Question

An equi-biconvex lens of focal power $10D$ is cut into two equal parts perpendicular to its axis. If its refractive index is $1.5$, the radius of curvature of one of the parts is

• A. $20cm$
• B. $10cm$
• C. $40cm$
• D. $5cm$

Answer 1

Answer: (B)

$10cm$

$\frac{1}{f}=10$

$f=10cm$

$\frac{1}{10}=\frac{1}{2}\times \frac{2}{R}$

$R=10cm$