Question

The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is

• A. greater than zero but less than 1.5
• B. greater than 1.5 but less than 2.0
• C. greater than 2.0 but less than 2.5
• D. greater than 2.5 but less than 2.0

Answer 1

Answer: (A)

Focal length of lens,
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
For equi-convex lens,
$R_{1}=+R, R_{2}=-R$
$\therefore \frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)$
$\frac{1}{f}=(\mu-1)\left(\frac{2}{R}\right)$
$f=\frac{R}{2(\mu-1)}$
$ f < R, $
so, $2(\mu-1) < 1$
i.e. $ (\mu-1) < \frac{1}{2} $
$(\mu-1) < 0.5 $
$\mu < 1.5 $
Focal length of convex lens is positive. So, $\mu$ cannot be negative, hence $\mu$ should be greater than zero but less than $1.5$