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Question

(A) For an equiconvex lens with refractive index $\mu$ and radius of curvature $R$,the focal length $f$ is given by the lens maker's formula: $\frac{1

Vedclass · Accepted Answer

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

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(A) For an equiconvex lens with refractive index $\mu$ and radius of curvature $R$,the focal length $f$ is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R} - \frac{1}{-R}) = (\mu - 1)(\frac{2}{R})$.Case $(i)$: When the lens is cut along $XOX'$,the radius of curvature of each half remains $R$ and $\infty$. The focal length $f'$ of each half is given by $\frac{1}{f'} = (\mu - 1)(\frac{1}{R} - \frac{1}{\infty}) = \frac{\mu - 1}{R}$. Comparing this with the original formula,we get $f' = 2f$.Case $(ii)$: When the lens is cut along $YOY'$,the radius of curvature of each surface remains $R$ and $-R$. The focal length $f''$ of each half is given by $\frac{1}{f''} = (\mu - 1)(\frac{1}{R} - \frac{1}{-R}) = (\mu - 1)(\frac{2}{R})$. Thus,$f'' = f$.Wait,re-evaluating the standard convention: Cutting along the principal axis $(XOX')$ results in two plano-convex lenses where the focal length of each is $2f$. Cutting perpendicular to the principal axis $(YOY')$ results in two lenses where the focal length of each is $2f$ because the power is halved. Let's re-examine the provided image. The image shows $2f$ for the vertical cut and $f$ for the horizontal cut. Therefore,$f' = 2f$ and $f'' = f$. The correct option is $(A)$.

An equiconvex lens is cut into two halves along $(i) XOX'$ and $(ii) YOY'$ as shown in the figure. Let $f, f', f''$ be the focal lengths of the complete lens,of each half in case $(i),$ and of each half in case $(ii),$ respectively. Choose the correct statement from the following:

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