Question

If an equiconvex lens is cut into two halves along $(i)XO{X}^{\prime }$ and $(ii)YO{Y}^{\prime }$ as shown in the figure. Let $f,f^{\prime },f^{″}$ be the focal lengths of the complete lens, of each lens in case $(i)$ and of each lens in case $(ii)$ respectively. Choose the correct statement from the following:

A) $f^{\prime }=f,f^{″}=f$
B) $f^{\prime }=2f,f^{″}=2f$
C) $f^{\prime }=f,f^{″}=2f$
D) $f^{\prime }=2f,f^{″}=f$

Answer 1

Hint:First calculate the initial focal length of the lens (before cutting the lens) and then calculate the focal length of the lens in each further case and relate it to the focal length of the original lens to get the answer.

Formulas Used:
${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$
Where, $f$ is the focal length of a lens, $\mu$ is the refractive index of a lens, $R_1$ is the radius of curvature of one side of the lens, $R_2$ is the radius of curvature of other side of the lens.

Complete step by step answer:
First, we calculate the radius of curvature of the original lens (without any cut)
We have, ${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$
Here, the lens is equiconvex. Therefore, both sides of the lens will have the same radius of curvature, only in opposite directions. Let it be $R$. We will use a minus $(-)$ sign to depict the opposite direction.
Therefore, the formula becomes, ${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R_{}}}-{\displaystyle \frac{1}{-R}})$
$\Rightarrow {\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1+1}{R}})$ i.e. ${\displaystyle \frac{1}{f}}={\displaystyle \frac{2(\mu -1)}{R}}$
Now, the lens is cut along $XO{X}^{\prime }$. Each half of the lens will still act as an equiconvex lens. Hence, we use the same concept used above to get the focal length of each of the pieces.
We have, ${\displaystyle \frac{1}{f^{\prime }}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$
On putting values, ${\displaystyle \frac{1}{f^{\prime }}}=(\mu -1)({\displaystyle \frac{1}{R_{}}}-{\displaystyle \frac{1}{-R}})$
$\Rightarrow {\displaystyle \frac{1}{f^{\prime }}}=(\mu -1)({\displaystyle \frac{1+1}{R}})$ i.e. ${\displaystyle \frac{1}{f^{\prime }}}={\displaystyle \frac{2(\mu -1)}{R}}$
Therefore, focal length of each piece in this case, $f^{\prime }=f$
Now, the lens is cut along $TO{Y}^{\prime }$. Hence, each of the pieces becomes a plano-convex lens.
Here, the radius of curvature for the plane side will be $\mathrm{\infty }$.
We have, ${\displaystyle \frac{1}{f^{{}^{″}}}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$
On putting values, ${\displaystyle \frac{1}{f^{{}^{″}}}}=(\mu -1)({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{\mathrm{\infty }}})$
This gives, ${\displaystyle \frac{1}{f^{{}^{″}}}}={\displaystyle \frac{\mu -1}{R}}$
Therefore, focal length of each piece in this case, $f^{″}=2f$

Hence, option C is the correct answer.

Note:Before solving these types of questions, carefully check which kind of lens is formed after cutting the original piece. Radius of curvature of a plane lens/mirror is always $\mathrm{\infty }$.