Question

A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is $ 4D $ , the power of a cut lens will be

• A. $ 2D $
• B. $ 3D $
• C. $ 4D $
• D. $ 5D $

Answer 1

Answer: (A)

Biconvex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens.
Focal length of biconvex lens

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$\frac{1}{f}=\left(n-1\right) \frac{2}{R}\quad\left(\because R_{1}=R, R_{2}=-R\right)$
$\Rightarrow f=\frac{R}{2\left(n-1\right)}\quad\ldots\left(i\right)$
For plano-convex lens
$\frac{1}{f_{1}}=\left(n-1\right)\left(\frac{1}{R}-\frac{1}{\infty}\right)$
$f_{1}=\frac{R}{\left(n-1\right)}\quad\ldots\left(ii\right)$
Comparing Eqs. $(i)$ and $(ii)$, we see that focal length becomes double.
As power of lens $P \propto\frac{1}{\text{focal length}}$
Hence, power will become half.
New power $=\frac{4}{2}=2D$