Choose the correct statement from the following:An equiconvex lens is cut two halves along (i) XOX' and (ii) YOY' as shown in the figure. Let $f, f^{\prime}…
Choose the correct statement from the following:- $f^{\prime}=f, f^{\prime \prime}=2 f$
- $f^{\prime}=2 f, f^{\prime \prime}=f$
- $f^{\prime}=f, f^{\prime \prime}=f$
- $f=2 f, f^{\prime \prime}=2 f$
Solution
Using lens maker formula
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Here $R_1=R$
$R_2=-R$ by convection
$\therefore \frac{1}{f}=(\mu-1) \frac{2}{\mathrm{R}}(\mu-1) \frac{1}{\mathrm{R}}=\frac{1}{2 f} \ldots(\mathrm{i})$
If we cut the lens along $\mathrm{XOX}^{\prime}$ then two halves of the lens will be having the same radii of curvalur and
$\begin{aligned}
& R_1=R, R_2=-R \\
& \therefore \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{2}{R}\right) \\
& \Rightarrow \frac{1}{f^{\prime}}=\frac{2}{2 f}=\frac{1}{f}
\end{aligned}$
From (i) it is clear that $f^{\prime}=f$
But when we cut along $\mathrm{YOY}$ ' then
$\begin{aligned}
& R_1=R \text { and } R_2=\infty \\
& \therefore \infty \frac{1}{f^{\prime \prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\
& =(\mu-1) \frac{1}{R}=\frac{1}{2 f} \\
& \Rightarrow \quad f^{\prime \prime}=2 f
\end{aligned}$Asked in: NEET 2003