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Q.
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
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Solution:
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 $p=\frac{1}{f} \dots(iv)$
$\therefore p'=\frac{1}{f'}=\frac{1}{2f} {\text{(Using}} (iii))$
$=\frac{p}{2}$
$p'=\frac{p}{2}=\frac{4D}{2}=2D$
${\text{(Given}} \,P=4D)$
