Question

A biconvexlens is cut intotwo 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. $2 D$
• B. $3 D$
• C. $4 D$
• D. $5 D$

Answer 1

Answer: (A)

$2 D$

Answer 2

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 $f=\frac{R}{2\left(\mu-1\right)}\ldots\left(i\right)$ 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 $\frac{1}{f'}=\left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ Here $R_{1}=+R, R_{2}=\infty$ $\therefore\frac{1}{f'}=\left(\mu-1\right)\left(\frac{1}{R}\right)$ or $f' =\frac{R}{\left(\mu-1\right)} \ldots\left(iii\right)$ From $\left(i\right)$ and $\left(ii\right)$, we get $f'=2f \dots(iii)$ Power of a lens $p=\frac{1}{f} \dots(iv)$ $\therefore p'=\frac{1}{f'}=\frac{1}{2f} {\text{(Using}} (iii))$ $=\frac{p}{2}$ $p'=\frac{p}{2}=\frac{4D}{2}=2D$ ${\text{(Given}} \,P=4D)$