Physics - Geometrical Optics Question with Solution | TestHub

Using the lens maker's formula for the initial equiconvex lens:

$\frac{1}{F}=(\mu -1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2(\mu -1)}{R}$

When the lens is cut into two equal parts along the vertical axis, each part becomes a plano-convex lens with radii of curvature $R$ and $\infty$.

The focal length $F^{\prime }$ of each part is given by:

$\frac{1}{F^{\prime }}=(\mu -1)\left(\frac{1}{R}-\frac{1}{\infty }\right)=\frac{\mu -1}{R}$

Comparing the two equations:

$\frac{1}{F^{\prime }}=\frac{1}{2}\left(\frac{2(\mu -1)}{R}\right)=\frac{1}{2F}$

$F^{\prime }=2F$

Answer: $2F$