An equiconvex lens is cut into two parts alon | vedclass

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(D) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Case $(i)$: Wh

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$f' = f, f'' = 2f$

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(D) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Case $(i)$: When the lens is cut along the $XOX'$ axis (perpendicular to the principal axis),the radius of curvature of each surface remains the same $(R_1 = R, R_2 = -R)$. Thus,the focal length of each half remains the same as the original lens. So,$f' = f$.Case $(ii)$: When the lens is cut along the $YOY'$ axis (parallel to the principal axis),the radius of curvature of one surface becomes infinite $(R_2 = \infty)$. The new focal length $f''$ is given by $\frac{1}{f''} = (n-1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{n-1}{R}$. Since $\frac{1}{f} = (n-1) \left( \frac{1}{R} - \frac{1}{-R} \right) = \frac{2(n-1)}{R}$,we have $\frac{1}{f''} = \frac{1}{2f}$,which implies $f'' = 2f$.Therefore,the correct relation is $f' = f$ and $f'' = 2f$.

An equiconvex lens is cut into two parts along its $(i) \ XOX'$ axis and $(ii) \ YOY'$ axis as shown in the figure. Let $f, f', f''$ be the focal lengths of the original lens,each half in the first case,and each half in the second case,respectively. Which of the following statements is correct?

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