The figure shows equiconvex lens of focal length ' $\mathrm{f}$ '. If the lens is cut along $\mathrm{AB}$,…
The figure shows equiconvex lens of focal length ' $\mathrm{f}$ '. If the lens is cut along $\mathrm{AB}$, the
focal length of each half will be
$2 \mathrm{f}$
$\mathrm{f}$
$3 \mathrm{f}$
$4 \mathrm{f}$
Solution
According to Lens Makers formula,
$\begin{aligned}
& \frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{R}\right) \\
& \Rightarrow \frac{1}{f}=(\mu-1)\left(\frac{2}{R}\right) \ldots \ldots \text { (i) }
\end{aligned}$
On cutting, $\frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)$
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R}\right) \ldots . \text { (ii) }$
From Eqs. (i) and (ii), we get
$f=2 f$
