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 principle axis, it will become a plano-convex lens.
Focal length of biconvex lens
$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R₁}-\frac{1}{R₂}\right)$
$\frac{1}{f}=\left(n-1\right)\frac{2}{R}(\because R_1=R,R_2=-R)$
$⟹f=\frac{R}{2\left(n-1\right)}\dots .\dots (i)$
For plano-convex lens
$\frac{1}{f₁}=\left(n-1\right)\left(\frac{1}{R}-\frac{1}{\in fty}\right)$
$f^1=\frac{R}{\left(n-1\right)}\dots \dots \dots \dots .(ii)$
Comparing Eqs. (i) and (ii), we see that focal length becomes double.
Power of lens P $\propto \frac{1}{focallength}$
Hence, power will become half.
New power = $\frac{4}{2}=2D$