Q. The radius of curvature of the convex face of a planoconvex lens is 15 cm and the refractive index of the material is 1.4. Then the power of the lens in diopter is
A) 1.6
B) 1.66
C) 2.6
D) 2.66
Explanation
(d) $\frac{1}{f} = \ (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
For planoconvex lens
$R₁ = \ \infty,\ R₂ = \ - R = \ - 1.5\ cm,\ \mu = 1.4$
$\therefore\ \frac{1}{f} = (1.4 - 1)\left( 0 + \frac{1}{15} \right)$
$or\frac{1}{f} = 0.4\ \times \frac{1}{15}$
Therefore, power of the lens in diopter
$P = \frac{100}{f} = \frac{40}{15} = 2.66\ D$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.