An equiconvex lens is cut two halves along (i) XOX' and (ii) YOY' as shown in the figure. Let f, f^prime, f^prime prime be the focal length of the…

Since the lens is equiconvex, the radius of curvature of each half is same.

Using lens maker formula $$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$f1​=(μ−1)(R1​1​−R2​1​) Here $R_1=R$R1​=R $R_2=-R$R2​=−R by convection $$\therefore \frac{1}{f}=(\mu-1) \frac{2}{\mathrm{R}}(\mu-1) \frac{1}{\mathrm{R}}=\frac{1}{2 f} \ldots(\mathrm{i})$$∴f1​=(μ−1)R2​(μ−1)R1​=2f1​…(i) If we cut the lens along $\mathrm{XOX}^{\prime}$XOX′ then two halves of the lens will be having the same radii of curvalur and $$\begin{aligned} & R_1=R, R_2=-R \\ & \therefore \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{2}{R}\right) \\ & \Rightarrow \frac{1}{f^{\prime}}=\frac{2}{2 f}=\frac{1}{f} \end{aligned}$$​R1​=R,R2​=−R∴f′1​=(μ−1)(R2​)⇒f′1​=2f2​=f1​​ From (i) it is clear that $f^{\prime}=f$f′=f But when we cut along $\mathrm{YOY}$YOY ' then $$\begin{aligned} & R_1=R \text { and } R_2=\infty \\ & \therefore \infty \frac{1}{f^{\prime \prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\ & =(\mu-1) \frac{1}{R}=\frac{1}{2 f} \\ & \Rightarrow \quad f^{\prime \prime}=2 f \end{aligned}$$​R1​=R and R2​=∞∴∞f′′1​=(μ−1)(R1​−∞1​)=(μ−1)R1​=2f1​⇒f′′=2f​