An equiconvex lens of focal length 10 cm is made up of material with refractive index 1.5. The radius of curvature of the each surface is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 10 cm
Lens maker's formula
$\frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$\frac{1}{f}=0.5\left[\frac{2}{R}\right]$
R = 10 cm