An equiconvex lens of focal length \(10\,\mathrm{cm}\) is cut into two identical planoconvex lenses. The focal length of each planoconvex lens will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(20\,\mathrm{cm}\)
Cutting an equiconvex lens in two along a plane perpendicular to its axis gives two planoconvex lenses, each with one curved surface of the original radius and one plane surface \( (R = \infty) \).
For the equiconvex lens \( (R_1 = R,\ R_2 = -R) \):
\( \displaystyle \frac{1}{f} = (n - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = \frac{2(n - 1)}{R} = \frac{1}{10\,\mathrm{cm}} \)
For each planoconvex half \( (R_1 = R,\ R_2 = \infty) \):
\( \displaystyle \frac{1}{f'} = (n - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{n - 1}{R} \)
\( \displaystyle f' = 2f = 20\,\mathrm{cm} \)