JEE Main Optics: Equivalent Power of Separated Lenses 💡
❓ Question Two thin convex lenses of focal lengths f 1 = 30 cm , f 2 = 10 cm f_1 = 30\text{ cm}, \quad f_2 = 10\text{ cm} are placed coaxially, 10 cm apart . Find the power of the combination . 🖼️ Question Image ✍️ Short Solution This is NOT the simple “lenses in contact” case. Because they are separated by distance d d d , we must use the correct combination formula 🔥 🔹 Step 1 — Formula for Two Lenses Separated by Distance Equivalent focal length: 1 F = 1 f 1 + 1 f 2 − d f 1 f 2 \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} Where: f 1 , f 2 f_1, f_2 in same units d d = separation 🔹 Step 2 — Convert Everything to Same Unit Given: f 1 = 30 cm , f 2 = 10 cm , d = 10 cm f_1 = 30\text{ cm},\quad f_2 = 10\text{ cm},\quad d = 10\text{ cm} All already in cm — good 👍 🔹 Step 3 — Substitute Values 1 F = 1 30 + 1 10 − 10 30 × 10 \frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{30\times 10} Ca...