Q66 / 75 · JEE Main 2026 (04 April Shift 2) · Physics
A convex lens is made from glass material having refractive index of $1.4$ with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is ______.
- A$0.5$
- B$2.5$
- C$0.8$
- Correct answerD$1.25$
Solution
Using the lens maker's formula:
$\dfrac{1}{f} = (\mu - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
For an equiconvex lens, $R_1 = R$ and $R_2 = -R$. Given $\mu = 1.4$:
$\dfrac{1}{f} = (1.4 - 1) \left( \dfrac{1}{R} - \dfrac{1}{-R} \right)$
$\dfrac{1}{f} = 0.4 \times \dfrac{2}{R}$
$\dfrac{1}{f} = \dfrac{0.8}{R}$
$\dfrac{f}{R} = \dfrac{1}{0.8} = 1.25$
Answer: $1.25$
$\dfrac{1}{f} = (\mu - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
For an equiconvex lens, $R_1 = R$ and $R_2 = -R$. Given $\mu = 1.4$:
$\dfrac{1}{f} = (1.4 - 1) \left( \dfrac{1}{R} - \dfrac{1}{-R} \right)$
$\dfrac{1}{f} = 0.4 \times \dfrac{2}{R}$
$\dfrac{1}{f} = \dfrac{0.8}{R}$
$\dfrac{f}{R} = \dfrac{1}{0.8} = 1.25$
Answer: $1.25$
Hence, the correct answer is option: 'D'.