A symmetric double convex lens of power 4D is cut in two halves by pl…

The correct answer is Option (1) → 2D

As the convex lens is symmetric i.e., $R_1=R_2=R$

Case I: When lens in not cut

$\frac{1}{f}=(μ-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ [Len's maker formula]

$\frac{1}{f}=(μ-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2}{R}(μ-1)$

and,

$\frac{2}{R}(μ-1)=4D$

$\frac{μ-1}{R}=2D$    ....(1)

Now,

Case II: When lens is cut into two half pieces

$R_1=R$ and $R_2=∞$

$\frac{1}{f'}=(μ-1)\left(\frac{1}{R}-\frac{1}{∞}\right)=\frac{μ-1}{R}$

$∴P'=2D$