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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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Verified Answer
The correct answer is:
$2 f$
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
$$
\begin{aligned}
& \frac{1}{f^{\prime}}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\
& \frac{1}{f_1}=(\mu-1)\left[\frac{1}{R}-\frac{1}{\infty}\right] \quad\left[\because R_1=R, R_2=\infty\right] \\
& f_1=\frac{R}{(\mu-1)} \\
& f_1=\frac{2 f(\mu-1)}{(\mu-1)} \\
& f_1=2 f
\end{aligned}
$$
$$
f=F
$$
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
$$
\begin{aligned}
& \frac{1}{f^{\prime}}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\
& \frac{1}{f_1}=(\mu-1)\left[\frac{1}{R}-\frac{1}{\infty}\right] \quad\left[\because R_1=R, R_2=\infty\right] \\
& f_1=\frac{R}{(\mu-1)} \\
& f_1=\frac{2 f(\mu-1)}{(\mu-1)} \\
& f_1=2 f
\end{aligned}
$$
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