Q. Consider an equiconvex lens of radius of curvature $R$ and focal length $f.\ $If $f > R$, the refractive index $\mu$ of the material of the lens
A) Is greater than zero but less than 1.5
B) Is greater than 1.5 but less than 2.0
C) Is greater than one but less than 1.5
D) None of the above
Explanation
(c) $\frac{1}{f} = (\mu - 1)\left( \frac{2}{R} \right)\ or\ f = \frac{R}{2(\mu - 1)}$
Now ,$f > R$
$\therefore\ \frac{R}{2(\mu - 1)} > R$
Or $\ \frac{1}{2(\mu - 1)} > 1\ $or$\ 2(\mu - 1) < 1$
Or $\mu - 1 < \frac{1}{2}\ \ $or$\ \ \mu < \left( 1 + \frac{1}{2} \right)$
Or $\ \mu < 1.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.