A bi convex lens of focal length 10cm is cut in two identical parts...

The focal length of the bi-convex lens $\left(f\right)$ can be calculated as follows:

$\begin{cases}\dfrac{1}{f} & = & ( \mu - 1 ) \left(\dfrac{1}{R} - \dfrac{1}{- R}\right) \\ & = & \dfrac{2\left(\mu - 1\right)}{R} . . . \left(1\right)\end{cases}$

When the lens is cut into two halves, the focal length \left(f '\right)$of each part can be calculated as follows:</p><p>$\begin{cases}\dfrac{1}{{f}^{'}} & = & ( \mu - 1 ) \left(\dfrac{1}{R} - \dfrac{1}{\infty}\right) \\ & = & \dfrac{\left(\mu - 1\right)}{R} . . . \left(2\right)\end{cases}$</p><p>Hence, the ratio of the powers for both the lenses can be given by</p><p>$\begin{cases}\dfrac{{P}^{'}}{P} & = & \dfrac{\dfrac{1}{f '}}{\dfrac{1}{f}} \\ & = & \dfrac{f}{f '} \\ & = & \dfrac{\dfrac{\left(\mu - 1\right)}{R}}{\dfrac{2\left(\mu - 1\right)}{R}} \\ & = & \dfrac{1}{2} \\ & ⇒ & P ' = \dfrac{P}{2} . . . \left(3\right)\end{cases}$</p><p>Substitute the value of the known parameter into equation (3) to calculate the required power.</p><p>$\begin{cases}P ' & = & \dfrac{10D}{2} \\ & = & 5D\end{cases}