Question

A biconvexlens is cut intotwo 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)

Focal length o f a biconvex lens,
$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Here, $R_1=+R,R_2=-R$

$\therefore \frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R}-\frac{1}{\left(-R\right)}\right)$
$\frac{1}{f}=\frac{2\left(\mu -1\right)}{R}$
or $f=\frac{R}{2\left(\mu -1\right)}\dots \left(i\right)$

When a biconvex lens is cut into two equal parts by a plane perpendicular to the principal axis, then each half behaves as a piano convex lens.
Focal length o f a piano convex lens
$\frac{1}{f^{\prime }}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Here $R_1=+R,R_2=\infty$
$\therefore \frac{1}{f^{\prime }}=\left(\mu -1\right)\left(\frac{1}{R}\right)$
or $f^{\prime }=\frac{R}{\left(\mu -1\right)}\dots \left(iii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
$f^{\prime }=2f\dots (iii)$
Power of a lens $p=\frac{1}{f}\dots (iv)$
$\therefore p^{\prime }=\frac{1}{f^{\prime }}=\frac{1}{2f}\text{(Using}(iii))$
$=\frac{p}{2}$
$p^{\prime }=\frac{p}{2}=\frac{4D}{2}=2D$
$\text{(Given}P=4D)$