A symmetric double convex lens of power 4D is cut in two halves by plane perpendicular to the principal axis. The power of each half is: |
2D 4D 8D zero |
2D |
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 $R_2=∞$ $\frac{1}{f'}=(μ-1)\left(\frac{1}{R}-\frac{1}{∞}\right)=\frac{μ-1}{R}$ $∴P'=2D$ |