An equiconvex lens of glass of focal length 0 | vedclass

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(A) For an equiconvex lens, the focal length $f$ is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \righ

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$1:1$

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(A) For an equiconvex lens, the focal length $f$ is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Since it is equiconvex, $R_1 = R$ and $R_2 = -R$, so $\frac{1}{f} = (n-1) \frac{2}{R}$.When the lens is cut perpendicular to the principal axis, each new lens is a plano-convex lens with one surface having radius $R$ and the other surface being flat (radius $\infty$).For the new lens, the focal length $f'$ is given by: $\frac{1}{f'} = (n-1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{n-1}{R}$.Comparing the two expressions, we get $\frac{1}{f'} = \frac{1}{2f}$, which means $f' = 2f$.Since both new lenses are identical, their focal lengths are equal $(f'_1 = f'_2 = 2f)$.Therefore, the ratio of their focal lengths is $2f : 2f = 1 : 1$.

An equiconvex lens of glass of focal length $0.1 \text{ m}$ is cut along a plane perpendicular to the principal axis into two equal parts. The ratio of the focal lengths of the new lenses formed is

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