Question

The equi-convex lens has a focal length ${}^{\prime }{f}^{\prime }$ 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 ?

• A. $\frac{f}{2}$
• B. $2f$
• C. $\frac{3f}{2}$
• D. $f$

Answer 1

Answer: (B)

Key Idea The lens Maker's formula is given as
$\frac{1}{f}=\left({\mu }_{\mathrm{med}}-1\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$
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,
$\frac{1}{f^{\prime }}=\left({\mu }_{\text{real }}-1\right)\left[\frac{1}{R}-\left(-\frac{1}{\infty }\right)\right]$
$\Rightarrow \frac{1}{f}=\left({\mu }_{\text{real }}-1\right)\left[\frac{1}{R}\right]$
So, from the Eqs. (i) and (ii), we get $f^{\prime }=2f$