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 $4 \, D$ , 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
$\frac{1}{f}=\left(n - 1\right)\left(\frac{1}{R ₁} - \frac{1}{R ₂}\right)$
$\frac{1}{f}=\left(n - 1\right)\frac{2}{R} \, \, \left(\right.\because R_{1}=R,R_{2}=-R\left.\right)$
$\Longrightarrow f=\frac{R}{2 \left(n - 1\right)} \, \ldots .\ldots \left(\right.i\left.\right)$
For plano-convex lens
$\frac{1}{f ₁}=\left(n - 1\right)\left(\frac{1}{R} - \frac{1}{\in fty}\right)$
$f^{1}=\frac{R}{\left(n - 1\right)} \, \, \ldots \ldots \ldots \ldots .\left(\right.ii\left.\right)$
Comparing Eqs. (i) and (ii), we see that focal length becomes double.
Power of lens P $ \propto \frac{1}{f o c a l \, l e n g t h}$
Hence, power will become half.
New power = $\frac{4}{2}=2 \, D$