Two thin convex lenses of focal length 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this…
- 5 D
- 1 D
- 20 D
- 10 D
Solution
$\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}-\frac{\mathrm{d}}{\mathrm{f}_1 \mathrm{f}_2}, \mathrm{~d}=$ distance between lens
$\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{0.3}+\frac{1}{0.1}-\frac{0.1}{(0.3)(0.1)}$
$\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{0.1}$
Power $=\frac{1}{\mathrm{f}_{\mathrm{eq}}}=10 \mathrm{D}$
Asked in: JEE Main 2025 (07 Apr Shift 1)