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Q.
The equi-convex lens has a focal length $'f'$ If the lens is cut along the line perpendicular to the principal axis and passing through the
pole, what will be the focal length of any half part ?
MHT CETMHT CET 2019Ray Optics and Optical Instruments
Solution:
Key Idea The lens Maker's formula is given as
$\frac{1}{f}=\left(\mu_{\operatorname{med}}-1\right)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$
where, $f=$ focal length of lens,
$R_{1}=$ radius of first curved part and
$R_{2}=$ Radius of second curved part
As, for equiconvex lens $R_{1}=R_{2}=R$ (say)
So, $\frac{1}{f}=\left(\mu_{\text {real }}-1\right) \frac{2}{R}$...(i)
Now, if lens is cut along the line perpendicular to the principal axis as shown in the figure,
The new cut part of the lens has, $R_{1}=R$ and $R_{2}=\infty$
Again by using the lens Maker's formula, focal length of the new part of the lens,
$\frac{1}{f'}=\left(\mu_{\text {real }}-1\right)\left[\frac{1}{R}-\left(-\frac{1}{\infty}\right)\right]$
$\Rightarrow \frac{1}{f}=\left(\mu_{\text {real }}-1\right)\left[\frac{1}{R}\right]$
So, from the Eqs. (i) and (ii), we get $f'=2 f$
