Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A biconvex lens of focal length $$20\text{ cm}$$ has curved surfaces whose radii are in the ratio $$1 \colon 2$$. Refractive index of the material of the lens is 1.5. Smaller of the radii of curvatures is
Let the radii be $$R$$ and $$2R$$. By the lens maker's formula for a biconvex lens, $$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R} + \frac{1}{2R}\right)$$, so $$\frac{1}{20} = 0.5 \times \frac{3}{2R} = \frac{3}{4R}$$. Hence $$R = 15\text{ cm}$$.
Was this solution helpful?
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.