f_1=2 f \end{aligned} $

A convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided…

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