Question

The power of a thin bi-convex lens is $P$. The lens is cut horizontally along its principle axis into two equal parts and then both the parts are placed in close contact as shown in the figure. The power of combination will be:

• A. $P$
• B. $2P$
• C. $\frac{P}{2}$
• D. Zero

Answer 1

The correct option is B $2P$

Let the initial focal length of the lens be $f$ corresponding to its given power $P$.

When the lens is cut longitudinally/horizontally, its focal length remain the same because refractive index and radius of curvature of the lens is not changed.
$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
${\mu }_2={\mu }_{\text{lens}},{\mu }_1={\mu }_{\text{air}}=1,|R_1|=|R_2|=R$

Thus,
$f_1=f_2=f$.

Now using the formula of power of combination for the equal halves of lens,
$P_e=P_1+P_2$
$\Rightarrow$Both the equal halves are supplementing the equivalent power of combination.

$\frac{1}{f_e}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}$

$\Rightarrow P_e=2P[\text{as,}\frac{1}{f}=P]$

Hence, option $\left(b\right)$ is correct.