Question

A thin, symmetric double convex lens of power $P$ is cut into three parts $\text{A, B}$ and $\text{C}$ as shown in figure.


Which of the following options is correct regarding the power of parts of lens?

• A. Power of part $\text{A}$ is $2P$
• B. Power of part $\text{B}$ is $2P$
• C. Power of part $\text{B}$ is $\frac{P}{2}$
• D. Power of part $\text{C}$ is $\frac{P}{4}$

Answer 1

The correct option is C Power of part $\text{B}$ is $\frac{P}{2}$

For part $\text{A}$:
The lens maker's formula is
$\frac{1}{f}=P=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\Rightarrow P=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R}-\frac{1}{-R})$
$\Rightarrow P=(\frac{{\mu }_2}{{\mu }_1}-1)\left(\frac{2}{R}\right).....\left(1\right)$

The radius of curvature and refractive index of the medium and the lens are not changed, hence the power of $\text{A}$ will remain unchanged, that is $P$.

For part $\text{B}$ and $\text{C}$:
In the case of part $\text{B}$, the radius of one surface will remain the same while that of the other plane surface will be $\infty$.

$\frac{1}{f^{\prime }}=P^{\prime }=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R}-\frac{1}{\infty })$
$\frac{1}{f^{\prime }}=P^{\prime }=(\frac{{\mu }_2}{{\mu }_1}-1)\left(\frac{1}{R}\right).....\left(2\right)$

From equation (1) and (2)

$P^{\prime }=\frac{P}{2}$

Hence, the focal length of part $\text{B}$ will be double that of the whole lens i.e. power of part $\text{B}$ is $\frac{P}{2}$.

Hence, option (c) is correct.