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. Power of part $\text{A}$ be $P_1$ , part $\text{B}$ be $P_2$ and part $\text{C}$ be $P_3$ . If the value of $(\frac{P_1}{P_3}\times P_2)$ is $\frac{P}{x}$, then find the value of $x$ (Give integer value).

Answer 1

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 $\frac{P}{2}$.

Thus, the value of $\frac{P_1}{P_3}\times P_2=P$

Comparing with the data given in the question we get, $x=1$.

Accepted answer: $1$