Question

The focal length of an equi biconvex lens of R.I.$=$ 1.5 is equal to

• A. half of its radius of curvature
• B. twice the radius of curvature
• C. radius of curvature
• D. $\frac{1}{4}th$ the radius of curvature

Answer 1

Answer: (C)

radius of curvature

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{f}=\frac{1}{2}\times \frac{2}{R}\Rightarrow f=R$