Physics - Geometrical Optics Question with Solution | TestHub

Since the lens is equiconvex, the radius of curvature of each half is same.
Using lens maker formula $$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$ Here $R_1=R$ $R_2=-R$ by convection $$\therefore \frac{1}{f}=(\mu -1)\frac{2}{\mathrm{R}}(\mu -1)\frac{1}{\mathrm{R}}=\frac{1}{2f}\dots (\mathrm{i})$$ If we cut the lens along ${\mathrm{XOX}}^{\prime }$ then two halves of the lens will be having the same radii of curvalur and $$\begin{array}{rl}& R_1=R,R_2=-R\\ & \therefore \frac{1}{f^{\prime }}=(\mu -1)\left(\frac{2}{R}\right)\\ & \Rightarrow \frac{1}{f^{\prime }}=\frac{2}{2f}=\frac{1}{f}\end{array}$$ From (i) it is clear that $f^{\prime }=f$ But when we cut along $\mathrm{YOY}$ ' then $$\begin{array}{rl}& R_1=R\text{ and }{R}_2=\infty \\ & \therefore \infty \frac{1}{f^{\prime \prime }}=(\mu -1)\left(\frac{1}{R}-\frac{1}{\infty }\right)\\ & =(\mu -1)\frac{1}{R}=\frac{1}{2f}\\ & \Rightarrow f^{\prime \prime }=2f\end{array}$$